text:
University of Florida/Egm4313/s12.teamboss/R5

Statement
Find the radius of convergence for:

 * {| style="width:100%" border="0"

$$\displaystyle r(x)=\sum_{k=0}^{\infty } (k + 1) k x^k $$     (1.0)
 * 1)
 * 1)
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle r(x)=\sum_{k=0}^{\infty }\frac{-1^{k}}{\gamma ^{k}} x^{2k} $$     (1.1)
 * 2)
 * 2)
 *
 * }

And find the radius of convergence for the Taylor series of 3) Sin(x) about x=0, 4) log(1+x) about x=0, and 5)log(1+x) about x=1.

Solution
1)--- First, we establish what $$ d_{k}$$ is. As found in the notes (section 7-c), we consider an infinite power series of the form:
 * {| style="width:100%" border="0"

r(x)=\sum_{k=0}^{\infty }d_kx^k $$     (1.2) The radius is then calculated using the formula below:
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{d_{k+1}}{d_{k}} \right | \right ] ^{-1} $$     (1.3)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

In the case of equation (1.0), $$ d_{k}$$ is equals to:
 * {| style="width:100%" border="0"

d_{k}= (k+1)(k) $$     (1.4)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This means, from equation (1.3):

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(k+2)(k+1)}{(k+1)(k)} \right | \right ] ^{-1} $$     (1.5)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This simplifies to:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(k+2)}{(k)} \right | \right ] ^{-1} $$     (1.6)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This limit becomes $$ \frac{\infty }{\infty} $$, and thus L'hopital's rule is necessary. The radius of convergence is then represented by:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{1}{1} \right | \right ] ^{-1} $$     (1.7)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This limit just becomes 1 and the radius of convergence is then established as 1.

2)--- In the case of equation (1.1), after making a bounds change where j=2k and then making k=j,$$ d_{k}$$ is equal to:
 * {| style="width:100%" border="0"

d_{k}= \frac {-1^{k/2}}{\gamma ^{k/2}} $$     (1.8) This means, from equation (1.3):
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{k/2+1/2})(\gamma^{k/2}) }{(\gamma ^{k/2+1/2})(-1^{k/2})} \right | \right ] ^{-1} $$     (1.9)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

$$ \gamma $$ is a constant in this case, and the radius of convergence simplifies to this form:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{1/2})}{(\gamma ^{1/2} )} \right | \right ] ^{-1} $$     (1.10)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

When taking the limit at infinity, the radius of convergence becomes $$ - \gamma^{1/2} $$.

3)--- The Taylor series that represents sin(x) about x=0 is:
 * {| style="width:100%" border="0"

\sin x=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} $$     (1.11)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

We put this in the form found in equation (1.2) first. This will allow us to get a $$ d_{k}$$. To put (1.11) in the form found in (1.2), we let k=2n+1:
 * {| style="width:100%" border="0"

r(x)=\sum_{k=1}^{\infty }\frac{-1^{ \frac{k+1}{2}}}{k!} x^k $$     (1.12)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This means $$ d_{k} $$ would equal:
 * {| style="width:100%" border="0"

\frac{-1^{ \frac{k+1}{2}}}{k!} $$     (1.13)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This means, from equation (1.3):

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{ (-1^{ \frac{k+2}{2}}) (k)! } {(k+1)! (-1^{ \frac{k+1}{2}})} \right | \right ] ^{-1} $$     (1.14)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This simplifies to:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{.5})}{(k+1)} \right | \right ] ^{-1} $$     (1.15)
 * $$\displaystyle
 * $$\displaystyle
 *
 * }

This inverted limit makes $$ R_{c}= \infty $$. This is the best possible case, as this means the series converges for all values of x. 4)--- The Taylor series that represents log(1+x) about x=0 is:
 * {| style="width:100%" border="0"

\sin x=\sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k $$     (1.16)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

So $$d_k $$ would be defined as shown below:
 * {| style="width:100%" border="0"

d_{k}= \frac{(-1)^k}{k} $$     (1.17) This means, from equation (1.3):
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1^{k+1})(k) }{(k+1)(-1^k)} \right | \right ] ^{-1} $$     (1.18)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

This simplifies to:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{(-1)(k) }{(k+1)} \right | \right ] ^{-1} $$     (1.19)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

This limit becomes $$ \frac{\infty }{\infty} $$, and thus L'hopital's rule is necessary. The radius of convergence is then represented by:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{-1}{1} \right | \right ] ^{-1} $$     (1.20)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

This absolute, inverted limit just becomes 1 and the radius of convergence is then established as 1.

5)--- The Taylor series that represents log(1+x) about x=1 is:
 * {| style="width:100%" border="0"

\sin x= - \sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k $$     (1.21)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

So $$d_k $$ would be defined as shown below:
 * {| style="width:100%" border="0"

d_{k}= - \frac{(-1)^k}{k} $$     (1.22) This means, from equation (1.3):
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | -\frac{(-1^{k+1})(k) }{(k+1)(-1^k)} \right | \right ] ^{-1} $$     (1.23)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

This simplifies to:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | -\frac{(-1)(k) }{(k+1)} \right | \right ] ^{-1} $$     (1.24)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

This limit becomes $$ \frac{\infty }{\infty} $$, and thus L'hopital's rule is necessary. The radius of convergence is then represented by:

 * {| style="width:100%" border="0"

R_{c} = \left [ \overset{lim}{k \rightarrow \infty } \left | \frac{1}{1} \right | \right ] ^{-1} $$     (1.25)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

This absolute, inverted limit just becomes 1 and the radius of convergence is then established as 1.

Author
Solved and Typed By -Egm4313.s12.team1.silvestri (talk) 17:26, 24 March 2012 (UTC) Reviewed By - --Egm4313.s12.team1.durrance (talk) 17:33, 30 March 2012 (UTC)

Statement
Determine whether the following functions are linear independent using the Wronskian.

 * {| style="width:100%" border="0"

f(x)=x^2, g(x)=x^4 $$     (2.0)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

f(x)=\cos(x), g(x)=\sin(3x) $$     (2.1) Using the Gramian over interval [a,b] = [1,1] to come to the same conclusion.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Wronskian Solution (1)
The Wronskian is defined as:
 * {| style="width:100%" border="0"

W(f,g):=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=fg'-gf' $$     (2.2) If the Wronskian does not equal zero, then the equations are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Values needed are:
 * {| style="width:100%" border="0"

f'=2x $$     (2.3)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

g'=4x^3 $$     (2.4) Substituting all of the conditions into Eq. (2.2):
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

W(f,g):=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=x^2(4x^3)-2x(x^4)=2x^{5}\neq 0 $$     (2.5) Because the Wronskian does not equal zero, Eqs. (2.0) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Wronskian Solution (2)
The Wronskian is defined as:
 * {| style="width:100%" border="0"

W(f,g):=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=fg'-gf' $$     (2.2) If the Wronskian does not equal zero, then the equations are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Values needed are:
 * {| style="width:100%" border="0"

f'=-\sin(x) $$     (2.6)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

g'=3\cos(3x) $$     (2.7) Substituting all of the conditions into Eq. (5.2):
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

W(f,g)=\det \begin{bmatrix} f & g\\f'&g'\end{bmatrix}=3\cos(x)(\cos(3x))-(-\sin(x))(\sin(3x))\neq 0 $$     (2.8) Because the Wronskian does not equal zero, Eqs. (2.1) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Gramian Solution (1)
The Gramian is defined as:
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix} $$     (2.9) Where the scalar product is defined as:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

\langle f, g \rangle := \int_a^b f(x) g(x) \,dx $$     (2.10) When the Gramian does not equal zero, the functions are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

So calculating the scalar products:
 * {| style="width:100%" border="0"

\langle f, f \rangle = \int_{-1}^1 (x^2)(x^2)\,dx = 2/5$$ $$\displaystyle \langle f, g \rangle = \int_{-1}^1 (x^2)(x^4)\,dx = 2/7$$ $$\displaystyle \langle g, f \rangle = \int_{-1}^1 (x^4)(x^2)\,dx = 2/7$$ $$\displaystyle \langle g, g \rangle = \int_{-1}^1 (x^4)(x^4)\,dx = 2/9 $$     (2.11) Substituting those values into Eq. (2.9) and calculating the determinant:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix}= (2/5)(2/9)-(2/7)(2/7)=16/2205\neq 0 $$     (2.12) Because the determinant does not equal zero, Eqs. (2.0) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Gramian Solution (2)
The Gramian is defined as:
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix} $$     (2.9) Where the scalar product is defined as:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

\langle f, g \rangle := \int_a^b f(x) g(x) \,dx $$     (2.10) When the Gramian does not equal zero, the functions are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

So calculating the scalar products:
 * {| style="width:100%" border="0"

\langle f, f \rangle = \int_{-1}^1 \cos(x)\cos(x)\,dx = 1.9998$$ $$\displaystyle \langle f, g \rangle = \int_{-1}^1 \cos(x)\sin(3x)\,dx = 0$$ $$\displaystyle \langle g, f \rangle = \int_{-1}^1 \sin(3x)\cos(x)\,dx = 0$$ $$\displaystyle \langle g, g \rangle = \int_{-1}^1 \sin(3x)\sin(3x)\,dx = 0.0018 $$     (2.13) Substituting those values into Eq. (2.9) and calculating the determinant:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix}= 1.998*0.0018-0=0.0036\neq 0 $$     (2.14) Because the determinant does not equal zero, Eqs. (2.1) are linearly independent.
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Author
Solved and Typed By - --Egm4313.s12.team1.durrance (talk) 18:35, 26 March 2012 (UTC)--128.227.12.77 18:17, 26 March 2012 (UTC) Reviewed By ---Egm4313.s12.team1.stewart (talk) 19:03, 26 March 2012 (UTC)

Statement
From [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec7c.djvu Lect. 7c Pg. 38] Verify that $$\mathbf{b_{1}},\mathbf{b_{2}}\!$$ are linearly independent using the Gramian.

