text:
University of Florida/Egm4313/s12.teamboss/R3

Statement
Given the double root $$ \lambda = 5\!$$ and the excitation $$ r(x)=7e^{5x}-2 x^2 \!$$, with the initial conditions $$ y(0) = 4, y'(0) = 5 \!$$ find the solution $$ y(x)\!$$. Plot this solution and the solution to the same problem except with the excitation $$ r(x) = 7e^{5x} \!$$.

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$. Using the given double root, we can find the equation for the characteristic equation which leads to the homogeneous solution:

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$$\displaystyle{(\lambda - 5)^2 = \lambda^2 - 10\lambda + 25 = 0}$$ (1.0)
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Homogeneous solution:
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$$\displaystyle{y'' - 10y' + 25y = r(x)}$$ (1.1)
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$$\displaystyle{y'' - 10y' + 25y = 7e^{5x} - 2x^2}$$ (1.2)
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First, by using the Modification Rule, we find that the general equation associated with the given double root is:

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$$\displaystyle{y_g(x) = C_1e^{5x} + C_2xe^{5x}}$$ (1.3)
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We need to find the particular solution to the excitation $$ r(x) = 7e^{5x} - 2x^2\!$$. In analyzing the excitation, it is found that the particular solution looks like this:

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$$\displaystyle{y_p(x) = Cx^2e^{5x} - [K_2x^2 + K_1x + K_0]}$$ (1.4)
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Now, we need to find the values for the constants $$ C, K_2, K_1, K_0\!$$, by taking the first and second derivatives of the particular solution:

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$$\displaystyle{y_p(x) = Cx^2e^{5x} - [K_2x^2 + K_1x + K_0]}$$ (1.4)
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$$\displaystyle{y_p(x)' = 2Cxe^{5x} + 5Cx^2e^{5x} - [2K_2x + K_1]}$$ (1.5)
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$$\displaystyle{y_p(x)'' = 2Ce^{5x} + 20Cxe^{5x} + 25Cx^2e^{5x} - [2K_2]}$$ (1.6)
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Now, plug these into the homogeneous solution (1.2) and simplify:

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$$\displaystyle{2Ce^{5x} - 2K_2 + 20K_2x + 10K_1 - 25K_2x^2 - 25K_1x - 25K_0 = 7e^{5x} - 2x^2}$$ (1.7)
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Comparing the coefficients for $$ e^{5x}, x^2 \!$$ and $$ x \!$$ allows us to solve for the unknown coefficients:

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$$\displaystyle{e^{5x}: 2C = 7 \rightarrow C = 7/2}$$ (1.8)
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$$\displaystyle{x^2: -25K_2 = -2 \rightarrow K_2 = 0.08}$$ (1.9)
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$$\displaystyle{x: 20K_2 - 25K_1 = 0 \rightarrow K_1 = 0.064}$$ (1.10)
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$$\displaystyle{-2K_2 + 10K_1 - 25K_0 = 0 \rightarrow K_0 = 0.0192}$$ (1.11)
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Plug into $$ y_p(x)\!$$ (1.4):

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$$\displaystyle{y_p(x) = \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.12)
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Now that we have the particular solution, we can use the initial conditions to solve for the unknown constants in the general solution, $$ y_g(x)\!$$, and then we will be able to solve for the final solution $$ y(x)\!$$:

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$$\displaystyle{y(x) = C_1e^{5x} + C_2xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.13)
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Take derivative of (1.13):

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$$\displaystyle {y(x)' = 5C_1e^{5x} + C_2e^{5x} +5C_2xe^{5x} + 7xe^{5x} + \frac{35}{2}x^2e^{5x} - [0.16x + 0.064]}$$ (1.14)
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Using the given initial conditions and equations (1.13) and (1.14):

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$$\displaystyle{y(0) = 4 \rightarrow 4 = C_1 - 0.0192 \rightarrow C_1 = 4.0192}$$ (1.15)
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$$\displaystyle{y'(0) = 5 \rightarrow 5 = 5C_1 + C_2 - 0.064 \rightarrow C_2 = -15.032}$$ (1.16)
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Plug these constants back into the solution (1.13) to obtain the final solution:

$$  \displaystyle y(x) = 4.0192e^{5x} - 15.032xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192] $$     (1.17)
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The solution to the same problem but with excitation $$ r(x) = 7e^{5x} \! $$ is:

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$$\displaystyle{y(x) = 4e^{5x}-25xe^{5x}+\frac{7}{2}x^{2}e^{5x}}$$ (1.18)
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This is the plot for (1.17) and (1.18):

Author
Solved and Typed By - --Egm4313.s12.team1.wyattling 22:00, 20 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.armanious 22:39, 21 February 2012 (UTC)

Statement
Perturbation method for double real root: Developing the 2nd homogeneous solution for the case of double real root as a limiting case of distinct roots (see [[media:iea.s12.sec7.djvu| Sec7 p. 7-5 ]]). Consider two distinct real roots of the form: $$\lambda_1= \lambda, \lambda_2= \lambda + \epsilon \!$$ 1) Find the homogeneous L2-ODE-CC having the above distinct roots. 2) Show that the following is a homogeneous solution: $$\frac{e^{(\lambda + \epsilon)x} - e^{\lambda x}}{\epsilon} \!$$ The fraction in (3) p.7-5, for small $$\epsilon\!$$, is a finite difference formula that approximates the derivative $$\frac{e^{(\lambda + \epsilon)x} - e^{\lambda x}}{\epsilon}\approx \frac{d}{d \lambda} e^{\lambda x} \!$$

In fact, $$ \frac{d}{d \lambda} e^{\lambda x} = \lim_{\epsilon \to 0}\frac{e^{(\lambda + \epsilon)x} - e^{\lambda x}}{\epsilon} \!$$

3)Find the limit of the homogeneous solution in (3) p.7-5 as [epsilon goes to 0] (think l'Hopital's rule)

4)Take the derivative of $$ e^{\lambda x}\!$$ with respect to $$\lambda\!$$ 5)Compare the results in Parts (3) and (4), and relate to the result by variation of parameters. 6)Numerical experiment: Compute (3) p.7-5 using at $$\lambda=5\!$$ and with $$\epsilon = 0.001\!$$, and compare to get the value obtained from the exact 2nd homogeneous solution.

