text:
User:Egm4313.s12.team14.pickett.report2

R2.8

2.8.8

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

{y}'' + y' + 3.25y = 0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Problem: Find the general solution for this homogeneous differential equation.

Solution: Assume that the solution is of the form
 * {| style="width:100%" border="0" align="left"

y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

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

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

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

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1 yields

$$\displaystyle \lambda^2 e^{\lambda x} + \lambda e^{\lambda x} + 3.25e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + \lambda + 3.25) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle \therefore (\lambda^2 + 4\lambda + 3.25) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(3.25)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{1 - 13)}}{2}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{-12}}{2}$$ $$\displaystyle \lambda = -0.5 \pm j{\sqrt{3}}$$ Where $$\displaystyle j = \sqrt{-1}$$

$$\displaystyle y_h = {C_1}e^{(-0.5 + j\sqrt{3})x} + {C_2}e^{(-0.5 - j\sqrt{3})x} $$ In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation Let $$\displaystyle a = -0.5, b = \sqrt{3}$$ $$\displaystyle y_h = {C_1}e^{(a + jb)x} + {C_2}e^{(a - jb)x} $$ $$\displaystyle y_h = {C_1}e^{ax}e^{jbx} + {C_2}e^{ax}e^{-jbx} $$ $$\displaystyle {y_h}' = {C_1}[e^{ax}({jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle + {C_2}[e^{ax}(-{jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle {y_h}'' = {C_1}[e^{ax}((jb)^2e^{jbx}) + ({jb}e^{jbx})({a}e^{ax}) + e^{jbx}((a)^2e^{ax}) + ({a}e^{ax})({jb}e^{jbx})]$$ $$\displaystyle + {C_2}[e^{ax}((jb)^2e^{-jbx}) + (-{jb}e^{-jbx})({a}e^{ax}) + e^{-jbx}((a)^2e^{ax}) + ({a}e^{ax})(-{jb}e^{-jbx})] $$

Plugging $$\displaystyle y_h, {y_h}' , {y_h}'' $$ into Eq. 1 and collecting terms gives $$\displaystyle 0 = {C_1}e^{ax}e^{jbx}[-b^2 + jab +a^2 + jab + jb + a + 3.25] $$ $$\displaystyle + {C_2}e^{ax}e^{-jbx}[-b^2 - jab +a^2 - jab - jb + a + 3.25] $$ $$\displaystyle 0 = {C_1}e^{-0.5x}e^{j{\sqrt{3} x}}[-3 - j{\sqrt{3}}/2 + 1/4 - j{\sqrt{3}}/2 + j{\sqrt{3}} - 1/2 + 3.25] $$ $$\displaystyle + {C_2}e^{-0.5x}e^{-j{\sqrt{3}}x}[-3 + j{\sqrt{3}}/2 + 1/4 + j{\sqrt{3}}/2 - j{\sqrt{3}} - 1/2 + 3.25] $$ $$\displaystyle 0 = {C_1}e^{-0.5x}e^{j{\sqrt{3}}x}(0) + {C_2}e^{-0.5x}e^{-j{\sqrt{3}}x}(0) $$

Therefore the solution holds for any values of $$\displaystyle C_1, C_2 , x $$

2.8.15

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

{y}'' + (0.54)y' + (0.0729 + \pi)y = 0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Problem: Find the general solution for this homogeneous differential equation.

Solution: Assume that the solution is of the form
 * {| style="width:100%" border="0" align="left"

y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

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

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

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

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1 yields

$$\displaystyle \lambda^2 e^{\lambda x} + (0.54)\lambda e^{\lambda x} + (0.0729 + \pi)e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + (0.54)\lambda + (0.0729 + \pi)) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle \therefore (\lambda^2 + (0.54)\lambda + (0.0729 + \pi)) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-(0.54) \pm \sqrt{(0.54)^2 - 4(1)(0.0729 + \pi)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-0.54 \pm j{3.545}}{2}$$ $$\displaystyle \lambda = -0.27 \pm j(1.772)$$ Where $$\displaystyle j = \sqrt{-1}$$