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{b_{1}}=2\mathbf{e}_{1}+7\mathbf{e}_{2} $$     (3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{b_{2}}=1.5\mathbf{e}_{1}+3\mathbf{e}_{2} $$     (3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
The Gramian:
 * {| style="width:100%" border="0"

\Gamma(f,g):=\det \begin{bmatrix} \langle f, f \rangle & \langle f, g \rangle \\ \langle g, f \rangle & \langle g, g \rangle \end{bmatrix} $$     (3.3)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

But for the vectors given the Gramian looks like this:

 * {| style="width:100%" border="0"

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}}):=\det \begin{bmatrix} \langle\mathbf{b_{1}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle \\ \langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle \end{bmatrix} $$     (3.4)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

And:

 * {| style="width:100%" border="0"

$$     (3.5)
 * $$\langle \mathbf{b_{i}}, \mathbf{b_{j}} \rangle \equiv \mathbf{b_{i}} \cdot \mathbf{b_{j}}
 * $$\langle \mathbf{b_{i}}, \mathbf{b_{j}} \rangle \equiv \mathbf{b_{i}} \cdot \mathbf{b_{j}}
 *

 * }

And for the scalar dot product:
 * {| style="width:100%" border="0"

$$  \displaystyle \langle \mathbf e_1, \mathbf e_2 \rangle = 0 $$     (3.6) Therefore:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

\langle \mathbf{b_{1}}, \mathbf{b_{1}} \rangle = (2\mathbf{e}_{1}+7\mathbf{e}_{2}) \cdot (2\mathbf{e}_{1}+7\mathbf{e}_{2})=4\mathbf{e}_{1}+49\mathbf{e}_{2}=4+49=53 $$     (3.7)
 *

 * }

 * {| style="width:100%" border="0"

\langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle = (2\mathbf{e}_{1}+7\mathbf{e}_{2}) \cdot (1.5\mathbf{e}_{1}+3\mathbf{e}_{2}) = 3\mathbf{e}_{1}+21\mathbf{e}_{2}=3+21=24 $$     (3.8)
 *

 * }

 * {| style="width:100%" border="0"

\langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle = (1.5\mathbf{e}_{1}+3\mathbf{e}_{2}) \cdot (2\mathbf{e}_{1}+7\mathbf{e}_{2})= 3\mathbf{e}_{1}+21\mathbf{e}_{2}=3+21=24 $$     (3.9)
 *

 * }

 * {| style="width:100%" border="0"

\langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle = (1.5\mathbf{e}_{1}+3\mathbf{e}_{2}) \cdot (1.5\mathbf{e}_{1}+3\mathbf{e}_{2})= 2.25\mathbf{e}_{1}+9\mathbf{e}_{2}=2.25+9=11.25 $$     (3.10)
 *

 * }

The determinant to solve for the Gramian is:

 * {| style="width:100%" border="0"

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}}):=\det \begin{bmatrix} \langle\mathbf{b_{1}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle \\ \langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle & \langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle \end{bmatrix}=\langle\mathbf{b_{1}}, \mathbf{b_{1}} \rangle\cdot \langle \mathbf{b_{2}}, \mathbf{b_{2}} \rangle -\langle \mathbf{b_{1}}, \mathbf{b_{2}} \rangle\cdot \langle \mathbf{b_{2}}, \mathbf{b_{1}} \rangle $$     (3.11)
 *

 * }

Plugging in values:
 * {| style="width:100%" border="0"

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}})=(53)\cdot (11.25)-(24)\cdot (24) $$     (3.12)
 *

 * }

 * {| style="width:100%" border="0"

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}})=596.25-576 $$     (3.13)
 *

 * }

 * {| style="width:100%" border="0"

\Gamma(\mathbf{b_{1}},\mathbf{b_{2}})=20.25 $$     (3.14) Qualifications for a linearly independent system of vectors:
 *

 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{\Gamma } \neq 0 \Rightarrow \mathbf{\Gamma^{-1} } exists \Rightarrow \mathbf{c}=\mathbf{\Gamma } ^{-1}\mathbf{d} $$     (3.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So the vectors $$\mathbf{b_{1}},\mathbf{b_{2}}\!$$ are linearly independent because:
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{\Gamma }= 20.25 \neq 0 $$     (3.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and Typed By - Egm4313.s12.team1.stewart (talk) 18:20, 25 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.rosenberg (talk) 05:20, 28 March 2012 (UTC)

Statement
Show that $$ y_p(x) = \sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''_{p,i} + p(x)y'_{p,i} + q(x)y_{p,i} = r_i(x)\!$$ with the excitation $$ r(x) = r_1(x) + r_2(x) + ... + r_n(x) = \sum_{i=0}^{n}r_i(x)\!$$. Discuss the choice of $$ y_p(x)\!$$, in example for $$ r(x) = Kcos(wx)\!$$. Why would you need to have both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ in $$ y_p(x)\!$$?

Solution
The following represent particular solutions and their derivatives, equated into a summation:
 * {| style="width:100%" border="0"

y_p(x) = y_{p,1}(x) + y_{p,2}(x) + ... + y_{p,n}(x) \rightarrow y_p(x) = \sum_{i=0}^{n}y_{p,i}(x) $$     (4.0)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

 * {| style="width:100%" border="0"

y_p'(x) = y_{p,1}'(x) + y_{p,2}'(x) + ... + y_{p,n}'(x) \rightarrow y_p'(x) = \sum_{i=0}^{n}y_{p,i}'(x) $$     (4.1)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

 * {| style="width:100%" border="0"

y_p(x) = y_{p,1}(x) + y_{p,2}''(x) + ... + y_{p,n}(x) \rightarrow y_p(x) = \sum_{i=0}^{n}y_{p,i}''(x) $$     (4.2)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Because the given ODE is in the form of L2-ODE-VC, it is linearly independent. Each $$ y_{p,i}\! $$ is a solution for each $$ r_i\! $$, and when there are multiple excitations, the solution to a sum of these excitations is the sum of the particular solutions. Using these particular solutions, we can show that they are the solutions to the following L2-ODE-VC (4.3) with the given excitation:

 * {| style="width:100%" border="0"

y_{p,i}'' + p(x)y_{p,i}' + q(x)y_{p,i} = r_i(x) $$     (4.3)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

 * {| style="width:100%" border="0"

\sum_{i=0}^{n}y_{p,i}''(x) + p(x)\sum_{i=0}^{n}y_{p,i}'(x) + q(x)\sum_{i=0}^{n}y_{p,i}(x) = \sum_{i=0}^{n}r_i(x) $$     (4.4)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

For $$ r(x) = Kcos(wx)\!$$. You need to have both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ in $$ y_p(x)\!$$ because when solving for the particular solution, it is necessary to take derivatives of the particular solution, and in the case of $$ r(x) = Kcos(wx)\!$$ the derivatives will produce extra terms since the derivative of $$ cos(wx)\!$$ will produce both $$ sin(wx)\!$$ and $$ cos(wx)\! $$ terms. Having both $$ cos(wx)\!$$ and $$ sin(wx)\!$$ is necessary to eliminate the extra terms.

Author
Solved and Typed By - --Egm4313.s12.team1.wyattling (talk) 19:17, 26 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.armanious (talk) 05:25, 30 March 2012 (UTC)

Statement
1. Show that $$\cos 7x$$ and $$\sin 7x$$ are linearly independent using the Wronskian and the Gramian(1 period). 2. Find 2 equations for the two unknowns M, N, and solve for M, N. 3. Find the overall solution $$y(x)$$ that corresponds to the initial condition (3b) p3-7. $$y(0)=1, y'(0)=0$$ Plot the solution over 3 periods.

Solution
Part 1a When using the Wronskian, if the solution does not equal to 0, then the two are linearly independent of each other. The Wronskian can be defined as:
 * {| style="width:100%" border="0"

\displaystyle W(f,g):=det\begin{bmatrix} f &g \\ f'&g' \end{bmatrix}=fg'-gf' $$     (5.1) Lets set $$f = \cos (7x)$$ and $$g = \sin (7x)$$ so that we can find $$f', g'$$
 *

 * }
 * {| style="width:100%" border="0"

\displaystyle f' = -7\sin (7x), g' = 7\cos (7x) $$     (5.2) Plugging these values into the Wronskian equation yields:
 *

 * }
 * {| style="width:100%" border="0"

\displaystyle W(f,g):=det\begin{bmatrix} \cos (7x) &\sin (7x) \\ -7\sin (7x)&7\cos (7x) \end{bmatrix}=7\cos ^2 (7x) - 7\sin ^2 (7x) $$     (5.3)
 *

 * }
 * {| style="width:100%" border="0"

\displaystyle 7\cos ^2 (7x) - 7\sin ^2 (7x) \neq 0 $$     (5.4)
 *

 * }

Thus f and g are in fact linearly independent of each other. Part 1b Now we need to solve it using the Gramian Gramian can be defined as:
 * {| style="width:100%" border="0"

\displaystyle \Gamma (f,g):=det\begin{bmatrix} <f,f> &<f,g> \\ <g,f> &<g,g> \end{bmatrix} $$     (5.5) Where like with the Wronskian, f and g are linearly independent of each other when $$\Gamma (f,g)\neq 0 $$ Integrating over one period implies that our boundaries will be $$(0, \frac{2\pi }{7})$$
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\displaystyle <f,f>=\int_{0}^{\frac{2\pi }{7}}\cos ^{2}(7x)dx $$     (5.6) Setting $$u=7x$$ gives us $$du=7dx$$ which also changes our integration factors to be $$(0,2\pi )$$ giving us:
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\displaystyle <f,f>=\frac{1}{7}\int_{0}^{2\pi }\cos ^{2}(u)du $$     (5.7) Integrating gives us:
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\displaystyle \frac{1}{7}[\frac{u}{2} +\frac{1}{4} \sin 2u\mathbf{\mid} _{0}^{2\pi }] $$     (5.8) Which equals:
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\displaystyle \frac{\pi }{7} $$     (5.9)
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\displaystyle <g,g>=\int_{0}^{\frac{2\pi }{7}}\sin ^{2}(7x)dx $$     (5.10) Setting $$u=7x$$ gives us $$du=7dx$$ which also changes our integration factors to be $$(0,2\pi )$$ giving us:
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\displaystyle <g,g>=\frac{1}{7}\int_{0}^{2\pi }\sin ^{2}(u)du $$     (5.11) Integrating gives us:
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\displaystyle \frac{1}{7}[\frac{u}{2} -\frac{1}{4} \sin 2u\mathbf{\mid} _{0}^{2\pi }] $$     (5.12) Which equals:
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\displaystyle \frac{\pi }{7} $$     (5.13)
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\displaystyle <f,g>=<g,f>=\int_{0}^{\frac{2\pi }{7}}\cos (7x)\sin (7x)dx $$     (5.14) Because cos and sin are orthogonal of each other, without going through with the integration, we can say that $$<f,g>=<g,f>=0$$ Plugging these values into our Gramian equation gives us:
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\displaystyle \Gamma (f,g):=det\begin{bmatrix} \frac{\pi }{7}&0 \\ 0 &\frac{\pi }{7} \end{bmatrix} =\frac {\pi ^2}{49} \neq 0 $$     (5.15) Thus, we again see that f and g are linearly independent of each other. Part 2 From the notes, we are given the following information:
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\displaystyle y''-3y'-10y=3\cos (7x) $$     (5.16)
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\displaystyle y_{p}(x)=M\cos (7x)+N\sin (7x) $$     (5.17)
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\displaystyle y'_{p}(x)=-M7\sin (7x)+N7\cos (7x) $$     (5.18)
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\displaystyle y''_{p}(x)=-M49\cos (7x)-N49\sin (7x) $$     (5.19) Plugging 5.17-19 back into our original equation in 5.16 yields:
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\displaystyle -M49\cos (7x)-N49\sin (7x)+M21\sin (7x)-N21\cos (7x)-M10\cos (7x)-N10\sin (7x)=3\cos (7x) $$     (5.20) Collecting like terms leaves us with:
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\displaystyle -59M\cos (7x)-59N\sin (7x)+21M\sin (7x)-21N\cos (7x)=3\cos (7x) $$     (5.21) Now we can equate coefficients to solve of M and N:
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\displaystyle -59M = 3,M=\frac{-3}{59} $$     (5.22)
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\displaystyle 21N = 3,N=\frac{1}{7} $$     (5.23) Part 3 The overall solution to the equation is expressed as $$ y(x)=y_{p}(x)+y_{h}(x)$$. We must now find the homogeneous equation. Writing our given equation in homogeneous form gives us:
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\displaystyle y''-3y'-10=0 $$     (5.24) rewriting in characteristic form:
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\displaystyle \lambda ^2-3\lambda -10 = 0 $$     (5.25) We can solve for our roots using simple factoring:
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\displaystyle (\lambda -5)(\lambda +2) = 0 $$     (5.26) Thus:
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\displaystyle \lambda_{1,2}=(5,-2) $$     (5.27) Yielding:
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\displaystyle y_{h}(x)=c_{1} e^{5x}+c_{2} e^{-2x} $$     (5.28) Now using the given initial conditions, we can solve for $$c_{1},c_{2}$$
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\displaystyle y'_{h}(x)=5c_{1}e^{5x}-2c_{2}e^{-2x} $$     (5.29) Plugging in initial conditions we have:
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\displaystyle y(0)=c_{1}+c_{2}=1, y'(0)=5c_{1}-2c_{2}=1 $$     (5.30) Solving for each gives us:
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\displaystyle c_{1}=\frac{2}{7}, c_{2}=\frac{5}{7} $$     (5.31) Therfore:
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\displaystyle y_{h}(x)=\frac{2}{7} e^{5x}+\frac{5}{7} e^{-2x} $$     (5.32) Now plugging in our answers from Step 2 into the general particular equation we get that:
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\displaystyle y_{p}(x)=\frac{-3}{59} \cos (7x) + \frac{1}{7} \sin (7x) $$     (5.33) so our final equation to is:
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\displaystyle y(x)=\frac{2}{7} e^{5x}+\frac{5}{7} e^{-2x} +\frac{-3}{59} \cos (7x) + \frac{1}{7} \sin (7x) $$     (5.34) To plot over 3 periods, we will plot from $$(0,\frac{6\pi }{7})$$
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Author
Solved and Typed By - Egm4313.s12.team1.rosenberg (talk) 18:45, 30 March 2012 (UTC) Reviewed By -Egm4313.s12.team1.silvestri (talk) 18:44, 30 March 2012 (UTC)