Solution
In order to find the corresponding L2-ODE-CC, the characteristic equation corresponding to the given solution must be found:
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$$  \displaystyle (\lambda'-\lambda)(\lambda'-\lambda-\epsilon)=0 $$     (2.0)
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$$  \displaystyle (\lambda')^2-2\lambda' \lambda + \lambda^2 - \lambda' \epsilon + \lambda \epsilon=(\lambda')^2-(2\lambda+\epsilon)\lambda'+(\lambda^2+\lambda\epsilon)=0 $$     (2.1)
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Therefore the corresponding L2-ODE-CC is:
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$$  \displaystyle y''-(2\lambda+\epsilon)y'+(\lambda^2+\lambda\epsilon)y=0 $$     (2.2) Note: if ε is equal to zero, the characteristic equation of 2.2 has a double real root at λ.
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To show that the following is a homogeneous solution, the first and second derivatives must be taken:
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$$  \displaystyle y(x)=\frac{e^{(\lambda + \epsilon)x}-e^{\lambda x}}{\epsilon} $$     (2.3)
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$$  \displaystyle y'(x)=\frac{(\lambda+\epsilon)e^{(\lambda + \epsilon)x}-\lambda e^{\lambda x}}{\epsilon} $$     (2.4)
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$$  \displaystyle y''(x)=\frac{(\lambda+\epsilon)^2e^{(\lambda + \epsilon)x}-\lambda^2 e^{\lambda x}}{\epsilon} $$     (2.5) Using these values in 2.2 yields:
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$$  \displaystyle \frac{(\lambda+\epsilon)^2e^{(\lambda + \epsilon)x}-\lambda^2 e^{\lambda x}}{\epsilon}-\frac{(2\lambda+\epsilon)[(\lambda+\epsilon)e^{(\lambda + \epsilon)x}-\lambda e^{\lambda x}]}{\epsilon}+\frac{(\lambda^2+\lambda\epsilon)[e^{(\lambda + \epsilon)x}-e^{\lambda x}]}{\epsilon}=0 $$     (2.6) Rearranging and simplifying yields:
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$$  \displaystyle \frac{[\lambda^2 +2\lambda\epsilon+\epsilon^2-2\lambda^2-3\lambda\epsilon-\epsilon^2+\lambda^2+\lambda\epsilon]e^{(\lambda + \epsilon)x}+[-\lambda^2+2\lambda^2+\lambda\epsilon-\lambda^2-\lambda\epsilon]e^{\lambda x}}{\epsilon}=0 $$     (2.7) All of the terms in the above equation cancel to yield:
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$$  \displaystyle \frac{0e^{(\lambda + \epsilon)x}+0e^{\lambda x}}{\epsilon}=0 $$     (2.8) Using l'Hopital's rule:
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$$  \displaystyle \lim_{\epsilon \to 0}\frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon}=\lim_{\epsilon \to 0}\frac{\frac{d}{d\epsilon}(e^{(\lambda+\epsilon)x}-e^{\lambda x})}{\frac{d}{d\epsilon} (\epsilon)}=\lim_{\epsilon \to 0}\frac{\frac{d}{d\epsilon}(e^{(\lambda+\epsilon)x})-0}{1} $$     (2.9)
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Simplifying further:
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$$  \displaystyle \lim_{\epsilon \to 0}\frac{d}{d\epsilon}(e^{(\lambda+\epsilon)x})=\lim_{\epsilon \to 0}xe^{(\lambda+\epsilon)x}=x\lim_{\epsilon \to 0}e^{(\lambda+\epsilon)x} $$     (2.10) This ultimately yields:
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$$  \displaystyle \lim_{\epsilon \to 0}\frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon}=xe^{\lambda x} $$ (2.11) The following should also be taken into consideration:
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$$  \displaystyle \frac{d}{d\lambda}(e^{\lambda x})=xe^{\lambda x} $$ (2.12) Clearly, the results of 2.11 and 2.12 are equivalent. This shows that $$y(x)=xe^{\lambda x} \!$$ is an appropriate solution to a homogeneous L2-ODE-CC having one double root.
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To test this, test values will be used to solve for the approximate (2.12) and exact (2.13) solutions of the ODE. For this test, $$\lambda=5\!$$ and $$\epsilon = 0.001\!$$
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$$  \displaystyle y_{approx}(x)=\frac{e^{(5 + 0.001)x}-e^{5 x}}{0.001}=1000(e^{0.001x}-1)e^{5x} $$     (2.13)
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$$  \displaystyle y_{exact}(x)=xe^{5x} $$     (2.14) If the above derivations are true, then the following approximation must also be true:
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$$  \displaystyle 1000(e^{0.001x}-1)\approx x $$ (2.15) Plotting both sides of 2.15 as a function of x shows that this is a valid approximation for most values of x. Figure 3.2-1
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Author
Solved and Typed By - Egm4313.s12.team1.armanious 05:51, 21 February 2012 (UTC) Reviewed By - --Egm4313.s12.team1.durrance 02:53, 22 February 2012 (UTC)

Statement
Find the complete solution for $$ y'' - 3y' + 2y = 4x^2\!$$, with the initial conditions $$ y(0) = 1, y'(0) = 0\!$$ Plot the solution $$y(x).\!$$

Solution
First we create the characteristic equation in standard form:
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$$\displaystyle{\lambda^2-3 \lambda +2=0}$$ (3.0) Then, by setting it equal to zero, we can find what $$ \lambda \!$$ equals:
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$$\displaystyle{(\lambda - 2)(\lambda -1) = 0}$$ (3.1)
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$$\displaystyle{\lambda = 2,\lambda =1}$$ (3.2)
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Given two, distinct, real roots, the general solution looks like this:
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$$  \displaystyle y_g(x)=C_1e^{2x} + C_2e^{x} $$     (3.3) By using the method of undetermined coefficients, the excitation $$ 4x^2\!$$ is analyzed to yield a particular solution: In assessing a polynomial with a second power, the form of the particular solution will look like this:
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$$  \displaystyle y_p (x)= A_2 x^2 + A_1 x + A_0 $$     (3.5)
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It's derivative would look like this:
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$$  \displaystyle y_p '(x)= 2 A_2 x + A_1 $$     (3.6)
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And the second derivative to follow would then become:
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$$  \displaystyle y_p ''(x)= 2 A_2 $$     (3.7)
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Based on the coefficients, the following system of equations exists:
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$$  \displaystyle 2A_2 =4 $$     (3.8)
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$$  \displaystyle -6A_2 + A_1=0 $$     (3.9)
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$$  \displaystyle 2A_2 -3 A_1 + A_0 =0 $$     (3.10)
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The results of this set of equations make the coefficients of A's:
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$$  \displaystyle A_2 =2 $$
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$$  \displaystyle A_1 =12 $$
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$$  \displaystyle A_0 =32 $$ The resulting particular equation looks like this:
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$$  \displaystyle y_p (x)= 2 x^2 + 12 x + 32 $$     (3.11) By adding the particular and general solutions, we get the complete solution:
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$$  \displaystyle 2 x^2 + 12 x + 32 + C_1e^{2x} + C_2e^{x} = y $$ (3.12)
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We consider the initial conditions by taking the first derivative of the complete solution:

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$$  \displaystyle 4 x + 12 + 2 C_1e^{2x} + C_2e^{x} = y' $$ (3.13)
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By plugging in 0 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
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$$  \displaystyle y(0)=1 =2 * 0^2 + 12 *0 + 32 + C_1e^{2*0} + C_2e^{0}=32 +C_1 +C_2=1 $$     (3.14)
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$$  \displaystyle y'(0)=0=4 *0 + 12 + 2 C_1e^{2*0} + C_2e^{0} = 12+ 2 C_1 + C_2=0 $$     (3.15)
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Solving the equations proves that $$C_1 = 19, C_2 = -50 \!$$: The resulting complete solution with consideration for initial conditions then becomes:
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$$  \displaystyle 2 x^2 + 12 x + 32 + 19 e^{2x} - 50e^{x} = y $$ (3.12)
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y plotted looks like this:

Author
Solved and Typed By -Egm4313.s12.team1.silvestri 15:56, 19 February 2012 (UTC) Reviewed By -Egm4313.s12.team1.armanious 03:02, 22 February 2012 (UTC)

Statement
From [[media:iea.s12.sec7b.djvu|R3.4 in Sec 3 p. 7-11]]; Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for
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$$  \displaystyle y''-3y'+2y=4x^{2}-6x^{5} $$     (4.1) is of the form
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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x_{j} $$     (4.2) i.e.,
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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{5}c_{j}x_{j} $$     (4.3)
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Basic Rule: Select $$y_{p}(x)\!$$ from the table and determine the coefficients by substituting $$y_{p}(x)\!$$ in

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$$  \displaystyle y''+ay'+by=r(x) $$     (4.4) Sum Rule: If $$r(x)\!$$ is the sum of the terms in the 1st column of table 2.1 then $$y_{p}(x)\!$$ is the sum of the corresponding terms in the 2nd column of this table. Table 2.1
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Solution
According to the table the Homogeneous equation has two $$r(x)\!$$ values of the form $$kx^{n}(n=0,1,2...)$$. Using the Basic Rule this means that there is a particular solution of the form $$K_{n}x^{n}+K_{n-1}x^{n-1}+...+K_{1}x+K_{0}\!$$ for each $$r(x)\!$$.

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$$  \displaystyle r_{1}(x)=4x^{2}\rightarrow r_{1}(x)=Kx^{n}(n=2) $$     (4.5)
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So the particular solution for this $$r_{1}(x)\!$$ should be:

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$$  \displaystyle y_{p1}= K_{2}x^{2}+K_{1}x^{1}+K_{0} $$     (4.6)
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Where $$K=4\!$$. Which simplifies to:

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$$  \displaystyle y_{p1}= \sum_{j=0}^{2}K_{j}x^{j} $$     (4.7)
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For the second $$r_{2}(x)\!$$:

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$$  \displaystyle r_{2}(x)=-6x^{5}\rightarrow r_{2}(x)=Cx^{n}(n=5) $$     (4.8)
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So the particular solution for this $$r_{2}(x)\!$$ should be:

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$$  \displaystyle y_{p2}= C_{5}x^{5}+C_{4}x^{4}+C_{3}x^{3}+C_{2}x^{2}+C_{1}x+C_{0} $$     (4.9)
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Where $$C=4\!$$. Which simplifies to:

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$$  \displaystyle y_{p1}= \sum_{j=0}^{5}C_{j}x^{j} $$     (4.10)
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Using the sum rule:

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$$  \displaystyle y_{p}=\sum_{j=0}^{2}K_{j}x^{j}+ \sum_{j=0}^{5}Cx^{j} $$     (4.11) Now using the Sum Rule which just states if there are two $$r(x)\!$$ values in any form on the left side of the table, then the particular solution is the sum of the solutions for $$r(x)\!$$ on the right side of the table.
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So the particular solution for $$y''-3y'+2y=4x^{2}-6x^{5}\!$$ where $$c=K+C$$

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$$  \displaystyle y_{p}=\sum_{j=0}^{5}c_{j}x^{j} $$     (4.12)
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Author
Solved and Typed By - User:Egm4313.s12.team1.stewart 22:47, 21 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri 04:57, 22 February 2012 (UTC)

Statement
Given:
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y''-3y'+2y=4x^2-6x^5 $$     (5.0)
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y_p(x)=\sum_{j=0}^5c_jx^j=4x^2-6x^5 $$     (5.1)
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Find: -Coefficients of particular solution using series expansion and matrix back-substitution -Solution $$\displaystyle y(x)$$ using initial conditions $$\displaystyle y(0)=1, y'(0)=0$$ Plot solution.