$$\displaystyle y_h = {C_1}e^{(-0.27 + j(1.772))x} + {C_2}e^{(-0.27 - j(1.772))x} $$ In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation Let $$\displaystyle a = -0.27, b = 1.772$$ $$\displaystyle y_h = {C_1}e^{(a + jb)x} + {C_2}e^{(a - jb)x} $$ $$\displaystyle y_h = {C_1}e^{ax}e^{jbx} + {C_2}e^{ax}e^{-jbx} $$ $$\displaystyle {y_h}' = {C_1}[e^{ax}({jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle + {C_2}[e^{ax}(-{jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle {y_h}'' = {C_1}[e^{ax}((jb)^2e^{jbx}) + ({jb}e^{jbx})({a}e^{ax}) + e^{jbx}((a)^2e^{ax}) + ({a}e^{ax})({jb}e^{jbx})]$$ $$\displaystyle + {C_2}[e^{ax}((jb)^2e^{-jbx}) + (-{jb}e^{-jbx})({a}e^{ax}) + e^{-jbx}((a)^2e^{ax}) + ({a}e^{ax})(-{jb}e^{-jbx})] $$

Plugging $$\displaystyle y_h, {y_h}' , {y_h}'' $$ into Eq. 1 and collecting terms gives $$\displaystyle 0 = {C_1}e^{ax}e^{jbx}[-b^2 + jab +a^2 + jab + (0.54)(jb + a) + 0.0729 + \pi]$$ $$\displaystyle + {C_2}e^{ax}e^{-jbx}[-b^2 - jab +a^2 - jab + (0.54)(-jb + a) + 0.0729 + \pi] $$ $$\displaystyle 0 = {C_1}e^{-0.27x}e^{j{1.772x}}[-(1.772)^2 + j(-0.27)(1.772) + (0.27)^2 $$ $$\displaystyle + j(-0.27)(1.772) + (0.54)(j(1.772) - 0.27) + 0.0729 + \pi] $$ $$\displaystyle + {C_2}e^{-0.27x}e^{-j1.772x}[-(1.772)^2 - j(-0.27)(1.772) + (0.27)^2 $$ $$\displaystyle - j(-0.27)(1.772) + (0.54)(-j(1.772) - 0.27) + 0.0729 + \pi] $$ $$\displaystyle 0 = {C_1}e^{-0.27x}e^{j1.772x}(0) + {C_2}e^{-0.27x}e^{-j1.772x}(0) $$

Therefore the solution holds for any values of $$\displaystyle C_1, C_2 , x $$

$$\displaystyle y = {C_1}e^{5x} + {C_2}{x}e^{5x} $$ $$\displaystyle y(x) = e^{5x} - {5x}e^{5x} $$

Given: $$\displaystyle {y_p}'' + p(x)y' +q(x)y = r(x) $$ (1) Where p(x) and q(x) are any functions of "x".

Find: Prove that $$\displaystyle y_p(x) = \sum_{i=1}^n y_{p,i}(x) $$ is the overall solution to (1) when $$\displaystyle r(x) = \sum_{i=1}^n r_{i}(x) $$ Also, discuss the need to have both sine and cosine functions in the particular solution "guess" when you only have a sine or cosine function as the input, r(x)

Solution: Let the indices "i" for the particular solutions correspond to the respective indices for the forcing functions such that $$\displaystyle {y_1}'' + p(x){y_1}' +q(x){y_1} = {r_1}(x) $$ (2) and $$\displaystyle {y_2}'' + p(x){y_2}' +q(x){y_2} = {r_2}(x) $$ (3)

Adding both equations together yields: $$\displaystyle {y_1} + p(x){y_1}' +q(x){y_1} + {y_2} + p(x){y_2}' +q(x){y_2} = {r_1}(x) + {r_2}(x) $$ By collecting terms and using summation notation we obtain: $$\displaystyle \sum_{i=1}^2 y_{p,i}''(x) + p(x){\sum_{i=1}^2 y_{p,i}'(x)} +q(x){\sum_{i=1}^2 y_{p,i}(x)} = \sum_{i=1}^2 r_{i}(x) $$ This concept is easily extended to an "n" number of excitation functions with a corresponding "n" number of particular solutions such that: $$\displaystyle \sum_{i=1}^n y_{p,i}''(x) + p(x){\sum_{i=1}^n y_{p,i}'(x)} +q(x){\sum_{i=1}^n y_{p,i}(x)} = \sum_{i=1}^n r_{i}(x) $$ This proves the concept of superposition for excitation functions and their particular solutions for second order differential equations with varying coefficients.