Statement
Consider the following L2-ODE-CC; see p.6-6:
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$$\displaystyle{y}''-4{y}'+13{y}=2e^{-2x}\cos(3x)$$ (6.0) Homogeneous solution:
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$$\displaystyle{y}_{h}(x)=e^{-2x}[A\cos3x+B\sin3x]$$ (6.1) Particular solution:
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$$\displaystyle{y}_{p}(x)=xe^{-2x}[M\cos3x+N\sin3x]$$ (6.2) Complete the solution for this problem.
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Find the overall solution $$y(x)$$ that corresponds to the initial condition (3b) p.3-7
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$$\displaystyle y(0)=1,y'(0)=0$$ (6.3)
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Solution
Start by finding $$y_{p}'$$ and $$y_{p}''$$.
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$$\displaystyle {y}_{p}'={{e}^{-2x}}[(-3Mx-2Nx+N)\sin 3x+(-2Mx+3Nx+M)\cos 3x]$$ (6.4)
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$$\displaystyle {y}_{p}''={{e}^{-2x}}[(12Mx-6M-5Nx-4N)\sin 3x+(-15Mx-12M-12Nx+6N)\cos 3x]$$ (6.5) Substitute $$y_p$$ and its derivatives into (6.0) to find $$M$$ and $$N$$.
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$$\displaystyle {y}_{p}''+4{y}_{p}'+13{y}_{p}={{e}^{-2x}}[(-6M)\sin 3x+(-10Mx-8M+6N)\cos 3x]$$ (6.6) Separating terms and setting equal to the excitation from (6.0):
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$$\displaystyle -6M{{e}^{-2x}}\sin 3x+(-8M+6N){{e}^{-2x}}\cos 3x-10Mx{{e}^{-2x}}\cos 3x=2{{e}^{-2x}}\cos 3x$$
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(6.7) From (6.7), we solve coefficients to get
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$$\displaystyle -6M=0$$ $$\displaystyle -8M+6N=2$$ $$\displaystyle -10M=0$$ (6.8) Solving for $$M$$ and $$N$$:
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$$\displaystyle M=0$$ $$\displaystyle N=\frac{1}{3}$$ (6.9) Which gives us the particular solution:
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$$\displaystyle {y}_{p}=\frac{1}{3}x{{e}^{-2x}}\sin 3x$$ (6.10) For the general solution,
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$$\displaystyle y={{y}_{h}}+{{y}_{p}}$$ (6.11)
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$$\displaystyle y=e^{-2x}[A\cos3x+B\sin3x]+\frac{1}{3}x{{e}^{-2x}}\sin 3x$$ (6.12)
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$$\displaystyle y=e^{-2x}[A\cos3x+(B+\frac{1}{3}x)\sin3x]$$ (6.13) To solve for $$A$$ and $$B$$, we use initial conditions from (6.3):
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$$\displaystyle y(0)=e^{-2(0)}[A\cos3(0)+(B+\frac{1}{3}(0))\sin3(0)]=1$$ (6.14) Which simplifies to:
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$$\displaystyle A=1$$ (6.15) For the second initial condition from (6.3):
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$$\displaystyle y'={{e}^{-2x}}[(-3A-2B+\frac{1}{3}+\frac{2}{3}x)\sin 3x+(3B+x-2)\cos 3x]$$ (6.16)
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$$\displaystyle y'(0)={{e}^{-2(0)}}[(-3A-2B+\frac{1}{3}+\frac{2}{3}(0))\sin 3(0)+(3B+(0)-2)\cos 3(0)]=0$$ (6.17)
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$$\displaystyle 3B-2=0$$ (6.18)
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$$\displaystyle B=\frac{2}{3}$$ (6.19) We can now write $$y$$ with all coefficients known.
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$$\displaystyle y={{e}^{-2x}}[(1)\cos 3x+(\frac{2}{3}+\frac{1}{3}x)\sin 3x]$$ (6.20) Final Equation
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$$\displaystyle y={{e}^{-2x}}[\cos 3x+(\frac{2}{3}+\frac{1}{3}x)\sin 3x]$$ (6.21)
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Author
Solved and Typed By - Egm4313.s12.team1.essenwein (talk) 00:28, 20 March 2012 (UTC) Reviewed By - --Egm4313.s12.team1.stewart (talk) 18:29, 25 March 2012 (UTC)

Statement
'''[https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec8b.djvu See R5.7 Lect. 8b pg. 11]:'''

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$$\mathbf{v}=4\mathbf e_1+2\mathbf e_2=c_1\mathbf b_1+c_2\mathbf b_2$$
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The oblique basis vectors b1,b2 are:
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$$\mathbf b_1 =2\mathbf e_1 + 7\mathbf e_2$$
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$$\mathbf b_2 =1.5\mathbf e_1 + 3\mathbf e_2$$ 1. Find the components c1, c2 using the Gramian matrix. 2. Verify the result found above.
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Solution
To find the components of the oblique basis vectors, the Gram matrix must be used. The Gram matrix is defined as such:
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$$  \displaystyle \boldsymbol \Gamma (\mathbf b_1, \mathbf b_2) := \begin{bmatrix} \langle \mathbf b_1, \mathbf b_1 \rangle & \langle \mathbf b_1, \mathbf b_2 \rangle \\ \langle \mathbf b_2, \mathbf b_1 \rangle & \langle \mathbf b_2, \mathbf b_2 \rangle \end{bmatrix} $$     (7.0)
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This matrix requires several scalar products to be found. one important feature of scalar products will be presented here and implicitly used throughout these calculations:
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$$  \displaystyle \langle \mathbf e_1, \mathbf e_2 \rangle = 0 $$     (7.1)
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Equation 7.1 states that the scalar product of two perpendicular vectors is zero. To find the Gram matrix, three scalar products must be found:
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$$  \displaystyle \langle \mathbf b_1, \mathbf b_1 \rangle = (2)(2)+(7)(7)=53 $$     (7.2)
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$$  \displaystyle \langle \mathbf b_1, \mathbf b_2 \rangle =\langle \mathbf b_2, \mathbf b_1 \rangle =(2)(1.5)+(7)(3)=24 $$     (7.3)
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$$  \displaystyle \langle \mathbf b_2, \mathbf b_2 \rangle = (1.5)(1.5)+(3)(3)=11.25 $$     (7.4) Using the above values the Gram matrix is found to be:
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$$  \displaystyle \boldsymbol \Gamma (\mathbf b_1, \mathbf b_2) = \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix} $$     (7.5) To find the components, the Gram matrix must be used to solve the following equation:
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$$  \displaystyle \begin{bmatrix} \langle \mathbf b_1, \mathbf b_1 \rangle & \langle \mathbf b_1, \mathbf b_2 \rangle \\ \langle \mathbf b_2, \mathbf b_1 \rangle & \langle \mathbf b_2, \mathbf b_2 \rangle \end{bmatrix}\begin{Bmatrix}c_1\\c_2 \end{Bmatrix}=\begin{Bmatrix}\langle \mathbf b_1, \mathbf v \rangle\\\langle \mathbf b_2, \mathbf v \rangle \end{Bmatrix} $$     (7.6)
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The right hand side of the equation can be found using:
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$$  \displaystyle \langle \mathbf b_1, \mathbf v \rangle =(2)(4)+(7)(2)=22 $$     (7.7)
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$$  \displaystyle \langle \mathbf b_2, \mathbf v \rangle =(1.5)(4)+(3)(2)=12 $$     (7.8) Using known values, 7.6 becomes:
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$$  \displaystyle \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}\begin{Bmatrix}c_1\\c_2 \end{Bmatrix}=\begin{Bmatrix}22\\12 \end{Bmatrix} $$     (7.9) To find the components, the Gramian (the determinant of the Gram matrix) and the inverse of the Gram must be found.
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$$  \displaystyle \Gamma = \begin{vmatrix} 53 & 24 \\ 24 & 11.25 \end{vmatrix}=(53)(11.25)-(24)(24)=20.25 $$     (7.10)
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$$  \displaystyle \boldsymbol \Gamma ^{-1} = \frac{1}{20.25}\begin{bmatrix} 11.25 & -24 \\ -24 & 53 \end{bmatrix}=\begin{bmatrix} \frac{5}{9} & -\frac{32}{27} \\ -\frac{32}{27} & \frac{212}{81} \end{bmatrix} $$     (7.11) This can be used in the matrix equation to solve for the components:
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$$  \displaystyle \begin{Bmatrix}c_1\\c_2 \end{Bmatrix}= \begin{bmatrix} \frac{5}{9} & -\frac{32}{27} \\ -\frac{32}{27} & \frac{212}{81} \end{bmatrix}\begin{Bmatrix}22\\12 \end{Bmatrix}=\begin{Bmatrix}-2\\ \frac{16}{3} \end{Bmatrix} $$     (7.12) Therefore the matrix v with respect to the oblique vectors is:
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$$  \displaystyle \mathbf v = -2 \mathbf b_1 + \frac{16}{3} \mathbf b_2 $$     (7.13) As a check, the definitions of each vectors with respect to the basis e1 and e2:
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$$  \displaystyle 4\mathbf e_1 + 2\mathbf e_2 = -2 (2\mathbf e_1 + 7\mathbf e_2) + \frac{16}{3} (1.5\mathbf e_1 + 3\mathbf e_2)=-4\mathbf e_1 -14\mathbf e_2 + 8\mathbf e_1 + 16\mathbf e_2 $$     (7.14) Which reduces to:
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$$  \displaystyle 4\mathbf e_1 + 2\mathbf e_2 = 4\mathbf e_1 + 2\mathbf e_2 $$     (7.15)
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Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 01:40, 25 March 2012 (UTC) Reviewed By - --Egm4313.s12.team1.stewart (talk) 18:31, 25 March 2012 (UTC)

Statement
'''[https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec8b.djvu see R5.8 Lect. 8b pg. 16]:''' Find the integral: $$\int x^n \log(1+x) dx\!$$