Solution
Using [http://upload.wikimedia.org/wikiversity/en/a/a6/Iea.s12.sec7b.djvu Sec7b-1 p.7-13 Eq. (1)]:

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\sum_{j=0}^3c_{j+2}(j+2)(j+1)x^j-3\sum_{j=0}^4c_{j+1}(j+1)x^j+2\sum_{j=0}^5c_jx^j=4x^2-6x^5 $$     (5.2) Which is a combined series expansion for the general coefficient series expansion of [http://upload.wikimedia.org/wikiversity/en/a/a6/Iea.s12.sec7b.djvu Sec7b-1 p.7-12 Eq. (4)]:
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$$\displaystyle \sum_{j=2}^5c_jj(j-1)x^{j-2}-3\sum_{j=1}^5c_jjx^{j-1}+2\sum_{j=0}^5c_jx^j=4x^2-6x^5 $$     (5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can find a simultaneous linear system of equations to solve for the coefficients of $$\displaystyle y_p(x)$$.

Note: by combining the summations in Eq. (5.2) by setting the upper bounds of the summations to a common value 3, the equation can be simplified to:
 * {| style="width:100%" border="0"

$$\displaystyle \sum_{j=0}^3[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_j]x^j+(2c_4-15c_5)x^4+2c_5x^5=4x^2-6x^5 $$     (5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore providing a system of equations by equating coefficients from the summations to the $$\displaystyle y_p(x)$$:
 * {| style="width:100%" border="0"

j=0: [c_2(2)(1)-3c_1(1)+2c_0]x^0=0x^0=[2c_0-3c_1+2c_2]x^0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=1: [c_3(3)(2)-3c_2(2)+2c_1]x^1=0x^1=[2c_1-6c_2+6c_3]x^1 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=2: [c_4(4)(3)-3c_3(3)+2c_2]x^2=4x^2=[2c_2-9c_3+12c_4]x^2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=3: [c_5(5)(4)-3c_4(4)+2c_3]x^3=0x^3=[2c_3-12c_4+20c_5]x^3 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

[2c_4-15c_5]x^4=0x^4 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

2c_5x^5=-6x^5 $$     (5.5) Eqs. (5.5) can be verified by solving for the coefficients by Eq. (5.2) to prove that the summations were combined correctly:
 * $$\displaystyle
 * $$\displaystyle
 *

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

j=0: 2c_0x^0=2c_0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=1: 2c_1x^1-3c_1(1)x^0=2c_1x-3c_1 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=2: 2c_2x^2-3c_2(2)x^1+c_2(2)(1)x^0=2c_2x^2-6c_2x+3c_2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=3: 2c_3x^3-3c_3(3)x^2+c_3(3)(2)x^1=2c_3x^3-9c_3x^2+6c_3x $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=4: 2c_4x^4-3c_4(4)x^3+c_4(4)(3)x^2=2c_4x^4-12c_4x^3+12c_4x^2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=5: 2c_5x^5-3c_5(5)x^4+c_5(5)(4)x^3=2c_5x^5-15c_5x^4+20c_5x^3 $$     (5.6) By summing the terms in (5.6) and grouping like terms, then equating it to the $$\displaystyle y_p(x)$$, we find the coefficient equations are the same as Eqs. (5.5):
 * $$\displaystyle
 * $$\displaystyle
 *

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

[2c_0-3c_1+2c_2]x^0+[2c_1-6c_2+6c_3]x^1+[2c_2-9c_3+12c_4]x^2+[2c_3-12c_4+20c_5]x^3+[2c_4-15c_5]x^4+2c_5x^5=0x^0+0x^1+4x^2+0x^3+0x^4-6x^5 $$     (5.7)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

The linear system of equations in Eqs. (5.5) can be placed in matrix form:
 * {| style="width:100%" border="0"

$$ \left[ \begin{array}{cccccc} 2 & -3 & 2 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0\\ 0 & 0 & 0 & 2 & -12 & 20\\ 0 & 0 & 0 & 0 & 2 & -15\\ 0 & 0 & 0 & 0 & 0 & 2\end{array} \right]$$ $$\left( \begin{array}{cccccc} {c}_{0}\\ {c}_{1}\\ {c}_{2}\\ {c}_{3}\\ {c}_{4}\\ {c}_{5}\end{array} \right)\ = \left[ \begin{array}{cccccc} 0\\ 0\\ 4\\ 0\\ 0\\ -6\end{array} \right]\ $$ (5.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving this system of equations by back-substitution yields the values of the coefficients for the $$\displaystyle y_p(x)$$:
 * {| style="width:100%" border="0"

c=\{-701.75, -691.5, -335.5, -105, -22.5, -3\} $$     (5.9)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Substituting these coefficients into Eq. (5.1):
 * {| style="width:100%" border="0"

y_p(x)=-701.75x^0-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5 $$     (5.10)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

The general solution is the sum of the homogeneous (found in R3.3) and particular solutions:
 * {| style="width:100%" border="0"

y(x)=-701.75-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5+C_1e^{2x}+C_2e^x $$     (5.11)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Considering the initial value condition $$\displaystyle y(0)=1$$ in the general solution:
 * {| style="width:100%" border="0"

y(0)=1=-701.75+C_1e^0+C_2e^0=-701.75+C_1+C_2 $$     (5.12) Taking the first derivative of Eq. (5.11) to consider the initial value condition $$\displaystyle y'(0)=0$$:
 * $$\displaystyle
 * $$\displaystyle
 *

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

y'(x)=-691.5-671x-315x^2-90x^3-15x^4+2C_1e^{2x}+C_2e^x $$     (5.13)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

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

y'(0)=0=-691.5+2C_1+C_2 $$     (5.14)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:
 * {| style="width:100%" border="0"

C_1=-11.25, C_2=714 $$     (5.15)
 * $$\displaystyle
 * $$\displaystyle
 *

 * }

This yields our final solution by substituting the coefficients into Eq. (5.11): Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:
 * {| style="width:100%" border="1"

$$\displaystyle y(x)=-701.75-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5-11.25e^{2x}+714e^x $$     (5.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

A plot of the solution: Figure 3.5-1

Author
Solved and Typed By - --Egm4313.s12.team1.durrance 22:56, 21 February 2012 (UTC)-- Reviewed By - Egm4313.s12.team1.silvestri 03:28, 22 February 2012 (UTC)

Statement
Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODEs-CC (see p.7-2b):
 * {| style="width:100%" border="0"

$$\displaystyle{y}_{p,1}''-3{y}_{p,1}'+2{{y}_{p,1}}={{r}_{1}}(x):=4{{x}^{2}}$$ (6.0)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle{y}_{p,2}''-3{y}_{p,2}'+2{{y}_{p,2}}={{r}_{2}}(x):=-6{{x}^{5}}$$ (6.1) The particular solution $$\displaystyle{y}_{p,1}$$ had been found in R3.3 p.7-11. Find the particular solution $$\displaystyle{y}_{p,2}$$, and then obtain the solution $$\displaystyle{y}$$ for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Compare the result with that obtained in R3.5.