The reason for having both sine and cosine in the particular solution when there is only either sine or cosine as the forcing function is due to the fact that the derivative of sine is cosine and the derivative of cosine is negative sine. Thus, without including both in the initial "guess" for the particular solution, we would be missing half of the solution. Or in other words: $$\displaystyle {y_p} = M\cos(wx) + N\sin(wx) $$ $$\displaystyle {y_p}' = -M\sin(wx) + N\cos(wx) $$ $$\displaystyle {y_p}'' = -M\cos(wx) - N\sin(wx) $$ So plugging these into an arbitrary second order differential equation would yield values for "N" and "M", when there is a forcing function of sine or cosine.

= Problem 8 =

Given

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

x^{n}log(1+x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

n=0, n=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Find the integral of EQ 1 using integration by parts for n=0 and n=1

Solution
For n=0

For n=1

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

\int xlog(1+x)dx$$ $$ Take u=(x+1) and du=dx
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\int (u-1)log(u)du$$
 * $$\displaystyle
 * $$\displaystyle

$$ This gives two separate integrals:
 * $$\displaystyle
 * }
 * }

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

\int u log(u)du-\int log(u)du$$ $$ Integrate the first integral (left hand) by parts, taking:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f=log(u), df=\frac{1}{u}du, dg=udu, g=\frac{u^{2}}{2}$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{1}{2}u^{2}log(u)-\frac{1}{2}\int udu$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\frac{1}{2}u^{2}log(u) - \frac{u^2}{4}$$ $$ Integrate the second integral (right hand) by parts, taking: :
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

f=log(u), df=\frac{1}{u}du, dg=du, g=u$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

ulog(u) - \int du $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

ulog(u)- u $$ $$ Substituting EQ.3 and EQ.4 into EQ.2 gives:
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

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

(\frac{1}{2}u^{2}log(u) - \frac{u^2}{4}) - (ulog(u)- u) + Constant (K) $$ $$ Since u=x+1
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\frac{1}{2}(x + 1)^{2}log(x + 1) - \frac{(x + 1)^2}{4} - (x + 1)log(x + 1) + (x + 1) + K$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

title: User:Egm4313.s12.team14.pickett.report2


text:
University of Florida/Egm4313/s12.team14.report2

Find
Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$\displaystyle r(x) $$

Consider no excitation:

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

r(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Plot the solution.

Solution
Characteristic equation:

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

(\lambda-\lambda_1)(\lambda-\lambda_2)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Substituting $$\displaystyle(Eq. 1a)$$ into $$\displaystyle(Eq. 3)$$:

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

(\lambda+2)(\lambda-5)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

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

\lambda^2-3\lambda-10=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

Non-homogeneous solution:

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

y''-3y'-10y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Homogeneous solution:

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

y_h(x)=C_1e^{-2x}+C_2e^{5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

Overall solution:

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

y(x)=C_1e^{-2x}+C_2e^{5x}+y_p(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

No excitation:

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

r(x)=0 => y_p(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

From intitial conditions:

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

y(0)=1=C_1+C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

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

y'(0)=0=-2C_1+5C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

Solving for $$\displaystyle C_1 $$ and $$\displaystyle C_2 $$:

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

C_1=\frac{5}{7} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and

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

C_2=\frac{2}{7} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

So, the final solution is:

$$\displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$

Find
Find and plot the solution for $$\displaystyle (Eq. 1) $$

Solution
Due to no excitation, $$\displaystyle (Eq. 1) $$ becomes:

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

y'' - 10y' +25y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Substituting $$\displaystyle \lambda $$ into $$\displaystyle (Eq. 4) $$:

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

\lambda^2-10\lambda+25=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

Factoring, and solving for $$\displaystyle \lambda $$:

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

(\lambda-5)^2=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

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

\lambda_2=\lambda_1=\lambda=5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since $$\displaystyle \lambda $$ is a double root, the general solution:

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

y=C_1e^{-5x}+C_2xe^{-5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

From intitial conditions:

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

y(0)=1=C_1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }

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

y'(0)=0=5C_1+C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }

Solving for $$\displaystyle C_1 $$ and $$\displaystyle C_2 $$:

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

C_1=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

and

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

C_2=-5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

So, the final solution is:

$$\displaystyle y(x)=e^{5x}-e^{5x} $$

Given
(a)
 * $$\displaystyle{y''+6y'+8.96y = 0}$$

(b)
 * $$\displaystyle{y''+4y'+(\pi^2+4)y = 0}$$

Find
General solution to the differential equation

Solution
(a)

Characteristic equation:

 * $$\displaystyle{\lambda^2+6\lambda+8.96 = 0}$$

Using the quadratic equation:

 * $$\displaystyle{x = \frac{-b\plusmn\sqrt{b^2-4ac}}{2a}\,}$$

where:

 * $$\displaystyle{a=1, b=6, c=8.96}$$

Place values into quadratic equation:

 * $$\displaystyle{x = \frac{-6\plusmn\sqrt{6^2-4*1*8.96}}{2*1}\,}$$

 * $$\displaystyle{x = \frac{-6\plusmn\sqrt{36-35.84}}{2}\,}$$

 * $$\displaystyle{x = \frac{-6\plusmn0.4}{2}\,}$$

 * $$\displaystyle{x = \frac{-6.4}{2}\,,\frac{-5.6}{2}}$$

 * $$\displaystyle{x = -3.2,-2.8}$$

$$\displaystyle{y = c_1e^{-3.2x}+c_2e^{-2.8x}}$$

(b)

Characteristic equation:

 * $$\displaystyle{\lambda^2+4\lambda+(\pi^2+4) = 0}$$

Using the quadratic equation:

 * $$\displaystyle{\lambda = \frac{-b\plusmn\sqrt{b^2-4ac}}{2a}\,}$$

where:

$$\displaystyle{a=1, b=4, c=(\pi^2+4)}$$

Place values into quadratic equation:

 * $$\displaystyle{\lambda = \frac{-4\plusmn\sqrt{4^2-4*1*(\pi^2+4)}}{2*1}\,}$$

 * $$\displaystyle{\lambda = \frac{-4\plusmn\sqrt{16-4\pi^2-16)}}{2}\,}$$

 * $$\displaystyle{\lambda = \frac{-4\plusmn2\pi i}{2}\,}$$

 * $$\displaystyle{\lambda = -2\plusmn\pi i}$$

$$\displaystyle{y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x)}$$

Given
(5)

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

 * $$\displaystyle {y}'' + 2\pi {y}' + {\pi}^2y = 0$$
 *
$$\displaystyle (Eq. 1) $$
 * }
 * }
 * }

(6)

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

 * $$\displaystyle {10{y}'' - 32{y}' + 25.6y = 0}$$
 *
$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

h

Find
For both (5) and (6), find a general solution. Then, check answers by substitution.

Solution
(5)

To obtain the general solution for (Eq. 1), we let

 * $$\displaystyle y = e^{\lambda x}$$

This yields:

 * $$\displaystyle e^{\lambda x} (\lambda^{2} + 2\pi\lambda +\pi^{2}) = 0$$

Therefore, we have:

 * $$\displaystyle \lambda^{2} + 2\pi\lambda +\pi^{2} = 0$$

Where $$\displaystyle \lambda$$ is a solution to the characteristic equation

 * $$\displaystyle \lambda^{2} + a\lambda + b = 0$$

We can see that

 * $$\displaystyle a = 2\pi$$ and
 * $$\displaystyle b = \pi^{2}$$.

Next, we must examine the quadratic formula to determine whether the system is over-, under-, or critically-damped. The discriminant is:

 * $$\displaystyle \triangle = a^{2} - 4b$$

 * $$\displaystyle \triangle = (2\pi^{2})^{2} - 4(\pi^{2})$$

 * $$\displaystyle \triangle = 0$$

Therefore, the equation is critically-damped.

The general solution is therefore represented by:

 * $$\displaystyle y = c_{1}e^{\lambda_{1}x} + c_{2}xe^{\lambda_{2}x}$$

Where $$\displaystyle \lambda_{1} = \lambda_{2} = \lambda$$.

We can determine the value of $$\displaystyle \lambda$$ by again using the quadratic equation, this time in full:

 * $$\displaystyle \lambda = \frac{1}{2} (-a + \sqrt{a^{2} - 4b})$$

for which we find that $$\displaystyle \lambda = -\pi$$.