Solution

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$$  \displaystyle \int x^n \log(1+x) dx $$ (8.0) To find this integral, integration by parts must be used. The formula for integration by parts is:
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$$  \displaystyle \int udv=uv-\int vdu + C $$ (8.1)
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$$  \displaystyle u = \log(1+x) $$     (8.2)
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$$  \displaystyle du=\frac{dx}{1+x} $$     (8.3)
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$$  \displaystyle dv=x^ndx $$     (8.4)
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$$  \displaystyle v=\frac{x^{n+1}}{n+1} $$     (8.5) Using these expressions in the formula yields:
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$$  \displaystyle \int x^n \log(1+x)dx=\frac{x^{n+1}}{n+1}\log(1+x)-\frac{1}{n+1}\int \frac{x^{n+1}}{1+x}dx + C $$ (8.6) Now, the integral $$\int \frac{x^{n+1}}{1+x} dx\!$$ must be found. To start, the fraction must be expanded using long division:
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$$  \displaystyle \frac{x^{n+1}}{1+x}=x^n-x^{n-1}+x^{n-2}-x^{n-3}+...+\frac{(-1)^{n+1}}{x+1}=\frac{(-1)^{n+1}}{x+1}+\sum_{k=0}^{n}(-1)^kx^{n-k} $$     (8.7) This expression can now be easily integrated to yield the following:
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$$  \displaystyle \int \frac{x^{n+1}}{1+x}dx=\int\frac{(-1)^{n+1}}{x+1}dx+\int\sum_{k=0}^{n}(-1)^kx^{n-k}dx $$     (8.8)
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$$  \displaystyle \int\frac{(-1)^{n+1}}{x+1}dx=(-1)^{n+1}\log(1+x) $$     (8.9)
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$$  \displaystyle \int\sum_{k=0}^{n}(-1)^kx^{n-k}dx=\sum_{k=0}^{n}\int(-1)^kx^{n-k}dx= \sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1} $$     (8.10) Therefore:
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$$  \displaystyle \int \frac{x^{n+1}}{1+x} dx= (-1)^{n+1}\log(1+x)+\sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1} $$     (8.11) Substituting into the original equation:
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$$  \displaystyle \int x^n \log(1+x)dx=\frac{x^{n+1}}{n+1}\log(1+x)-\frac{1}{n+1}((-1)^{n+1}\log(1+x)+\sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1}) + C $$ (8.12) Simplifying this yields:
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$$  \displaystyle \int x^n \log(1+x)dx=\frac{1}{n+1}[(x^{n+1}-(-1)^{n+1})\log(1+x)-\sum_{k=0}^{n}\frac{(-1)^k}{n-k+1}x^{n-k+1}]+C $$     (8.13) To illustrate this, two test cases with n=0 and n=1 will be used. For n=0
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$$  \displaystyle \int x^0 \log(1+x)dx=\frac{1}{0+1}[(x^{0+1}-(-1)^{0+1})\log(1+x)-\frac{(-1)^0}{0-0+1}x^{0-0+1}]+C $$     (8.14) This simplifies to:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle \int \log(1+x)dx=(x+1)\log(1+x)-x+C $$     (8.15) For n=1
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
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$$  \displaystyle \int x^1 \log(1+x)dx=\frac{1}{1+1}[(x^{1+1}-(-1)^{1+1})\log(1+x)-\frac{(-1)^0}{1-0+1}x^{1-0+1}-\frac{(-1)^1}{1-1+1}x^{1-1+1}]+C $$     (8.16) This simplifies to:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle \int x \log(1+x)dx=\frac{1}{2}[(x^{2}-1)\log(1+x)-\frac{1}{2}x^{2}+x]+C $$     (8.17) In fact, this formula can be further generalized for any $$\int x^n \log(r+x)dx \!$$ where r is any real number. The most notable step that changes is the long division expansion. Each term in the expansion increases by a factor of $$r^{k} \!$$. The result is:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{x^{n+1}}{r+x}=x^n-rx^{n-1}+r^2x^{n-2}-r^3x^{n-3}+...+\frac{(-r)^{n+1}}{x+r}=\frac{(-r)^{n+1}}{x+r}+\sum_{k=0}^{n}(-r)^kx^{n-k} $$     (8.18) It is important to note that this particular step will fail when r=0 because $$0^{0} \!$$ is undefined. A special case with r=0 will also be shown for full generality. The rest of the process is identical to that shown above, with every $$\log (1+x) \!$$ term replaced with $$\log (r+x) \!$$. The final result of the integration yields:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="1"

$$  \displaystyle \int x^n \log(r+x)dx=\frac{1}{n+1}[(x^{n+1}-(-r)^{n+1})\log(r+x)-\sum_{k=0}^{n}\frac{(-r)^k}{n-k+1}x^{n-k+1}]+C; r\neq 0 $$     (8.19) When r=0, the integral is of the form $$\int x^n \log(x)dx \!$$, which can be integrated using integration by parts.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
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$$  \displaystyle u = \log(x) $$     (8.20)
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 *

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$$  \displaystyle du=\frac{dx}{x} $$     (8.21)
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 *

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$$  \displaystyle dv=x^ndx $$     (8.22)
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 *

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$$  \displaystyle v=\frac{x^{n+1}}{n+1} $$     (8.23) Using these in (8.1) yields:
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 *

 * }
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$$  \displaystyle \int x^n \log(x)dx=\frac{x^{n+1}}{n+1}\log(x)-\frac{1}{n+1}\int \frac{x^{n+1}}{x}dx + C $$ (8.24) Now, the integral $$\int \frac{x^{n+1}}{x} dx\!$$ must be found. This is simply:
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 *

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$$  \displaystyle \int \frac{x^{n+1}}{x}dx=\int x^ndx=\frac{x^{n+1}}{n+1}+C $$     (8.25) Substituting (8.25) into (8.24) yields:
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 *

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$$  \displaystyle \int x^n \log(x)dx=\frac{x^{n+1}}{n+1}\log(x)-\frac{x^{n+1}}{(n+1)^2} + C $$ (8.26) Thus, the overall solution for any real value of r is
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 *

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$$  \displaystyle \int x^n \log(r+x)dx=\begin{cases} \frac{1}{n+1}[(x^{n+1}-(-r)^{n+1})\log(r+x)-\sum_{k=0}^{n}\frac{(-r)^k}{n-k+1}x^{n-k+1}]+C & \text{ if } r\neq 0\\ \frac{x^{n+1}}{n+1}\log(x)-\frac{x^{n+1}}{(n+1)^2} + C& \text{ if } r=0 \end{cases} $$     (8.27)
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 *

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Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 02:51, 25 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri (talk) 18:29, 30 March 2012 (UTC)

Statement
Consider the following L2-ODE-CC with log(1+x) as the excitation:
 * {| style="width:100%" border="0"

r(x)=y''-3y'+2y $$     (9.0)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

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r(x)=log(1+x) $$     (9.1)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Also, consider the initial conditions:
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y(- \frac{3}{4})=1, y'(- \frac{3}{4})=0 $$     (9.2) 1)Project the excitation r(x) on the polynomial basis
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

 * {| style="width:100%" border="0"

\left \{ b_{j}(x) = x^j, j=0,1,...n \right \} $$     (9.3) i.e., find $$ d_{j} $$ such that:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

r(x)\approx r_n (x) = \sum_{j=0}^{n} d_jx^j $$     (9.4) for x in [-.75, 3], and for n= 0,1. Plot $$ r(x) $$ and $$ r_n(x) $$ to show uniform approximation and convergence. Note that $$ <x^i, r>=\int_{a}^{b} x^i\log(1+x) dx $$ In a separate series of plots, compare the approximation of the function log(1+x) by 2 methods:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

B. Taylor series expansion about $$ \hat x =0 $$
 * A. Projection on Polynomial basis (1) p8-17
 * A. Projection on Polynomial basis (1) p8-17

 * }

Observe and discuss the pros and cons of each method.

2) Find $$ y_n (x) $$ such that:
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y''_n +ay'_n + by_n = r_n(x) $$     (9.5)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

With the same initial conditions (9.2). Plot $$ y_n(x)$$ for n=0,1 for x in [-.75,3] In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Solution
Part 1 First to project the excitation r(x)=log(1+x) onto the polynomial basis. We know that $$ <b_i,b_j> \cdot d_j=<b_i,r> $$. $$ <b_0, b_0>$$ is the only term of concern for the $$ \gamma $$ matrix when n=0.
 * {| style="width:100%" border="0"

<b_0,b_0> = \int_{-3/4}^{3} x^0x^1 dx=(3+3/4)=3.75 $$     (9.6)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

$$<b_i,r>$$ when n=0 is calculated below:
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<b_0,r> = \int_{-3/4}^{3} x^0log(1+x)dx=2.141751035 $$     (9.7) This means that, as stated in the opening sentence:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

<b_i,b_j> \cdot d_0=<bi,r> $$ $$\displaystyle 3.75 \cdot d_0=2.141751035 $$ $$\displaystyle d_0=.571136093 $$     (9.8) With that in mind, $$ r_n(x)$$ is developed from equation (9.4) as shown below:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

r(x)\approx r_n (x) = \sum_{j=0}^{n} d_jx^j= \sum_{j=0}^{0} d_jx^0=.571136093 $$     (9.9) Now projecting the excitation onto the polynomial basis with n=1: We know that (in matrix form) $$ \gamma \cdot d_j=<b_i,r> $$. Beginning with the $$ \gamma $$ matrix:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

\gamma = \begin{bmatrix} <b_0,b_0> & <b_0,b_1>\\ <b_1,b_0>& <b_1,b_1> \end{bmatrix} $$     (9.9) This matrix becomes:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

\gamma = \begin{bmatrix} 3.75 & 4.21875\\ 4.21875& 9.140625 \end{bmatrix} $$     (9.10) The matrix containing $$<b_i,r>$$ is shown below, defined as matrix c:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

c =\begin{Bmatrix} <b_o,r>\\ <b_1,r> \end{Bmatrix} ==== \begin{Bmatrix} \int_{-3/4}^{3} x^0log(1+x))\\ \int_{-3/4}^{3} x^1log(1+x)) \end{Bmatrix} ==== \begin{Bmatrix} 2.141751035\\ 5.007550553 \end{Bmatrix} $$     (9.10) We then find the $$ d_j $$ matrix in the following way:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

\gamma ^{-1} \cdot c=d_j =\begin{Bmatrix} -.0939750342\\ .5912076831 \end{Bmatrix} $$     (9.11) With that in mind, $$ r_n(x)$$ is developed from equation (9.4) as shown below:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0"

r(x)\approx r_n (x) = \sum_{j=0}^{n} d_jx^j= \sum_{j=0}^{1} d_jx^j=-.0939750342+.5912076831x $$     (9.12) Graphed out compared to the actual excitation of log(1+x) [in red], the projections with n=0[in blue],1[in green] are compared below:
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Part 2 First we create the characteristic equation in standard form:
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$$\displaystyle{\lambda^2-3 \lambda +2=0}$$ (9.13) Then, by setting it equal to zero, we can find what $$ \lambda \!$$ equals:
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 *

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$$\displaystyle{(\lambda - 2)(\lambda -1) = 0}$$ (9.14)
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 *

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$$\displaystyle{\lambda = 2,\lambda =1}$$ (9.15)
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 *

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Given two, distinct, real roots, the homogeneous solution looks like this:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h(x)=C_1e^{2x} + C_2e^{x} $$     (9.16) By using the method of undetermined coefficients, for n=0, the excitation $$ 5.571136093$$ is analyzed to yield a particular solution: In assessing a polynomial with a power of 0, the form of the particular solution will look like this:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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$$  \displaystyle y_p (x)= A_0 $$     (9.17)
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 * style="width:95%" |
 *