Solution
Beginning with equation (6.1), we find that a particular solution has the form
 * {| style="width:100%" border="0"

$$\displaystyle{{y}_{p,2}}(x)=\sum\limits_{j=0}^{n}$$ (6.2) where $$\displaystyle{n=5}$$. That is,
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {{y}_{p,2}}(x)={{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}}$$
 * style="width:95%" |
 * style="width:95%" |

(6.3) Differentiating twice:
 *

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

$$\displaystyle{y}_{p,2}'(x)={{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}}$$ (6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle{y}_{p,2}''(x)=2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}}$$ (6.5) Substitute (6.3-5) into (6.1) to obtain
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle (2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}})-3({{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}})$$ $$\displaystyle+2({{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}})=-6{{x}^{5}}$$
 * style="width:95%" |
 * style="width:95%" |

(6.6) Rearranging terms with respect to $$\displaystyle{x}$$ power:
 *

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

$$\displaystyle (2{{K}_{2}}-3{{K}_{1}}+2{{K}_{0}})+x(6{{K}_{3}}-6{{K}_{2}}+2{{K}_{1}})+{{x}^{2}}(12{{K}_{4}}-9{{K}_{3}}+2{{K}_{2}})$$ $$\displaystyle+{{x}^{3}}(20{{K}_{5}}-12{{K}_{4}}+2{{K}_{3}})+{{x}^{4}}(-15{{K}_{5}}+2{{K}_{4}})+{{x}^{5}}(2{{K}_{5}})=-6{{x}^{5}}$$ (6.7) In matrix form:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \left[ \begin{array}{cccccc} 2 & -3 & 2 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0\\ 0 & 0 & 0 & 2 & -12 & 20\\ 0 & 0 & 0 & 0 & 2 & -15\\ 0 & 0 & 0 & 0 & 0 & 2\end{array} \right]$$ $$\left( \begin{array}{cccccc} {K}_{0}\\ {K}_{1}\\ {K}_{2}\\ {K}_{3}\\ {K}_{4}\\ {K}_{5}\end{array} \right)\ = \left[ \begin{array}{cccccc} 0\\ 0\\ 0\\ 0\\ 0\\ -6\end{array} \right]\ $$ (6.8) Solving for $$\displaystyle K$$, using MATLAB, yields
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle K=\{-708.75,-697.5,-337.5,-105,-22.5,-3\}$$ (6.9) which means that
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {y}_{p,2}=-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}}$$ (6.10) From R3.3:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {{y}_{p,1}}=32+12x+2{{x}^{2}}$$ (6.11) Summing for the final solution:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y={{C}_{1}}{{y}_{p,1}}+{{C}_{2}}{{y}_{p,2}}$$ (6.12)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y={{C}_{1}}(32+12x+2{{x}^{2}})$$ $$\displaystyle +{{C}_{2}}(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})$$ (6.13) For initial condition $$\displaystyle y(0)=-5$$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y(0)=-5={{C}_{1}}(32)+{{C}_{2}}(-708.75)$$ (6.14) Similarly, for initial condition $$\displaystyle y'(0)=2$$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y'(0)=2={{C}_{1}}(12)+{{C}_{2}}(-697.5)$$ (6.15) Equations (6.14) and (6.15) yield the following matrix equation:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \left[ \begin{array}{cc} 32 & -708.75\\ 12 & -697.5\end{array} \right]$$ $$\left( \begin{array}{cc} {C}_{1}\\ {C}_{2}\end{array} \right)\ = \left[ \begin{array}{cc} -5\\ 2\end{array} \right]\ $$ (6.16) Solving (6.16) in MATLAB yields
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle C=\{-0.355,-0.009\}$$ (6.17) Therefore the combined solution of $$\displaystyle y$$ is
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y(x)=-0.355(32+12x+2{{x}^{2}})$$ $$\displaystyle -0.009(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})$$ (6.18) Which can be simplified to Final Equation
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y(x)=-4.98125+2.0175x+2.3275{{x}^{2}}+0.945{{x}^{3}}+0.2025{{x}^{4}}+0.027{{x}^{5}}$$ (6.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and Typed By - Egm4313.s12.team1.essenwein 00:40, 18 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri 04:29, 22 February 2012 (UTC)

Statement
Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities. These equalities are:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{j=2}^{5} C_{j} j(j-1) x^{^{j-2}}=\sum_{j=0}^{3} C_{j+2} (j+2)(j+1) x^{^{j}} \!$$ (7.0)
 *

 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{j=1}^{5} C_{j} jx^{^{j-1}}=\sum_{j=0}^{4} C_{j+1} (j+1) x^{^{j}} \!$$ (7.1)
 *

 * }

Solution
This transition occurs through a mid-level variable change. In this particular case, for (7.0), j-2 is represented by k:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{k=0}^{3} C_{k+2} (k+2)(k+1) x^{^{k}} \!$$ (7.3)
 *

 * }

We can then represent the variable k with a j. This new summation looks like this:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{j=0}^{3} C_{j+2} (j+2)(j+1) x^{^{j}} \!$$ (7.4)
 *

 * }

Expanding both sides, (7.0 and 7.4) yields the same result:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$C_{5}(5)(4)x^{^{3}} + C_{4}(4)(3)x^{^{2}}+C_{3}(3)(2)x^{^{1}} +C_{2}(2)(1)x^{^{0}}\!$$ (7.5) We follow the same process for equality (7.1): This transition occurs through a mid-level variable change. In this particular case, for (7.1), j-1 is represented by k:
 *

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

 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{k=0}^{4} C_{k+1} (k+1) x^{^{k}} \!$$ (7.6)
 *

 * }

We can then represent the variable k with a j. This new summation looks like this:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{j=0}^{4} C_{j+1} (j+1) x^{^{j}} \!$$ (7.7)
 *

 * }

Expanding both sides, (7.1 and 7.7)yields the same result:
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$C_{5}(5)x^{^{4}} + C_{4}(4)x^{^{3}}+C_{3}(3)x^{^{2}} +C_{2}(2)x^{^{1}}+C_{1}\!$$ (7.8)
 *

 * }

Author
Solved and Typed By - Egm4313.s12.team1.silvestri 17:20, 19 February 2012 (UTC) Reviewed By - --128.227.113.77 17:05, 22 February 2012 (UTC)

Statement
Find a general solution for the following two problems:

Given
Homogeneous solution: $$ y'' + 4y' + 4y = e^{-x}cosx\!$$

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$. First, we will find the general solution, $$ y_g(x) \!$$, by finding the roots of the characteristic equation:

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

$$\displaystyle{\lambda^2 + 4\lambda + 4\lambda = 0 \rightarrow (\lambda + 2)^2}$$ (8.0)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which means that the characteristic equation has a double root of $$ \lambda = -2 \!$$. Based on the double root, then, the general solution is:

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

$$\displaystyle{y_g(x) = (C_1 + C_2x)e^{-\frac{ax}{2}} \rightarrow y_g(x) = (C_1 + C_2x)e^{-2x}}$$ (8.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Next, we need to find the particular solution, and based on analysis of the excitation, $$ r(x) = e^{-x}cosx \!$$, we find that the particular solution is the following:

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

$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, we need to find the values for the constants $$ K, M\!$$, by taking the first and second derivatives of the particular solution (8.2):

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

$$\displaystyle{y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{y_p(x)' = -e^{-x}Ksinx - e^{-x}Kcosx + e^{-x}Mcosx - e^{-x}Msinx}$$ (8.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{y_p(x)'' = -e^{-x}Kcosx + e^{-x}Ksinx + e^{-x}Ksinx + e^{-x}Kcosx - e^{-x}Msinx - e^{-x}Mcosx - e^{-x}Mcosx + e^{-x}Msinx}$$ (8.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, plug these into the homogeneous solution and simplify:

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

$$\displaystyle{-e^{-x}(Kcosx - Ksinx - Ksinx - Kcosx + Msinx + Mcosx + Mcosx - Msinx) + 4[-e^{-x}(Ksinx + Kcosx - Mcosx + Msinx)] + 4[e^{-x}(Kcosx + Msinx)] = e^{-x}cosx}$$ (8.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{-e^{-x}[-2Ksinx + 2Mcosx] - e^{-x}[4Ksinx + 4Kcosx - 4Mcosx + 4Msinx] + e^{-x}[4Kcosx + 4Msinx] = e^{-x}cosx}$$ (8.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Pull $$ e^{-x} \! $$ out of equation (8.6) and simplify:

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

$$\displaystyle{e^{-x}[4Kcosx + 4Msinx - 4Ksinx - 4Kcosx + 4Mcosx - 4Msinx + 2Ksinx - 2Mcosx] = e^{-x}cosx}$$ (8.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{e^{-x}[-2Ksinx + 2Mcosx] = e^{-x}cosx}$$ (8.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, solve for the unknown coefficients:

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

$$\displaystyle{2M = 1 \rightarrow M = \frac{1}{2}}$$ (8.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{-2K = 0 \rightarrow K = 0}$$ (8.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Plug into $$ y_p(x)\!$$ (8.2):

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

$$\displaystyle{y_p(x) = \frac{1}{2}e^{-x}sinx}$$ (8.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, we have both the particular solution and the general solution, which allows for us to solve for the final solution, $$ y(x) \! $$

$$  \displaystyle y(x) = (C_1 + C_2x)e^{-2x} + \frac{1}{2}e^{-x}sinx $$     (8.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Given
Homogeneous solution: $$ y'' + y' (\pi^2 + \frac{1}{4}) = e^{-\frac{x}{2}}sin{\pi x}\!$$

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$. First, we will find the general solution, $$ y_g(x) \!$$, by finding the roots of the characteristic equation, using the quadratic equation:

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

$$\displaystyle{\lambda^2 + \lambda + (\pi^2 + \frac{1}{4}) = 0 \rightarrow \lambda = \frac{-1 \pm \sqrt{1-4(1)(\pi^2 + \frac{1}{4})}}{2}}$$ (8.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$\displaystyle{\lambda = -\frac{1}{2} + i\pi}$$ (8.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The roots of the characteristic equation are complex conjugates, meaning that the general solution, $$ y_g(x) \!$$, is the following:

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

$$\displaystyle{y_g(x) = e^{-\frac{x}{2}}(Acos{\pi x} + Bsin{\pi x})}$$ (8.15)
 * style="width:95%" |
 * style="width:95%" |
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 * }

Next, we need to find the particular solution, and based on analysis of the excitation, $$ r(x) = e^{-\frac{x}{2}}sin{\pi x} \!$$, we find that the particular solution is the following:

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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-\frac{x}{2}}Kcos{\pi x} + e^{-\frac{x}{2}}Msin{\pi x}}$$ (8.16)
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Now, we need to find the values for the constants $$ K, M \!$$, by taking the first and second derivatives of the particular solution (8.16):

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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-\frac{x}{2}}Kcos{\pi x} + e^{-\frac{x}{2}}Msin{\pi x}}$$ (8.16)
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$$\displaystyle{y_p(x)' = {-\pi e^{\frac{-x}{2}}Ksin{\pi x} - \frac{{}e^{\frac{-x}{2}}}{2}Kcos{\pi x}} + {\pi e^{\frac{-x}{2}}Mcos{\pi x} + \frac{{}e^{\frac{-x}{2}}}{2}Msin{\pi x}}}$$ (8.17)
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$$\displaystyle{y_p(x)'' = {-\pi^2 e^{\frac{-x}{2}}Kcos{\pi x} +  \pi \frac{e^{\frac{-x}{2}}}{2}Ksin{\pi x}}  +  \pi \frac{e^{\frac{-x}{2}}}{2}Ksin{\pi x}  + \frac{e^{\frac{-x}{2}}}{4}Kcos{\pi x}  - \pi^2 {e^{\frac{-x}{2}}}Msin{\pi x} - \pi \frac{e^{\frac{-x}{2}}}{2}{}Mcos{\pi x} - \pi \frac{e^{\frac{-x}{2}}}{2}{}Mcos{\pi x} + \frac{e^{\frac{-x}{2}}}{4}{}Msin{\pi x}}$$ (8.18)
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Now, we plug these into the homogeneous solution and simplify but we find that everything cancels out, meaning the unknown constants, $$ M, K \!$$ are both equal to $$ 0 \! $$. This means that the final solution is equal to just the general solution:

$$  \displaystyle y(x) = e^{\frac{-x}{2}}(Acos{\pi x} + Bsin{\pi x}) $$     (8.19)
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Author
Solved and Typed By ---Egm4313.s12.team1.wyattling 22:10, 20 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.rosenberg 03:07, 22 February 2012 (UTC)

Statement
K 2011 page 85 problems 13 and 14 Problem 13: Find the complete solution for $$ 8y'' - 6y' + y = 6\cosh x\!$$, with the initial conditions $$ y(0) = 0.2, y'(0) = 0.05\!$$ Problem 14: Find the complete solution for $$ y'' + 4y' + 4y = e^{-2x}\sin 2x\!$$, with the initial conditions $$ y(0) = 1, y'(0) = -1.5\!$$

Solution
Problem 13: In order to solve the equation we must use the definition $$ y=y_{h} + y_{p}\!$$. We start with the given equation:
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$$\displaystyle{8y''-6y'+y=6\cosh x}$$ (9.1) From this we get the homogeneous characteristic equation:
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$$\displaystyle{8 \lambda ^2-6 \lambda +1=0 x}$$ (9.2) Factoring the characteristic equation gives us:
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$$\displaystyle{(4 \lambda -1)(2 \lambda -1)=0}$$ (9.3) This give us the roots:
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$$\displaystyle{\lambda = \frac {1}{4}, \frac {1}{2}}$$ (9.4) Plugging in the roots gives us the general homogeneous solution:
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$$\displaystyle{y_{h} = c_{1} e^{\frac {1}{4} x}+c_{2} e^{\frac {1}{2} x}}$$ (9.5) Now we must solve for the particular solution. So far we know:
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$$\displaystyle{\cosh x = \frac {e^x +e^{-x}}{2}}$$ (9.6) Now using the sum rule, $$\displaystyle{y_{p} = y_{p1} + y_{p2}}$$, gives us:
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$$\displaystyle{y_{p} = Ce^x + Ke^{-x}}$$ (9.7) Now we must take the first and second derivatives of equation 9.7:
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$$\displaystyle{y_{p}' = Ce^x - Ke^{-x}}$$ (9.8)
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$$\displaystyle{y_{p}'' = Ce^x + Ke^{-x}}$$ (9.9) Substituting 9.8 and 9.9 back into our original equation gives us:
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$$\displaystyle{8(Ce^x+Ke^{-x})-6(Ce^x-Ke^{-x})+Ce^x+Ke^{-x}=6(\frac {e^x+e^{-x}}{2})}$$ (9.10) Simplifying equation 9.10 give us:
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$$\displaystyle{3Ce^x+15Ke^{-x}=3(e^x+e^{-x})}$$ (9.11) From this we can deduce that:
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$$\displaystyle{3C=3, 15K=3}$$ (9.12) Thus we get that:
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$$\displaystyle{C=1, K= \frac{1}{5}}$$ (9.13) We now have our particular equation:
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$$\displaystyle{y_{p} = e^x + \frac{1}{5} e^{-x}}$$ (9.14) Thus:
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$$\displaystyle{y= c_{1} e^{\frac {1}{4} x}+c_{2} e^{\frac {1}{2} x}+e^x + \frac{1}{5} e^{-x}}$$ (9.15) Now we can solve for $$ c_{1}\!$$ and $$ c_{2}\!$$ using the given initial values. We are given that at $$ y(0)=0.2\!$$, we can now plug this into our equation to get:
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$$\displaystyle{0.2= c_{1} e^{\frac {1}{4} (0)}+c_{2} e^{\frac {1}{2} (0)}+e^{(0)} + \frac{1}{5} e^{-(0)}}$$ (9.16) This simplifies to:
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$$\displaystyle{0.2= c_{1}+c_{2}+1+ \frac{1}{5}}$$ (9.17) Giving us that:
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$$\displaystyle{c_{1}+c_{2} = -1}$$ (9.18) For our second condition, $$y'(0)=0.05\!$$, we must we must take the derivative of our general solution:
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$$\displaystyle{y'= \frac {1}{4} c_{1} e^{\frac {1}{4} x}+ \frac {1}{2} c_{2} e^{\frac {1}{2} x}+e^x - \frac{1}{5} e^{-x}}$$ (9.19) At our initial condition gives us:
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$$\displaystyle{0.05= \frac {1}{4} c_{1} + \frac {1}{2} c_{2} +1 - \frac{1}{5}}$$ (9.20) Thus:
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$$\displaystyle{c_{1}+2c_{2}=-3}$$ (9.21) Combining 9.18 and 9.21 gives us that:
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$$\displaystyle{c_{1}=1, c_{2}=-2}$$ (9.22) Plugging these coefficients in gives us our final solution:
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$$\displaystyle{y= e^{\frac {1}{4} x}-2e^{\frac {1}{2} x}+e^x + \frac{1}{5} e^{-x}}$$ (9.23)
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Problem 14: In order to solve the equation we must use the definition $$ y=y_{h} + y_{p}\!$$. We start with the given equation:
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$$\displaystyle{y''+4y'+4y=e^{-2x} \sin 2x}$$ (9.24) From this we get the homogeneous characteristic equation:
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$$\displaystyle{\lambda ^2 +4 \lambda +4 = 0}$$ (9.25) Factoring the characteristic equation gives us:
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$$\displaystyle{(\lambda +2)(\lambda +2)=0}$$ (9.26) This give us the double root:
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$$\displaystyle{\lambda = -2}$$ (9.27) Plugging in the roots gives us the general homogeneous solution:
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$$\displaystyle{y_{h} = c_{1} e^{-2x}+c_{2} xe^{-2x}}$$ (9.28) Now we must solve for the particular solution. So far we know:
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$$\displaystyle{e^{-2x}\sin 2x = Cxe^{-2x}\cos 2x+Kxe^{-2x}\sin 2x}$$ (9.29) Now we must take the first and second derivatives of equation 9.29:
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$$\displaystyle{y_{p}' = -2e^{-2x}(Cx\cos 2x+Kx\sin 2x)+e^{-2x}(C\cos 2x-2Cx\sin 2x+K\sin 2x+2Kx\cos 2x)}$$ (9.30)
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$$\displaystyle{y_{p}'' = (-4C+4K)e^{-2x}\cos 2x+(-4k-4C)e^{-2x}\sin 2x}$$ (9.31) Substituting 9.30 and 9.31 back into our original equation gives us:
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$$\displaystyle{(-3C+4K)e^{-2x}\cos 2x+(-3k-4C)e^{-2x}\sin 2x=e^{-2x}\sin 2x}$$ (9.32) From this we can deduce that:
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$$\displaystyle{-3C+4K=0,-3K-4C=1}$$ (9.33) Thus we get that:
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$$\displaystyle{C=\frac{-4}{25}, K= \frac{-3}{25}}$$ (9.34) We now have our particular equation:
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$$\displaystyle{y_{p} = e^{-2x}(\frac {-4}{25}x\cos 2x-\frac {3}{25}x\sin 2x)}$$ (9.35) Thus:
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$$\displaystyle{y=c_{1} e^{-2x}+c_{2} xe^{-2x}+e^{-2x}(\frac {-4}{25}x\cos 2x-\frac {3}{25}x\sin 2x)}$$ (9.36) Now we can solve for $$ c_{1}\!$$ and $$ c_{2}\!$$ using the given initial values. We are given that at $$ y(0)=1\!$$, we can now plug this into our equation to get:
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$$\displaystyle{y(0)=(c_{1} +c_{2} x)e^{-2(0)}+0=1}$$ (9.37) Giving us that:
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$$\displaystyle{c_{1} = 1}$$ (9.38) For our second condition, $$y'(0)=-1.5\!$$, we must we must take the derivative of our general solution:
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$$\displaystyle{y'= -2c_{1}e^{-2x}+c_{2} (-2xe^{-2x}+e^{-2x})+\frac {-4}{25} e^{-2x}\cos 2x - \frac {3}{25} e^{-2x}\sin 2x+\frac {2}{25}e^{-2x}x\cos 2x + \frac {14}{25}e^{-2x}x\sin 2x}$$ (9.39) At our initial condition we have:
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$$\displaystyle{y'(0)=-2+c_{2} -\frac{4}{25} -0 +0 +0 =0}$$ (9.40) Thus:
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$$\displaystyle{c_{2}=\frac {54}{25} }$$ (9.41) Plugging these coefficients in gives us our final solution:
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$$\displaystyle{y=1 e^{-2x}+\frac {54}{25} xe^{-2x}+e^{-2x}(\frac {-4}{25}x\cos 2x-\frac {3}{25}x\sin 2x)}$$ (9.42)
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Author
Solved and Typed By - Egm4313.s12.team1.rosenberg 03:04, 22 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.essenwein 02:39, 22 February 2012 (UTC)