 * }

The first and second derivatives of $$ A_0 $$, being a constant, are 0. So when plugging in $$ y_p (x)$$ into (9.0), $$ A_0 $$ is determined to be:
 * {| style="width:100%" border="0"

.571136093=2 A_0 $$ $$\displaystyle .2855680465= A_0 $$     (9.18)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

The general solution, after adding $$ y_h $$ and $$ y_p $$ then becomes:
 * {| style="width:100%" border="0"

y_g(x)=C_1e^{2x} + C_2e^{x} + .2855680465 $$     (9.18)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

We consider the initial conditions by taking the first derivative of the general solution:

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$$  \displaystyle 2C_1e^{2x} + C_2e^{x} = y'_g(x) $$     (9.19)
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 * style="width:95%" |
 *

 * }

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
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$$  \displaystyle y_g(-3/4)=1 =C_1e^{2(-3/4)} + C_2e^{-3/4} + .2855680465 $$     (9.20)
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 *

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$$  \displaystyle y'_g(-3/4)=0= 2C_1e^{2(-3/4)} + C_2e^{-3/4} $$     (9.21)
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 * style="width:95%" |
 *

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Solving the equations proves that $$C_1 = -3.201861878, C_2 = 3.024904915 \!$$: The resulting complete solution with consideration for initial conditions then becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle y_g(x)=(-3.024904915)e^{2x} + (3.024904915)e^{x} + .2855680465 $$     (9.22) By using the method of undetermined coefficients, for n=1, the excitation $$ 5.571136093$$ is analyzed to yield a particular solution: In assessing a polynomial with a power of 0, the form of the particular solution will look like this:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle y_p (x)= A_1 x+ A_0 $$     (9.23)
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 * style="width:95%" |
 *

 * }

The first derivative of $$ y_p $$ is simply $$ A_1 $$ and the second derivative is 0. So when plugging in $$ y_p (x)$$ into (9.0), $$ A_0, A_1 $$ are determined to be:
 * {| style="width:100%" border="0"

.5912076831=2 A_1 $$ $$\displaystyle .2956038416= A_1 $$ $$\displaystyle -.0939750342= -3A_1+2A_0 $$ $$\displaystyle -.0939750342= -3(.2956038416)+2A_0 $$ $$\displaystyle .3964182452=A_0 $$     (9.24)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

The general solution, after adding $$ y_h $$ and $$ y_p $$ then becomes:
 * {| style="width:100%" border="0"

y_g(x)=C_1e^{2x} + C_2e^{x} + .2956038416x + .3964182452 $$     (9.25)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

We consider the initial conditions by taking the first derivative of the general solution:

 * {| style="width:100%" border="0"

$$  \displaystyle 2C_1e^{2x} + C_2e^{x} + .2956038416 = y'_g(x) $$     (9.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By plugging in -3/4 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
 * {| style="width:100%" border="0"

$$  \displaystyle y_g(-3/4)=1 =C_1e^{2(-3/4)} + C_2e^{-3/4} + .2956038416(-3/4) + .3964182452 $$     (9.27)
 * style="width:95%" |
 * style="width:95%" |
 *

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$$  \displaystyle y'_g(-3/4)=0= 2C_1e^{2(-3/4)} + C_2e^{-3/4} + .2956038416 $$     (9.28)
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 * style="width:95%" |
 *

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Solving the equations proves that $$C_1 = -13.92092674, C_2 = 12.54701279 \!$$: The resulting complete solution with consideration for initial conditions then becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle y_g(x)=(-13.92092674)e^{2x} + (12.54701279)e^{x} + .2956038416x + .3964182452 $$     (9.29) Graphed below, the approximations are shown. In green, the approximation with n=1, in blue, the approximation with n=0, and in red, the truncated Taylor series as n=1 is shown over the interval of [-3/4 to 3].
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 *

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Author
Solved and Typed By - Egm4313.s12.team1.silvestri (talk) 21:26, 29 March 2012 (UTC) Reviewed By - Egm4313.s12.team1.durrance (talk) 19:02, 30 March 2012 (UTC)


text:
University of Florida/Egm4313/s12.teamboss/R6

Statement
Find the (smallest) period of $$ \cos n \omega x $$ and $$ \sin n \omega x $$. Show that these functions also have period $$ p $$. Show that the constant $$ a_0 $$ is also a periodic function with period $$ p $$.

Solution
First, we identify the property of periodic functions, as shown on p9-2 of the notes, equation (1)

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$$\displaystyle{f(x+np)=f(x)}$$ (1.0)
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 *
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So for $$ \cos n \omega x $$, the following determines the smallest period.

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$$\displaystyle{\cos (n\omega x})$$ (1.1) We then add $$ 2 \pi $$ to see a final expression that will be equal to $$\displaystyle{f(x+np)=f(x)} $$
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 *
 * }
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$$\displaystyle{\cos (n\omega x + 2 \pi)}$$ (1.2)
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 *
 * }

We then pull out a term that leaves an x plus a number (our period) inside the trigonometric function, cosine.
 * {| style="width:100%" border="0"

$$\displaystyle{\cos[n\omega ( x + \frac{2 \pi}{n \omega})]}$$ (1.3) This means the smallest period is $$ \frac{2 \pi}{n \omega} $$.
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 *
 * }

For $$ \sin n \omega x $$, the following determines the smallest period.

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$$\displaystyle{\sin (n\omega x)}$$ (1.4) We then add $$ 2 \pi $$ to see a final expression that will be equal to $$\displaystyle{f(x+np)=f(x)} $$
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 *
 * }
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$$\displaystyle{\sin (n\omega x + 2 \pi)}$$ (1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We then pull out a term that leaves an x plus a number (our period) inside the trigonometric function, cosine.
 * {| style="width:100%" border="0"

$$\displaystyle{\sin[n\omega ( x + \frac{2 \pi}{n \omega})]}$$ (1.6) This means the smallest period is $$ \frac{2 \pi}{n \omega} $$, just as for its cosine equivalent.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From the notes (page 9-5, (1)), it can be stated that $$ \omega = \frac{2 \pi}{p} $$ where $$ p $$ is the period. Both of these functions have $$ \frac{2 \pi}{n \omega} $$ as their period. If we put in 1 for n, and $$ \frac{2 \pi}{p} $$ for $$\omega $$, the period simplifies to simply $$ \displaystyle p$$.

As for the constant $$ a_0 $$, a constant is, almost by definition, a periodic function. The property of periodic functions as referenced in (1.0) supports this.

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$$\displaystyle{f(x+np)=f(x)}$$ (1.0) For a function of f that is a constant, x is not a factor effecting its value. This means, no matter what is added to x, the function will return the same constant number. This added value can just as well be $$ p $$, and thus, $$ a_o $$ is a periodic function with period $$\displaystyle p $$.
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 *
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Author
Solved and Typed By -Egm4313.s12.team1.silvestri (talk) 02:19, 7 April 2012 (UTC) Reviewed By - --Egm4313.s12.team1.rosenberg (talk) 01:19, 11 April 2012 (UTC)

Statement
Kreyszig pg 491 pbs. 11,12 Is the given function odd or even or neither even nor odd? Show details of your work. -11)$$ f(x) = x^2 (-1 < x <1), p=2 $$ -12)$$ f(x) = 1- \frac{x^2}{4} (-2 < x <2), p=4 $$ -

Continued)Find the Fourier series expansion for $$ f(x) $$ on p.9-8 as follows: 1. Develop the Fourier series expansion of $$f(\bar{x})$$ Plot $$f(\bar{x})$$ and the truncated Fourier series $$f_n(\bar{x})$$

dsfawf

for n=0,1. Observe the values of $$f(\bar{x})$$ at the points of discontinuities, and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of $$ f(x) $$. Hint: $$f(\bar{x})$$ is an even function; what would become of the coefficients $$\bar{b}_k$$? 2. Do the same as above, but using $$f(\tilde{x})$$ to obtain the Fourier series expansion of $$ f(x) $$; compare to the result obtained above. Hint: Transform $$f(\tilde{x})$$ into an odd function, and expand this odd function into Fourier series; observe the coefficients of this series.

Solution
'''Part 1: K 2011 p.491 pb. 11'''
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$$  \displaystyle f(x)=x^2, (-1<x<1), p=2 $$     (2.0) The preceding function is an even function. Even functions are functions where $$f(-x)=f(x)\!$$. Odd functions on the other hand are functions where $$f(-x)=-f(x)\!$$. The Fourier series of even and odd functions can be found more easily than the Fourier series of a function that is neither even nor odd. Normal Fourier series contain both sine and cosine terms. Because of the nature of even functions, the coefficient of all sine terms in their series is zero. Likewise, all cosine terms in odd functions have a coefficient of zero. Therefore its Fourier series expansion will be of the form:
 * style="width:95%" |
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 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=a_0+\sum_{n=1}^{\infty}a_n \cos \frac{n\pi}{L}x $$     (2.1) Where:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_0=\frac{1}{L} \int_0^Lf(x)dx $$     (2.2) And
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 *
 * }
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$$  \displaystyle a_n=\frac{2}{L} \int_0^Lf(x) \cos \frac{n\pi x}{L}dx $$     (2.3) To get the constant term of the expansion, equation (2.0) will be substituted into equation (2.2) with L=1:
 * style="width:95%" |
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 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a_0=\frac{1}{1} \int_0^1x^2dx=\left [ \frac{x^3}{3} \right ]^1_0=\frac{1}{3} $$     (2.4) To get the coefficients of the rest of the terms in the expansion, equation (2.0) will be substituted into equation (2.3) with L=1:
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 *
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$$  \displaystyle a_n=\frac{2}{1} \int_0^1x^2 \cos \frac{n\pi x}{1}dx=2\left [ \frac{(\pi^2n^2x^2-2)\sin (\pi n x)+2\pi n x \cos(\pi nx)}{\pi^3n^3} \right ]^1_0 $$     (2.5) Simplifying, this yields:
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 *
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 * {| style="width:100%" border="0"