text:
User talk:Howardkuo

Your articles
Hi Howard,

I just noticed the two articles you created and want to inform you that they are now nominated for deletion. The reason for this is that you're trying to write articles about subjects that might belong in Wikipedia, the free encyclopedia, but not on wikiversity. If you need help with Wikiversity, please try the help pages we offer or ask here. Thank you for understanding, nl:Mark W (Mwpnl) ¦ talk 19:20, 18 February 2012 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:Howardkuo


text:
User:Howardkuo

Mount Cotopaxi It is a stratovalcano

title: User:Howardkuo


text:
User:Numiri/KinderCalculus/nouns, verbs, adjectives

prerequisite
Basic definitions of fraction is a prerequisite here. This concept can be covered at a Kindergarten level. We are not requiring any fraction arithmetic yet. The terms "numerator" and "denominator" can be a mouthful, so we opt for "tops" and "bottoms" instead. This adds more context to the term "improper fractions" (now called "top heavy") and "proper fractions" (now called "bottom heavy").

nouns, verbs, adjectives
nouns

simple nouns are numbers and variables.

variables: are "blank numbers", or containers for amounts such as your age or test score. Variables are written as letters or shapes, such as

$$ \begin{align} x         &= \textrm{Heather's\ age}\\ \heartsuit &= \textrm{Chandler's\ age\ } = x + 3\\ \square   &= \textrm{Chandler's\ test score} \end{align} $$

compound nouns, aka layers, are combinations of verbs and nouns, and they are usually written with parentheses or shapes. Examples of nouns (simple and compound) are: 1, x, $$\heartsuit$$, x + 1, y * ( x + 1 ).

anti-nouns are negative, reciprocal, log, and root

phrases are mathematical expressions. We prefer this term for syllabic brevity.

forward-nouns are positive numbers, integers (not reciprocals), and leaves

verbs

verbs are operations like +, -, *, /, ^, 0--, --0, cos, etc.

anti-verbs are minus, divide, log and root

forward-verbs are plus, times, powers

seed & repeater verbs: in the context of self-plus and self-times, the primitive operation is called the seed verb and the resulting operation from the repetition is the repeater verb. For instance, in 3x = x + x + x, the seed verb is "+" and the repeater verb is times. Likewise, x^3 = x * x * x has the seed verb "*" and repeater verb as "^".

The table below shows that most arithmetic traces back to the + verb. The Blead section of this book explains these verbs in detail.

adjectives adjectives are the negative & reciprocal noun modifier. They can be introduced as a reflection of a noun about the verb's center. (the center of a verb + or * is the group-theoretic identity 0 or 1 respectively)


text:
VendingMachine

Idea
Vending machines are sometimes used to dispense parts used in DIY projects at some colleges. They are also used for ordinary reasons. So the idea is to purchase one and use it for a different purpose than it was designed for.

Goals

 * Find used/old vending machine.
 * Get it working.
 * Put DIY project kits or parts in it.

Tutorials

 * Mechanical coin detection system
 * Dollar bill changer system

Current Status

 * Just started

title: VendingMachine


text:
Category:Howard Community College fall2012

Here are the personal and team web pages that were worked on during the fall of 2012 at Howard Community College, Columbia Maryland USA.


text:
User:Egm6322.s12.TEAM1.KIM/HW3

Problem 3: Pose Question to be solved
 Solved without assistance

Given
Power series of tan x: $$\displaystyle x \in ]-\pi /2,\pi/2 [$$ $$\displaystyle tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + ... + \frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}x^{2n-1} + ...$$ Continued fraction of tan x: $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -\frac{x^2}{\displaystyle 5 -\frac{x^2}{\displaystyle 7 -\frac{x^2}{\ddots}}}}}$$

Find
Take $$x = \pi / 4$$ and compare the term -by-term convergence of the power series (2) p.48-25b and the continued fraction (1). Provide a plot to visualize the convergence.

Solution
the exact solution of $$\displaystyle tan\left(\frac{\pi}{4}\right )$$ is 1.0 x = $$\frac{\pi}{4}$$ For power series, when n = 1 $$\displaystyle tanx = x = \frac{\pi}{4}=0.7853$$ When n = 2, $$\displaystyle tanx = x + \frac{1}{3}x^3= 0.9467$$ When n = 3, $$\displaystyle tanx = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 = 0.9865$$ When n = 4, $$\displaystyle tanx = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 = 0.99649$$ When n =5, $$\displaystyle tanx = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \frac{62}{2835}x^9 = 0.9989$$ For continued fraction method, n = 1 $$\displaystyle tan x = \frac{x}{\displaystyle 1} = \frac{\pi}{4} = 0.7853 $$ When n = 2, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -{x^2}}= 2.0498$$ When n =3, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -{x^2}}} = 1.059685$$ When n =4, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -\frac{x^2}{\displaystyle 5 -{x^2}}}} = 1.00144$$ When n = 5, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -\frac{x^2}{\displaystyle 5 -\frac{x^2}{\displaystyle 7 -{x^2}}}}}= 1.000018$$ These results show that if n goes to lager, the convergence speed of continued fraction method will be faster than power series.

Problem 7: Legendre homogeneous solution
 Solved without assistance

Given
$$\displaystyle A_n = \frac{2n+1}{2} \int_{-1}^{1} f(\mu)P_n(\mu)d\mu $$ $$\displaystyle f(\mu) = T_0 \sinh(1-\mu^2) $$

Find
Compute the first 3 non-zero coefficients using Gauss-Legendre quadrature to within 5% error.

Solution
$$\displaystyle \begin{align} &P_0\left(\mu\right)=1 ,\\ &P_2\left(\mu\right)=\frac{1}{2}\left(3\mu^2-1\right) ,\\ &P_4\left(\mu\right)=\frac{1}{8}\left(35\mu^4-30\mu^2+3\right) ,\\ &P_6\left(\mu\right)=\frac{1}{16}\left(231\mu^6-315\mu^4+105\mu^2-5\right) .\\ \end{align} $$

Using the above values and plugging into eq 7.2,  $$\displaystyle A_n = \frac {2(n) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_n(\mu)d\mu$$ $$\displaystyle A_0 = \frac {2(0) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_0(\mu)d\mu = \frac {1}{2} * 1.492T_0 = 0.746T_0$$ $$\displaystyle A_2 = \frac {2(2) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_2(\mu)d\mu = -0.7997T_0$$ $$\displaystyle A_4 = \frac {2(4) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_4(\mu)d\mu = 0.066713T_0$$

Using Matlab, I compute the coefficient.

Problem 1: Computing non-zero coefficients in the Legendre Series
 Solved without assistance

Given: Expectation of 1% accuracy
Reference: Problem 7, Report 10 $$\displaystyle f(\mu) = T_0 \sinh(1-\mu^2) $$ $$\displaystyle A_n = \frac {2(n) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_n(\mu)d\mu$$

Find
A) Compute the first non-zero coefficient in the Legendre Series with increasing number of integration points until convergence to within 1% accuracy is achieved. B) Provide the expression for the exact solution if possible (use WA) C) Plot the integration error versus the number of integration points.

part1
According to the homework 10.7, the coefficient $$\displaystyle A_n$$is non-zero term when n = even number. Thus A0, A2, A4, A6.... will be non-zero term. The first non-zero term coefficient is $$\displaystyle A_0 = 0.7459T_0$$ $$I_n(f) = \sum_{j=1}^n w_jf(x_j)\approx A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0 \sinh{(1 - x^2)}dx $$ When j =0, the above equation is following that $$\displaystyle A_0 = 0.7459T_0 = I_n(f) = \sum_{j=1}^n w_jf(x_j)$$ Where $$\displaystyle x_j$$ is the roots of polynomials When integration point is 1, $$\displaystyle A_0 = 1/2*\sum_{j=1}^1 w_1f(x_1)=1/2*(1.175*2)=1.175T_0$$ When integration point is 2, $$\displaystyle A_0 = 1/2*(w_1*f(x_1)+w_2*f(x_2))=0.7172T_0$$ When integration point is 3, $$\displaystyle A_0 = 1/2*(w_1*f(x_1)+w_2*f(x_2)+w_3*f(x_3))=0.7505T_0$$ When integration point is 4, $$\displaystyle A_0 = 1/2*(w_1*f(x_1)+w_2*f(x_2)+w_3*f(x_3)+w_4*f(x_4))=0.7459T_0$$ These results show that n=4, the accuracy is almost 0%. Now see the lecture note 45-7. Equation in lecture note 45-7 is that $$E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{f^{2n}(\eta)}{(2n)!}, \eta \in [-1,1]$$ $$I(f) = I_n(f) + E_n(f)$$ $$I(f) - I_n(f) = E_n(f)$$ When n goes to oo, E_n(f) will be zero. Assume the exact A_0 is 0.7459.For the accuracy, 0.007459 will be allowed in $$\displaystyle I(f)-I_n(f)$$ $$\displaystyle E_n(f)= 0.007459$$ using the matlab, the results showed that

Thus the answer is n = 4.

part2
By using Wlofram Alpha, exact expression should be

The exact solution is 0.74601369.....(this value was estimated by n=5)

part3
The above table is from wikipedia. $$E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{f^{2n}(\eta)}{(2n)!}, \eta \in [-1,1]$$ When $$\displaystyle \eta $$ is 0.9, 9, -0.9,

Problem 8: Verification
 Solved without assistance

Find
Verify equation 8.1 and show that for $$ i=1,...,n \ $$ and thus

Part1
$$\ell_j(x_i) = \prod_{m=0,\, m\neq j}^{k} \frac{x_i-x_m}{x_j-x_m} $$ From above equation, when $$m = j$$, the product skips And if $$i = j$$ the above equation's all terms are $$\frac{x_j-x_m}{x_j-x_m} = 1$$ except where $$x_j = x_m$$ if $$i \neq j$$ then since $$m \neq j$$ doesn't preclude it, one term in the product will be for $$m=i$$, $$\frac{x_i-x_i}{x_j-x_i} = 0$$ Thus, $$\ell_j(x_i) = \delta_{ji} = \begin{cases} 1, & \text{if } j=i  \\ 0, & \text{if } j \ne i \end{cases} $$ Where $$\displaystyle \delta_{ji}$$ is that The value is 1 if they are equal $$ \delta_{1 \, 1} = 1 $$

And 0 otherwise. $$\delta_{1 \, 2} = 0. $$

Part2
$$\displaystyle p^h(x) = \sum_{j=1}^n p(x_j)\ell_j(x) $$ $$\displaystyle d(x)=p(x) - p^h(x) = p(x) - \sum_{j=1}^n p(x_j)\ell_j(x) $$ $$\displaystyle d(x_i)=p(x_i) - p^h(x_i) = p(x_i) - \sum_{j=1}^n p(x_j)\ell_j(x_i) $$ $$\displaystyle d(x_i)=p(x_i) - \sum_{j=1}^n p(x_j)\ell_j(x_i) = p(x_i)-[p(x_1)\ell_1(x_i) + p(x_2)\ell_2(x_i)+ p(x_3)\ell_3(x_i)+ .......+ p(x_n)\ell_n(x_i)] $$ When j = i, $$\displaystyle \ell_j(x_i) = \delta_{ji} = 1$$ $$\displaystyle p(x_i)\ell_i(x_i) = p(x_i)$$ The other terms are always zero. Hence, $$\displaystyle d(x_i) = p(x_i) - p(x_i) = 0$$

Problem 5: Integrate and Find Number of Integration Points
 Solved without assistance

Find: A & B below
A) Integrate Equation 5.1 analytically (not numerically) exactly over $$ [-1,1] \ $$ B) Determine the least nmber of integration points to integrate Eq5.1 exactly with the GL Quadrature. Carry out the integration and compare to the exact result.

part A
$$\displaystyle p(x) = \frac{x^7}{8} + \frac{x^6}{7} + \frac{x^2}{3} + 5$$ $$\displaystyle \int_{x=-1}^{x=1} p(x)dx = \int_{x=-1}^{x=1} (\frac{x^7}{8} + \frac{x^6}{7} + \frac{x^3}{3} + 5)dx $$ $$\displaystyle = [\frac {x^8}{8^2} + \frac{x^7}{7^2} + \frac{x^3}{3^2} +5x]_{-1}^{1} = \frac{492}{4} $$

part B
p(x) is 7 order equation. If n = 4, $$\displaystyle \int_{-1}^{1} f[x]dx = w_1f[x_1] + w_2f[x_2] + w_3f[x_3]+w_4f[x_4] $$ Where $$\displaystyle f_1[x] =1,f_2[x] =x,f_3[x]=x^2, ...... ,f_8[x] = x^7$$ $$\displaystyle \int_{-1}^{1} f_1[x]dx = w_1f_1[x_1] + w_2f_1[x_2] + w_3f_1[x_3]+w_4f_1[x_4] $$ $$\displaystyle \int_{-1}^{1} f_2[x]dx = w_1f_2[x_1] + w_2f_2[x_2] + w_3f_2[x_3]+w_4f_2[x_4] $$ $$\displaystyle \int_{-1}^{1} f_3[x]dx = w_1f_3[x_1] + w_2f_3[x_2] + w_3f_3[x_3]+w_4f_3[x_4] $$ $$\displaystyle \int_{-1}^{1} f_4[x]dx = w_1f_4[x_1] + w_2f_4[x_2] + w_3f_4[x_3]+w_4f_4[x_4] $$ $$\displaystyle \int_{-1}^{1} f_5[x]dx = w_1f_5[x_1] + w_2f_5[x_2] + w_3f_5[x_3]+w_4f_5[x_4] $$ $$\displaystyle \int_{-1}^{1} f_6[x]dx = w_1f_6[x_1] + w_2f_6[x_2] + w_3f_6[x_3]+w_4f_6[x_4] $$ $$\displaystyle \int_{-1}^{1} f_7[x]dx = w_1f_7[x_1] + w_2f_7[x_2] + w_3f_7[x_3]+w_4f_7[x_4] $$ $$\displaystyle \int_{-1}^{1} f_8[x]dx = w_1f_8[x_1] + w_2f_8[x_2] + w_3f_8[x_3]+w_4f_8[x_4] $$ From the above equations, $$\displaystyle w_1 + w_2 + w_3 + w_4= 2$$

$$\displaystyle w_1x_1 + w_2x_2 + w_3x_3 + w_4x_4= 0$$

$$\displaystyle w_1x_1^2 + w_2x_2^2 + w_3x_3^2 + w_4x_4^2= \frac{2}{3}$$

$$\displaystyle w_1x_1^3 + w_2x_2^3 + w_3x_3^3 + w_4x_4^3= 0$$

$$\displaystyle w_1x_1^4 + w_2x_2^4 + w_3x_3^4 + w_4x_4^4= \frac{2}{5}$$

$$\displaystyle w_1x_1^5 + w_2x_2^5 + w_3x_3^5 + w_4x_4^5= 0$$

$$\displaystyle w_1x_1^6 + w_2x_2^6 + w_3x_3^6 + w_4x_4^6= \frac{2}{7}$$

$$\displaystyle w_1x_1^7 + w_2x_2^7 + w_3x_3^7 + w_4x_4^7= 0$$

As a result of the above equations, the solution is that $$\displaystyle x_1 = -0.86113$$ $$\displaystyle x_2 = -0.33998$$ $$\displaystyle x_3 = 0.33998$$ $$\displaystyle x_4 = 0.86113$$ $$\displaystyle w_1 = 0.34785$$ $$\displaystyle w_2 = 0.65214$$ $$\displaystyle w_3 = 0.65214$$ $$\displaystyle w_4 = 0.34785$$