$$  \displaystyle a_n=\frac{4}{n^2 \pi^2} \cos n\pi $$     (2.6) Therefore, the Fourier expansion of f(x) is:
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 *
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$$  \displaystyle x^2=\frac{1}{3}+\sum_{n=1}^{\infty}\left (\frac{4}{n^2 \pi^2} \cos n\pi \right ) \cos n\pi x $$ (2.7) '''Part 2: K 2011 p.491 pb. 12'''
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$$  \displaystyle f(x)=1-\frac{x^2}{4}, (-2<x<2), p=4 $$     (2.8) The preceding function is an even function. Therefore, equations (2.1)-(2.3) are applicable. To get the constant term of the expansion, equation (2.8) will be substituted into equation (2.2) with L=1:
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$$  \displaystyle a_0=\frac{1}{2} \int_0^21-\frac{x^2}{4}dx= \frac{1}{2}\left[ x-\frac{x^3}{12} \right ]^2_0=\frac{4}{3} $$     (2.9) To get the coefficients of the rest of the terms in the expansion, equation (2.0) will be substituted into equation (2.3) with L=1:
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$$  \displaystyle a_n=\frac{2}{2} \int_0^2\left (1-\frac{x^2}{4} \right ) \cos \frac{n\pi x}{2}dx=\left [ \frac{(8-\pi^2n^2(x^2-4))\sin\frac{n\pi x}{2}-4n\pi x\cos\frac{n\pi x}{2}}{2\pi^3n^3} \right ]^2_0 $$     (2.10) Simplifying, this yields:
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$$  \displaystyle a_n=\frac{-4\cos(n\pi)}{n^2 \pi^2} $$     (2.11) Therefore, the Fourier expansion of f(x) is:
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$$  \displaystyle 1-\frac{x^2}{4}=\frac{4}{3}+\sum_{n=1}^{\infty}\left (\frac{-4\cos(n\pi)}{n^2 \pi^2} \right ) \cos \frac{n\pi x}{2} $$     (2.12) Part 3 The Fourier series expansion of the function f(x) in Figure 1 can be found in one of two ways. The first way involves transforming x in the following way:
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$$  \displaystyle \bar x=x-\frac54 $$     (2.13) The resulting function is now an even function that alternates between the constant value A and zero and has period 4. Using the same methods used above, the Fourier series of the function can be found. The constant coefficient of the series is found by:
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$$  \displaystyle \bar a_0=\frac12 \int_0^2f(x)dx=\frac12 \int_0^1Adx+\frac12 \int_1^20dx=1/2\left [ Ax \right ]_0^1=\frac A2 $$ (2.14) Note that because the function is discontinuous at the midpoint, the integral must be split into two different integrals. The coefficients of the cosine terms of the series can be found by:
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$$  \displaystyle \bar a_n=\frac22 \int_0^2f(x)\cos \frac{n\pi x}{2}dx= \int_0^1A\cos \frac{n\pi x}{2}dxdx+\frac12 \int_1^20dx=\left [ \frac{2A}{n\pi}\sin\frac{n\pi x}{2} \right ]_0^1=\frac{2A}{n\pi}\sin \frac{n\pi}{2} $$     (2.15) Therefore, the Fourier series for the transformed function is:
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$$  \displaystyle f(\bar x)=\frac A2 + \sum_{n=0}^{\infty} \left ( \frac{2A}{n\pi}\sin\frac{n\pi}{2} \right )\left ( \cos \frac{n\pi\bar x}{2} \right ) $$     (2.16) Through a change in variables, the Fourier series of the original function, f(x), can be found:
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$$  \displaystyle f(x)=\frac A2 + \sum_{n=0}^{\infty} \left ( \frac{2A}{n\pi}\sin\frac{n\pi}{2} \right )\left ( \cos \frac{n\pi(x-1.25)}{2} \right ) $$     (2.17) The following figure shows a plot of $$f(\bar x)\!$$ with its truncated Fourier series for n=0,1.
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Figure 2 Figure 3 shows the same plot with more values of n:

Note that all plots assume that A=1. It should be noted that the accuracy of the expansion increases with an increase in the number of terms in the expansion. Also, there is strong fluctuation at points of discontinuity. These fluctuations are apparent even at high values of n, but they decrease in size and move closer to the points of discontinuity. This is a property of Fourier series known as the Gibbs phenomenon. The other method of developing the Fourier series is to transform the function into an odd function. This can be done by transforming both the independent variable and overall function as follows:
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$$  \displaystyle \tilde x=x-\frac14 $$     (2.18)
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$$  \displaystyle f(\hat x)=f(\tilde x)-\frac A2 $$ (2.19) An odd function has a Fourier series expansion of the form:
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$$  \displaystyle f(x)=\sum_{n=0}^{\infty}b_n\sin \frac{n\pi x}{L} $$     (2.20) Where:
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$$  \displaystyle b_n=\frac2L\int_{0}^{L}f(x)\sin \frac{n\pi x}{L}dx $$     (2.21) Applying this to the transformed function, with L=2:
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$$  \displaystyle \hat b_n=\frac22\int_{0}^{2}f(x)\sin \frac{n\pi \hat x}{2}d\hat x=\int_{0}^{2}\frac A2\sin \frac{n\pi \hat x}{2}d\hat x=\left [ -\frac{A}{n\pi}\cos\frac{n\pi\hat x}{2} \right ]^2_0=\frac{A}{n\pi}(1-\cos n\pi) $$     (2.22) The Fourier series is therefore:
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$$  \displaystyle f(\hat x)=\sum_{n=0}^{\infty}\left (\frac{A}{n\pi}(1-\cos n\pi) \right )\left ( \sin\frac{n\pi\hat x}{2} \right ) $$     (2.23)
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$$  \displaystyle f(\tilde x)=\frac A2+\sum_{n=0}^{\infty}\left (\frac{A}{n\pi}(1-\cos n\pi) \right )\left ( \sin\frac{n\pi\tilde x}{2} \right ) $$     (2.24)
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$$  \displaystyle f(x)=\frac A2+\sum_{n=0}^{\infty}\left (\frac{A}{n\pi}(1-\cos n\pi) \right )\left ( \sin\frac{n\pi(x-0.25)}{2} \right ) $$     (2.25) The following figure shows a plot of $$f(\tilde x)\!$$ with its truncated Fourier series for n=0,1.
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Figure 4 Note that all plots assume A=1. The accuracy of the expansion as shown on the plot is expected given the low number of terms used in the expansion. As Figure 3 shows, if the number of terms used to form the plot in Figure 4 is increased, the accuracy of the expansion will greatly increase (but more slowly at points of discontinuity because of the Gibbs phenomenon). This level of accuracy is shown in Figure 5, which contains the expansion to 100 and 200 terms.

Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 07:10, 11 April 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri (talk) 18:30, 11 April 2012 (UTC)

Statement
Kreyszig pg 491 pbs 15,17 Is the given function even or odd or neither even not odd? Find its Fourier series. Show details of your work. -15)

-17)

- Plot the truncated Fourier series for n=2,4,8

Solution
By definition, a function, $$ f(x)\!$$ is even if $$ f(-x) = f(x)\!$$, and its Fourier series reduces to a Fourier cosine series:
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$$\displaystyle{f(x) = a_{0} + \sum_{n=1}^{\infty}a_{n}cos{\frac{n\pi}{L}}x}$$ (3.0)
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A function, $$ f(x)\!$$ is odd if $$ f(-x) = -f(x)\!$$, and its Fourier series reduces to a Fourier sine series:
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$$\displaystyle{f(x) = \sum_{n=1}^{\infty}b_{n}sin{\frac{n\pi}{L}}x}$$ (3.1) -15) By inspection, it is evident that the given function in problem 15 is odd because for every $$ x\!$$ value that is negative, $$ f(x)\!$$ is negative. Therefore, $$ f(-x) = -f(x)\!$$.
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The following are the defining functions for the given graph:
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$$\displaystyle{f(x) = \begin{cases} -\pi - x & \text{ if } x= -\pi < x < -\frac{\pi}{2}\\ x & \text{ if } x= -\frac{\pi}{2} < x < \frac{\pi}{2}\\ \pi - x & \text{ if } x= \frac{\pi}{2} < x < \pi \end{cases}}$$ (3.2)
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Because the function is odd, we use equation (3.1) to develop the Fourier series, where the period is $$2\pi\!$$ and $$ L\!$$ equals $$\pi\!$$:
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$$\displaystyle{f(x) = \sum_{n=1}^{\infty}b_{n}sin{\frac{n\pi}{L}}x = \sum_{n=1}^{\infty}b_{n}sin(nx)}$$ (3.3)
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Now we need to find $$ b_{n}\!$$, which is of the following form for odd functions:
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$$\displaystyle{b_{n} = \frac{2}{L}\int_{0}^{L}f(x)\sin{\frac{n{\pi}x}{L}}dx}$$ (3.4)
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$$\displaystyle{b_{n} = \frac{2}{\pi}\int_{0}^{\pi}sin(nx)dx = \frac{2}{\pi}[\int_{0}^{\pi/2}xsin(nx)dx + \int_{\pi/2}^{\pi}(\pi-x)sin(nx)dx]}$$ (3.5)
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$$\displaystyle{b_{n} = \frac{2}{\pi}[[\frac{sin(nx)-nxcos(nx)}{n^2}]_{0}^{\pi/2}+[\frac{-sin(nx)+n(\pi-x)cos(nx)}{n^2}]_{\pi/2}^{\pi}]}$$     (3.6)
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Evaluating and simplifying the definite integral gives the following:

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$$\displaystyle{b_{n} = \frac{2}{n^2\pi}[2sin(\frac{n\pi}{2})-sin(n\pi)]}$$ (3.7)
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Evaluating equation (3.7) at $$ n = 1, 2, 3, 4, 5, 6, 7, 8\!$$

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$$\displaystyle{b_{1} = \frac{4}{\pi}, b_{2} = 0, b_{3} = -\frac{4}{9\pi}, b_{4} = 0, b_{5} = \frac{4}{25\pi}, b_{6} = 0, b_{7} = -\frac{4}{49\pi}, b_{8} = 0}$$ (3.8)
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Thus:

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$$ f(x) = \frac{4}{\pi}(sinx - \frac{1}{9}sin3x + \frac{1}{25}sin5x - \frac{1}{49}sin7x +...) $$ (3.9)
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Plots of truncated Fourier series: n = 2: n = 4: n = 8:

-17) By inspection, it is evident that the given function in problem 17 is even because for every $$ x\!$$ value that is negative, $$ f(x)\!$$ is positive. Therefore, $$ f(-x) = f(x)\!$$.

The following are the defining functions for the given graph:
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$$\displaystyle{f(x)=\begin{cases}x+1 & \text{ if } -1<x<0\\ 1-x & \text{ if } 0<x<1 \end{cases}}$$ (3.10)
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Because the function is even, we use equation (3.0) to develop the Fourier series, where the period is $$2\!$$ and $$ L = 1\!$$:

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$$\displaystyle{f(x) = a_{0} + \sum_{n=1}^{\infty}a_{n}cos{\frac{n\pi}{L}x} = a_{0} + \sum_{n=1}^{\infty}a_{n}cos(n\pi x)}$$ (3.11)
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Now we need to find $$a_{0}\!$$ using Euler's Formula:
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$$\displaystyle{a_{0} = \frac{1}{2L}\int_{-L}^{L}f(x)dx}$$ (3.12)
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$$\displaystyle{a_{0} = \frac{1}{2}\int_{-1}^{1}f(x)dx = \frac{1}{2}[\int_{-1}^{0}(x+1)dx + \int_{0}^{1}(1-x)dx]}$$ (3.13)
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$$\displaystyle{a_{0} = \frac{1}{2}[[\frac{x^2}{2}+x]_{-1}^{0}+[x-\frac{x^2}{2}]_{0}^{1}]}$$     (3.14)
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Evaluating the definite integral in equation (3.14) yields $$ a_{0}\!$$:

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$$\displaystyle{a_{0} = \frac{1}{2}}$$ (3.15)
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Now we find $$ a_{n}\!$$ using Euler's Formula:

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$$\displaystyle{a_{n} = \frac{1}{L}\int_{-L}^{L}f(x)cos{\frac{n{\pi}x}{L}}dx = \int_{-1}^{1}f(x)cos(n\pi x)dx}$$ (3.16)
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$$\displaystyle{a_{n} = \int_{-1}^{0}(x+1)cos(n\pi x)dx + \int_{0}^{1}(1-x)cos(n\pi x)dx}$$ (3.17)
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$$\displaystyle{a_{n} = [\frac{\pi n(x+1)sin(\pi nx) + cos(\pi nx)}{n^2\pi^2}]_{-1}^{0} + [-\frac{\pi n(x-1)sin(\pi nx)+cos(\pi nx)}{n^2\pi^2}]_{0}^{1}}$$ (3.18)
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Evaluating the definite integral in equation (3.18) yields the following equation for $$a_n\!$$:

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$$\displaystyle{a_{n} = \frac{2(1-cos(\pi))}{n^2\pi^2}}$$ (3.19)
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Evaluating equation (3.19) at $$ n = 1, 2, 3, 4, 5, 6, 7, 8\!$$ yields:

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$$\displaystyle{a_{1} = \frac{4}{\pi^2}, a_{2} = 0, a_{3} = \frac{4}{9\pi^2}, a_{4} = 0, a_{5} = \frac{4}{25\pi^2}, a_{6} = 0, a_{7} = \frac{4}{49\pi^2}, a_{8} = 0}$$ (3.20)
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Thus, the Fourier series is:

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$$ f(x) = \frac{1}{2} + \frac{4}{\pi^2}(cos(\pi x) + \frac{1}{9}cos(3\pi x) + \frac{1}{25}cos(5\pi x) + \frac{1}{49}cos(7\pi x) + ...) $$ (3.21)
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Plots of truncated Fourier series: n = 2: n = 4: n = 8:

Author
Solved and Typed By ---Egm4313.s12.team1.wyattling (talk) 06:19, 11 April 2012 (UTC) Reviewed By -Egm4313.s12.team1.armanious (talk) 15:18, 11 April 2012 (UTC)

Statement
Consider the L2-ODE-CC (5) p7b-7 with the window function f(x) p.9-8 as excitation:
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$$\displaystyle{y''-3y'+2y=r(x)}$$ $$r(x)=f(x)$$ (4.0) and the initial conditions
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$$\displaystyle{y(0)=1, y'(0)=0}$$ (4.1) 1. Find $$ y_n(x) $$ such that:
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$$\displaystyle{y''_n +ay'_n + by_n=r_n(x)}$$ (4.2) with the same initial conditions (4.1). Plot $$ y_n(x)$$ for n=2,4,8, for x in [0,10]. 2.Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution. Level 1:n=0,1
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Part 1 Solution
First, we shift the excitation f(x) to the left by introducing a new independent variable, t. This allows the period to start at zero.
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$$\displaystyle t=x-\frac{1}{4}$$ (4.3) The piecewise representation of the window function is now (in terms of t) as follows:
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$$\displaystyle f(t)=A$$ for $$\displaystyle t=\left[ 0,2 \right]$$ $$\displaystyle f(t)=0$$ for $$\displaystyle t=\left[ 2,4 \right]$$ (4.4) To find the Fourier transform, the period of oscillation is determined:
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$$\displaystyle p=2L=4$$ (4.5) And the frequency of oscillation:
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$$\displaystyle \omega =\frac{2\pi }{p}=\frac{\pi }{2}$$ (4.6) The general form of a Fourier transform is
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$$\displaystyle f(t)={{a}_{0}}+\sum\limits_{1}^{n}{\left[ {{a}_{n}}\cos (n\omega t)+{{b}_{n}}\sin (n\omega t) \right]}$$ (4.7) The following equations are given values of the constants in (4.7), evaluated for the function in (4.4):
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$$\displaystyle {{a}_{0}}=\frac{1}{2L}\int\limits_{0}^{2L}{f(t)dt}=\frac{1}{4}\int\limits_{0}^{2}{Adt}=\frac{A}{4}(2-0)=\frac{A}{2}$$ (4.8)
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$$\displaystyle {{a}_{n}}=\frac{1}{L}\int\limits_{0}^{2L}{f(t)\cos (n\omega t)dt} =\frac{1}{2}\int\limits_{0}^{2}{A\cos (\frac{n\pi }{2}t)dt} =\frac{A}{n\pi }(\sin n\pi -\sin 0)$$ (4.9)
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$$\displaystyle {{b}_{n}}=\frac{1}{L}\int\limits_{0}^{2L}{f(t)\sin (n\omega t)dt}=\frac{1}{2}\int\limits_{0}^{2}{A\sin (\frac{n\pi }{2}t)dt}=-\frac{A}{n\pi }(\cos n\pi -\cos 0)=\frac{A}{n\pi }(1-\cos n\pi )$$ (4.10) In their simplest forms:
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$$\displaystyle {{a}_{n}}=0$$ (4.11)
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$$\displaystyle {{b}_{n}}=\left\{ 0,\frac{A}{\pi },0,\frac{A}{2\pi },0,\frac{A}{3\pi },0,\frac{A}{4\pi },...,0,\frac{2A}{n\pi } \right\}$$ (4.12) Plugging in r(x)=f(t) in (4.0),
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$$\displaystyle y''-3y'+2y=\frac{1}{2}+\frac{2}{n\pi }\sin (\frac{n\pi }{2}t)$$ (4.13) From the general form, a particular solution to (4.13) and its derivatives are as follows:
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$$\displaystyle {{y}_{n}}={{A}_{n}}\cos (\frac{n\pi }{2}t)+{{B}_{n}}\sin (\frac{n\pi }{2}t)$$ (4.14)
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$$\displaystyle {{y}_{n}}'=-{{A}_{n}}\frac{n\pi }{2}\sin (\frac{n\pi }{2}t)+{{B}_{n}}\frac{n\pi }{2}\cos (\frac{n\pi }{2}t)$$ (4.15)
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$$\displaystyle {{y}_{n}}''=-{{A}_{n}}\frac{4}\cos (\frac{n\pi }{2}t)-{{B}_{n}}\frac{4}\sin (\frac{n\pi }{2}t)$$ (4.16)
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By plugging in (4.14), (4.15), and (4.16) into (4.13), we find An and Bn in terms of another constant, Cn:

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$$\displaystyle {{A}_{n}}=6n\pi {{C}_{n}}$$ (4.17)
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$$\displaystyle {{B}_{n}}=(8-{{n}^{2}}{{\pi }^{2}}){{C}_{n}}$$ (4.18)
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$$\displaystyle {{C}_{n}}={{\left[ \tfrac{1}{8}n\pi ({{(8-{{n}^{2}}{{\pi }^{2}})}^{2}}+{{(6n\pi )}^{2}}) \right]}^{-1}}$$ (4.19) After substituting t with (4.3), the solutions are shown for n=2,4,8.
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$$\displaystyle {{y}_{2}}={{y}_{1}}+{{y}_{2}}$$ (4.20)
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$$\displaystyle {{y}_{4}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}$$ (4.21)
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$$\displaystyle {{y}_{8}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}+{{y}_{5}}+{{y}_{6}}+{{y}_{7}}+{{y}_{8}}$$ (4.22)
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Part 2 Solution
The excitation for n=0 and n=1 are the same, because $$a_1=b_1=0$$.

The homogeneous solution to (4.0) is
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$$\displaystyle {{y}_{h}}={{C}_{1}}{{\operatorname{e}}^{2t}}+{{C}_{2}}{{\operatorname{e}}^{t}}$$ (4.23) For $$r(x)=a_0$$, the particular solution is
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$$\displaystyle {{y}_{p}}={{C}_{3}}$$ (4.24) Evaluating, we find that
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$${{C}_{1}}=-\tfrac{3}{4}$$ $${{C}_{2}}=\tfrac{3}{2}$$ $${{C}_{3}}=\tfrac{1}{4}$$ (4.25) Thus, the complete solution $$y_0$$ is
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$$ {{y}_{0}}=\tfrac{1}{4}-\tfrac{3}{4}{{\operatorname{e}}^{2(x-\tfrac{1}{4})}}+\tfrac{3}{2}{{\operatorname{e}}^{x-\tfrac{1}{4}}}$$ (4.26) Integrating $$y_0$$ using MATLAB's ode45 command, the following plot is obtained for y:
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Author
Solved and Typed By - Egm4313.s12.team1.essenwein (talk) 00:00, 11 April 2012 (UTC) Reviewed By - --Egm4313.s12.team1.wyattling (talk) 14:02, 11 April 2012 (UTC)

Statement
Continuation of R4.2, R4.3, R4.4. R4.2,p.7c-26:$$y_n(x)=y_{h,n}(x) + y_{p,n}(x)$$ For each value of n=3,5,9, redisplay the expressions for the 3 functions $$ y_{p,n}(x),y_{h,n}(x),y_n(x) $$, and plot these 3 functions separately over the interval $$ [0,20 \pi] $$. Exact solution: $$ y(x) = y_h(x) + y_p(x)$$ Redisplay the expressions for $$y_p(x),y_h(x), y(x)$$ Superimpose each of the above plots with that of the exact solution. R4.3,p.7c-28: Level 1: Understand and run the TA's code to produce a similar plot, but over a larger interval [0,10]. Do zoom-in plots about points $$ x=-0.5,0,+0.5 $$ and comment on the accuracy of different approximations. R4.4,p.7c-29: Level 1: Understand and run the TA's code to produce a similar plot, but over a larger interval [0.9,10], and for n=4,7. Do zoom-in plots about x=1,1.5,2,2.5 and comments on the accuracy of the approximations.

R4.2,p.7c-26
This first part of the problem is to redo 4.2:

In order to find the overall solution for the L2-ODE-CC corresponding to the Taylor series expansion of the sine function, both the homogenous and particular solutions must be found. The homogenous equation can be found through this method:
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$$  \displaystyle y''-3y'+2y=0 $$     (5.0)
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$$  \displaystyle \lambda ^2-3\lambda+2=0 \rightarrow \lambda = 1,2 $$     (5.1)
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$$  \displaystyle y_{h,n}(x)=C_1e^x+C_2e^{2x} $$     (5.2) Next, the excitation must be expanded to the desired n.  The following shows the excitation expanded to n=3, 5, and 9:
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$$  \displaystyle r_3(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} $$     (5.3)
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$$  \displaystyle r_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!} $$     (5.4)
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$$  \displaystyle r_9(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!}-\frac{x^{15}}{15!}+\frac{x^{17}}{17!}-\frac{x^{19}}{19!} $$     (5.5)
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Next, the particular solution must be found. The particular solution will be of the form:
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$$  \displaystyle y_{p,n}=\sum_{i=0}^{n}k_ix^i $$     (5.6) Using the derivation discussed in R4.1, a matrix equation in the form Ak = d can be found, where k is the matrix containing the coefficients of the particular solution and d is the matrix containing the coefficients of each power of x in the expansion of the excitation. The general formula for A is:
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$$  \displaystyle \begin{bmatrix} b& a & 2 & &  & &\\ & b & 2a & 6 & & &\\ & & b & 3a & 12 & &\\ & &  & ... & ... & ... &\\ &  &  &  & b & a(n-1) & n(n-1)\\ & &  &  &  &b &an\\ & &  &  &  & &b \end{bmatrix} $$     (5.7) n=3 The matrix A is found as:
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$$  \displaystyle A = \begin{bmatrix} 2 & -3 & 2 & &  &  &  & \\ & 2 & -6 & 6 &  &  &  & \\ &  & 2 & -9 & 12 &  & &  \\ &  &  & 2 &  -12& 20 &  & \\ &  &  &  & 2 & -15 & 30 & \\ &  &  &  &  & 2 & -18 & 42\\ &  &  &  &  &  & 2 &-21 \\ &  &  &  &  &  &  & 2 \end{bmatrix} $$     (5.8) Therefore, the matrix equation is:
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$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & &  &  &  & \\ & 2 & -6 & 6 &  &  &  & \\ &  & 2 & -9 & 12 &  & &  \\ &  &  & 2 &  -12& 20 &  & \\ &  &  &  & 2 & -15 & 30 & \\ &  &  &  &  & 2 & -18 & 42\\ &  &  &  &  &  & 2 &-21 \\ &  &  &  &  &  &  & 2 \end{bmatrix} \begin{bmatrix} k_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7 \end{bmatrix}= \begin{bmatrix} 0\\ 1\\ 0\\ -\frac{1}{6}\\ 0\\ \frac{1}{120}\\ 0\\ -\frac{1}{5040} \end{bmatrix} $$     (5.9)
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$$  \displaystyle \begin{bmatrix} k_0\\ k_1\\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7 \end{bmatrix}= \begin{bmatrix} -0.1992\\ -0.3984\\ -0.3984\\ -0.0990\\ -0.0078\\ -0.0031\\ -0.0010\\ -9.9206\times 10^{-5} \end{bmatrix} $$     (5.10) Therefore the particular solution is:
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$$  \displaystyle y_{p,3}(x)=-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (5.11) And the overall solution for n=3 is:
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$$  \displaystyle y_{3}(x)=C_1e^x+C_2e^{2x}-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (5.12) Using the initial conditions:
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$$  \displaystyle 1=C_1+C_2-0.1992$$ &  $$ 0=C_1+2C_2-0.3984 $$     (5.13) Solving yields $$C_1=2 $$  &  $$ C_2=-0.8008 \!$$. Therefore the overall solution is:
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$$  \displaystyle y_{3}(x)=2e^x-0.8008e^{2x}-9.9206\times 10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x-0.1992 $$     (5.14) n=5 The matrix equation is: Therefore the particular solution is:
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$$  \displaystyle y_{p,5}(x)=-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$     (5.15) And the overall solution for n=5 is:
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$$  \displaystyle y_{5}(x)=C_1e^x+C_2e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2000 $$     (5.16) Using the initial conditions:
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$$  \displaystyle 1=C_1+C_2-0.2 \& 0=C_1+2C_2-0.3999 $$     (5.17) Solving yields $$C_1=2.0001 \& C_2=-0.8001 \!$$. Therefore the overall solution is:
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$$  \displaystyle y_{5}(x)=2.0001e^x-0.8001e^{2x}-1.2526\times 10^{-8}x^{11}-2.0668\times 10^{-7}x^{10}-1.0334\times 10^{-6}x^9-4.6503\times 10^{-6}x^8-1.1781\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.2 $$     (5.18) n=9 The matrix equation is: Therefore the particular solution is:
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$$  \displaystyle y_{p,9}(x)=-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (5.19) And the overall solution for n=9 is:
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$$  \displaystyle y_{9}(x)=C_1e^x+C_2e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (5.20) Using the initial conditions:
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$$  \displaystyle 1=C_1+C_2-0.2 \& 0=C_1+2C_2-0.4 $$     (5.21) Solving yields $$C_1=2 \& C_2=-0.8 \!$$. Therefore the overall solution is:
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$$  \displaystyle y_{9}(x)=2e^x-0.8e^{2x}-4.1103\times 10^{-18}x^{19}-1.1714\times 10^{-16}x^{18}-1.0543\times 10^{-15}x^{17}-8.9615\times 10^{-15}x^{16}-4.5405\times 10^{-13}x^{15}-9.1408\times 10^{-12}x^{14}-6.3985\times 10^{-11}x^{13}-4.1590\times 10^{-10}x^{12}-1.5021\times 10^{-8}x^{11}-2.2040\times 10^{-7}x^{10}-1.1020\times 10^{-6}x^9-4.9591\times 10^{-6}x^8-1.1904\times 10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000 $$     (5.22) The following plot shows the overall solution of the ODE at n=3,5,9 over the domain [0,4π]: Figure 4.2-2
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The near perfect overlap between all three graphs shows that the approximations converge quickly for very low values of n.

Next, the exact ODE will be calculated to find the accuracy of the above approximations. The homogeneous solution is the same as the homogeneous solution above, only with different coefficients. The particular solution will be in the form:
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$$  \displaystyle y_p(x)=K\cos x + M\sin x $$ (5.23)
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$$  \displaystyle y_p'(x)=-K\sin x + M\cos x $$ (5.24)
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$$  \displaystyle y_p''(x)=-K\cos x - M\sin x $$ (5.25) Plugging these values into the ODE:
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$$  \displaystyle -K\cos x - M\sin x + 3K\sin x - 3M\cos x + 2K\cos x + 2M\sin x =\sin x $$ (5.26) Separating sine and cosine terms yield two linear equations which can be used to solve for the unknown coefficients:
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$$  \displaystyle K-3M=0 $$ &  $$  3K+M=1 $$     (5.27) Solving these equations yield K = 0.3 and M = 0.1.
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$$  \displaystyle y_p(x)=0.3\cos x + 0.1\sin x $$ (5.28) Therefore the exact overall solution is:
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$$  \displaystyle y(x)=C_1e^x + C_2e^{2x}+0.3\cos x + 0.1\sin x $$ (5.29)
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$$  \displaystyle y'(x)=C_1e^x + 2C_2e^{2x}-0.3\sin x + 0.1\cos x $$ (5.30) Using the initial conditions yield the equations:
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$$  \displaystyle 1=C_1 + C_2 + 0.3 $$  &  $$  0=C_1 + 2C_2 + 0.1 $$     (5.31)
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Solving these two equations yield $$C_1=1.5 \& C_2=-0.8 \!$$. Therefore the exact overall solution is:
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$$  \displaystyle y(x)=1.5e^x - 0.8e^{2x}+0.3\cos x + 0.1\sin x $$ (5.32)
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The following figure shows the plot of the exact solution over the plot in Fig. 4.2-2: Figure 4.2-3

R4.3,p.7c-28
In this problem we are given the TA's Matlab code but must change the interval on which we are plotting. This give us the following code:

x=-1:0.01:1.5; y=log(1+x); x1=-1.5:0.01:1.5; y1=zeros(1,301); for i=1:4 for j=1:301 y1(j)=y1(j)-((-x1(j))^i)/i; end end y2=zeros(1,301); for i=1:7 for j=1:301 y2(j)=y2(j)-((-x1(j))^i)/i; end end y3=zeros(1,301); for i=1:11 for j=1:301 y3(j)=y3(j)-((-x1(j))^i)/i; end end y4=zeros(1,301); for i=1:16 for j=1:301 y4(j)=y4(j)-((-x1(j))^i)/i; end end h=plot(x,y); orange = [1 0.5 0.2]; set(h,'Color',orange); hold on; plot(x1,y1,'r'); plot(x1,y2,'g'); plot(x1,y3,'b'); plot(x1,y4,'c'); axis([0 10 -10 10]); legend('log(1+x)','T_4','T_7','T_1_1','T_1_6'); grid on; Zooming at -0.5 This graph shows fairly accurate results with slight deviation, with largest deviation occurring at the T4 term. Zooming about point 0 In this graph, only T16 shows, thus meaning that it is the best approximation. Zooming at 0.5 All lines appear to be fairly good approximations with the exception of T4 which deviates from the rest.

R4.4,p.7c-29
Again we must adjust the TA's Matlab code:

syms x f=log(x+1); fT1=taylor(f,5,1); fT2=taylor(f,8,1); fT3=taylor(f,12,1); X=.9:0.1:10; Y(:,1)=subs(fT1,'x',X); Y(:,2)=subs(fT2,'x',X); Y(:,3)=subs(fT3,'x',X); Y(:,4)=log(1+X); figure plot(X,Y); axis([.9,10,-10,10]);

Zooming at 1 Since the three lines are all stacked, both n=4 and n=7 are good approximations at x=1. Zooming at 1.5 Again, since the three lines are all stacked, both n=4 and n=7 are good approximations at x=1.5. Zooming at 2 Here, both can again be said to be good approximations, however, you can see some very slight deviation in n=4. Zooming at 2.5 In this plot, n=7 is still a pretty strong approximation, while n=4 is beginning to deviate at a fairly rapid rate and thus will soon be a poor approximation.

Author
Solved and Typed By - --Egm4313.s12.team1.rosenberg (talk) 01:19, 11 April 2012 (UTC) Reviewed By ---Egm4313.s12.team1.durrance (talk) 19:12, 11 April 2012 (UTC)

Statement
Verify (2) p.10-7 as follows: Given the ODE:
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$$  \displaystyle y''_{p}+4y'_{p}+13y_{p}=2e^{-2x}cos(3x) $$     (6.1) 1)Simplify the 1st term $$y''_p$$ on the lhs of (2) p.10-4. 2)Simplify the 2nd term $$4y'_p $$, and combine with the simplified 1st term. 3)Finally, add the 3rd term $$13y_p$$ 4)Find the final expression for $$y_p(x)$$
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Solution
From [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec10-1.djvu Lecture 10 Pg. 9]

To find the particular solution of an ODE we can use the method of undetermined coefficients. In report 3 we used the table 2.1 below to find the standard form of the particular solution given the form of the excitation $$r(x)\!$$. Table 2.1 So the general form of the particular solution is:
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$$  \displaystyle y_{p}(x)=xe^{-2x}[Mcos3x+Nsin3x] $$     (6.2)
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Task 1 asks to simplify they first term of the given ODE which is $$y_{p}''(x)\!$$. This is done by taking the second derivative of $$y_{p}(x)\!$$ and plugging it into equation 6.1
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$$  \displaystyle y''_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.3)
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$$y_{p}''(x)\!$$ simplifies to:
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$$  \displaystyle y''_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.4)
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Now we must do the same for the second term $$4y_{p}'(x)\!$$.
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$$  \displaystyle 4y'_{p}(x)=\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.5)
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$$  \displaystyle 4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)] $$     (6.6)
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Add the second term to the first to continue building the original ODE (equation 6.1).
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$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}+\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.7)
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$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.8)
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Finally add the third term $$13y_{p}(x)\!$$ to the other two to have the lhs of the ODE 6.1
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$$  \displaystyle 13y_{p}(x)=13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.9)
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$$  \displaystyle y''_{p}(x)+4y'_{p}(x)+13y_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)])+4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.10)
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$$  \displaystyle y_{p}(x)=xe^{-2x}[Mcos3x+Nsin3x]=6e^{-2x}[Ncos(3x)-Msin(3x)] $$     (6.11)
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This simplifies to the particular solution below:
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$$  \displaystyle y_{p}(x)=6e^{-2x}[Ncos(3x)-Msin(3x)] $$     (6.12)
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Author
Solved and Typed By ---Egm4313.s12.team1.stewart (talk) 23:50, 10 April 2012 (UTC) Reviewed By - Egm4313.s12.team1.essenwein (talk) 18:12, 11 April 2012 (UTC)

Statement
Find the separated ODEs for the heat equation:
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$$ \frac{\partial u}{\partial t}=\kappa \frac{\partial^2 u}{\partial x^2} $$     (7.0) Where $$\kappa$$ is the coefficient of heat conductivity.
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Solution
Assume:
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$$ u(x,t)=F(x) \cdot G(t) $$     (7.1) So, by take partial derivatives of Eq. (7.1) with respect to the individual variables:
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$$ \frac{\partial u(x,t)}{\partial t}=F(x) \cdot \dot G(t) $$     (7.2)
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$$ \frac{\partial^2 u(x,t)}{\partial x^2}=F''(x) \cdot G(t) $$     (7.3)
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By substituting the partial derivatives back into Eq. (7.0):
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$$ F(x) \cdot \dot G(t)=\kappa F''(x) \cdot G(t) $$     (7.4)
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Separating the equations to similar variables yields:
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$$ \frac{\dot G(t)}{G(t)}=\kappa \frac{F''(x)}{F(x)}=C \text{ (constant)} $$     (7.4a)
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Therefore the separated ODEs are:
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$$ \dot G(t)-CG(t)=0$$ $$ F''(x)-\frac{C}{\kappa}F(x)=0 $$     (7.5)
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Author
Solved and Typed By - --Egm4313.s12.team1.durrance (talk) 02:13, 9 April 2012 (UTC) Reviewed By - --Egm4313.s12.team1.stewart (talk) 23:48, 10 April 2012 (UTC)

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