These results show that $$\displaystyle \int_{x=-1}^{x=1} p(x)dx = \sum_{j=1}^{n}w_jp(x_j) $$ Actually, $$\displaystyle \int_{x=-1}^{x=1} p(x)dx = \frac{492}{49} = 10.0408 $$ However, using the above values, $$\displaystyle \sum_{j=1}^{n}w_jp(x_j) = 10.2630$$ I think it is round-off error and truncation error.

title: User:Egm6322.s12.TEAM1.KIM/HW3


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title: Talk:Radiation astronomy/Greens


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VendingMachine/Howard Community Collge/Spring2012

 * /p1501ghi/


text:
VendingMachine/Howard Community Collge/Spring2012/p1501ghi

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 * sfoerster

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text:
User:Egm4313.s12.team11.arrieta/Report 3

Problem 3.6
Solved by Francisco Arrieta

Problem Statement

 * Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODE-CC (see p. 7-2b)

 * $$ y_{p,1}''-3y_{p,1}'+2y_{p,1}=r_1(x)=4x^2 \!$$

 * $$ y_{p,2}''-3y_{p,2}'+2y_{p,2}=r_2(x)=-6x^5 \!$$

 * The particular solution to $$ y_{p,1} \!$$ had been found in R3.3 p.7-11.
 * Find the particular solution $$ y_{p,2} \!$$, and then obtain the solution $$ y \!$$ for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.

Given

 * First particular solution

 * $$ y_{p,1}=2x^2+6x+7 \!$$

 * Initial Conditions

 * $$ y(0)=1 \!$$
 * $$ y'(0)=0 \!$$

Particular Solution

 * Since the specific excitation $$ r_2(x)=-6x^5 \!$$, using table 2.1 from K 2011 p.82, the choice for the particular solution is $$ y_p(x)=\sum_{j=0}^{n}K_jx^j \!$$

 * Then

 * $$ y_{p,2}=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 \!$$
 * $$ y_{p,2}'=c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4 \!$$
 * $$ y_{p,2}''=2c_2+6c_3x+12c_4x^2+20c_5x^3 \!$$

 * Plugging these equations back in the original $$ y_{p,2}''-3y_{p,2}'+2y_{p,2}=-6x^5 \!$$

 * $$ (2c_2+6c_3x+12c_4x^2+20c_5x^3)-3(c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4)+2(c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5)=-6x^5 \!$$

 * Using coefficient matching of left hand side to the right hand side of the equation, the following equations are obtained

 * $$ c:  2c_2-3c_1+2c_0=0 \!$$
 * $$ x:  6c_3-6c_2+2c_1=0 \!$$
 * $$ x^2:  12c_4-9c_3+2c_2=0 \!$$
 * $$ x^3:  20c_5-12c_4+2c_3=0 \!$$
 * $$ x^4:  -15c_5+2c_4=0 \!$$
 * $$ x^5:  2c_5=-6 \!$$

 * Which is equivalent to the following matrix

 * $$ \begin{bmatrix}

2&-3 & 2 & 0 & 0 & 0\\ 0& 2 &-6 & 6 & 0 & 0\\  0& 0 & 2 &-9 &12 & 0\\  0& 0 & 0 & 2 &-12& 20\\  0& 0 & 0 & 0 & 2 & -15\\  0& 0 & 0 & 0 & 0 & 2 \end{bmatrix}

\begin{bmatrix} c_0\\ c_1\\ c_2\\ c_3\\ c_4\\ c_5 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ -6 \end{bmatrix} \!$$

 * Using back substitution method to solve for every coefficient, starting with $$ c_5 \!$$

 * $$ 2c_5=-6 \rightarrow c_5=-3 \!$$
 * $$ 2c_4-15(-3)=0 \rightarrow c_4=-\frac{45}{2} \!$$
 * $$ 2c_3-12(-\frac{42}{2})+20(-3)=0 \rightarrow c_3=-105 \!$$
 * $$ 2c_2-9(-105)+12(-\frac{42}{2})=0 \rightarrow c_2=-\frac{675}{2} \!$$
 * $$ 2c_1-6(-\frac{675}{2})+6(-105)=0 \rightarrow c_1=-\frac{1395}{2} \!$$
 * $$ 2c_0-3(-\frac{1395}{2})+2(-\frac{675}{2})=0 \rightarrow c_0=-\frac{2835}{4} \!$$

 * Plugging these values back into $$ y_{p,2}=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 \!$$ gives

$$ y_{p,2}=-\frac{2835}{4}-\frac{1395}{2}x-\frac{675}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$

 * Since these equations are L2-ODE-CC, the superposition principle applies and $$ y_p=y_{p,1}+y_{p,2} \!$$ and the general particular solution becomes

$$ y_p(x)=-\frac{2807}{4}-\frac{1383}{2}x-\frac{671}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$

Homogeneous Solution

 * $$ y_{h}''-3y_{h}'+2y_{h}=0 \!$$

 * Due to the linearity of this homogeneous equation, a linear combination of two linear independent solutions is also a solution

 * $$ y_{h,1}''-3y_{h,1}'+2y_{h,1}=0 \!$$
 * $$ y_{h,2}''-3y_{h,2}'+2y_{h,2}=0 \!$$

 * With solutions

 * $$ y_{h,1}=e^{\lambda_1x} \!$$
 * $$ y_{h,2}=e^{\lambda_2x} \!$$

 * In order to determine the value of $$ \lambda_{1,2} \!$$, the characteristic equation must be determine from the homogeneous equation

 * $$ \lambda^2-3\lambda+2=0 \!$$

 * $$ \lambda_{1,2}=\frac{-a\pm \sqrt{a^2-4b}}{2} \!$$
 * $$ \lambda_{1,2}=\frac{-(-3)\pm \sqrt{(-3)^2-4(2)}}{2} \!$$
 * $$ \lambda_1=2 \; \; \; \; \; \; \; \lambda_2=1\!$$

 * Then the solutions for each distinct linearly independent homogeneous equation becomes

 * $$ y_{h,1}=e^{2x} \!$$
 * $$ y_{h,2}=e^{x} \!$$

 * Because of linearity, the linear combination of the previous two equations times two constants that satisfy 2 initial conditions, is also a solution

$$ y_h(x)=C_1e^{2x}+C_2e^x \!$$

General Solution

 * The overall solution for the L2-ODE-CC

 * $$ y(x)=y_h+y_p \!$$
 * $$ y(x)=C_1e^{2x}+C_2e^x-\frac{2807}{4}-\frac{1383}{2}x-\frac{671}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$
 * $$ y'(x)=2C_1e^{2x}+C_2e^x-\frac{1383}{2}-671x-315x^2-90x^3-15x^4 \!$$

 * Using the initial conditions to solve for $$ C_1 \!$$ and $$ C_2 \!$$

 * $$ 1=C_1+C_2-\frac{2807}{4} \!$$
 * $$ 0=C_1+C_2-\frac{1383}{2} \!$$
 * $$ C_1=-\frac{45}{4} \; \; \; C_2=714 \!$$

 * The general solution becomes

$$ y(x)=-\frac{45}{4}e^{2x}+714e^x-\frac{2807}{4}-\frac{1383}{2}x-\frac{671}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$

--Egm4313.s12.team11.arrieta 20:26, 19 February 2012 (UTC)


text:
User:1sfoerster/enes100/myWorkOnVendingMachine

link to vending machine project

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text:
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Licensing
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