Therefore, the general solution to the equation is:

$$\displaystyle y = c_{1}e^{-\pi x} + c_{2}xe^{-\pi x}$$

Verification by substitution:

 * $$\displaystyle {y}' = -\pi c_{1} e^{-\pi x} + c_{2} e^{-\pi x} - \pi c_{2} x e^{-\pi x}$$

 * $$\displaystyle {y}'' = {\pi}^2 c_{1} e^{-\pi x} - \pi c_{2} e^{-\pi x} + {\pi}^2 c_{2} x e^{-\pi x} - \pi c_{2} e^{-\pi x}$$

Plugging into Eq. 1 and combining terms, it is found:

 * $$\displaystyle c_{1}e^{-\pi x} ({\pi}^2-2{\pi}^2+{\pi}^2) + c_{2} e^{-\pi x}(-\pi-\pi+2\pi)+c_{2}xe^{-\pi x}({\pi}^2-2{\pi}^2+{\pi}^2) = 0$$

All the terms cancel, so the statement is true, and the solution is verified.

(6)

To obtain the general solution for (Eq. 2), let

 * $$\displaystyle y = e^{\lambda x}$$

So then, we have

 * $$\displaystyle 10\lambda^{2} - 32\lambda + 25.6 = 0$$

as a solution to the characteristic equation. All terms must be divided by 10, to put the equation in standard form:

 * $$\displaystyle \lambda^{2} - 3.2\lambda + 2.56 = 0$$

Examining the discriminant, we find that

 * $$\displaystyle \triangle = a^{2} - 4b = (-3.2)^{2} - 4(2.56) = 0$$

Therefore, the equation is critically-damped.

The general solution is represented by:

 * $$\displaystyle y = c_{1}e^{\lambda_{1}x} + c_{2}xe^{\lambda_{2}x}$$

and $$\displaystyle \lambda_{1} = \lambda_{2} = \lambda$$.

To determine the value of $$\displaystyle \lambda$$, we set:

 * $$\displaystyle \lambda = \frac{1}{2} (-a + \sqrt{a^{2} - 4b})$$

for which we find that $$\displaystyle \lambda = 1.6$$.

Therefore, the general solution to the equation is:

$$\displaystyle y = c_{1}e^{1.6x} + c_{2}xe^{1.6x}$$

Verification by substitution:

 * $$\displaystyle {y}'=1.6c_{1}e^{1.6x}+c_{2}e^{1.6x}+1.6c_{2}xe^{1.6x}$$

 * $$\displaystyle {y}'' = (1.6^2)c_{1}e^{1.6x}+3.2c_{2}e^{1.6x}+(1.6^2)c_{2}xe^{1.6x}$$

Plugging into Eq. 1 and combining terms, it is found:

 * $$\displaystyle c_{1}e^{1.6x}(0)+c_{2}e^{1.6x}(0)+c_{2}xe^{1.6x}(0)=0$$

All the terms cancel, so the statement is true, and the solution is verified.

Given

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

{y}'' + a{y}' + by = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Given the basis:

(a)
 * {| style="width:100%" border="0" align="left"

e^{2.6x},e^{-4.3x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

(b)
 * {| style="width:100%" border="0" align="left"

e^{-\sqrt{5}x},xe^{-\sqrt{5}x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
For the given Information above, Find an ODE for both (a) and (b)

Solution
(a)The general solution can be written as:
 * {| style="width:100%" border="0" align="left"

y=C_{1}e^{2.6x}+C_{2}e^{-4.3x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The characteristic equation can be written in the following way:
 * {| style="width:100%" border="0" align="left"

(\lambda-2.6)(\lambda+4.3)=\lambda^{2}+1.7\lambda-11.18=0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Giving the ODE:
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle {y}''+1.7{y}'-11.18y=0 $$
 * }
 * }

(b) The general solution can be written as:
 * {| style="width:100%" border="0" align="left"

y=C_{1}e^{-\sqrt{5}x}+C_{2}xe^{-\sqrt{5}x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The characteristic equation can be written in the following way:
 * {| style="width:100%" border="0" align="left"

(\lambda+\sqrt{5})(\lambda+\sqrt{5})=\lambda^{2}+2\sqrt{5}\lambda+5=0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Giving the ODE:
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle {y}''+2\sqrt{5}{y}'+5y=0 $$
 * }
 * }

Given

Spring-dashpot equation of motion from sec 1-5
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle{my{_{k}}''+m\frac{k}{c}y{_{k}}'+ky_{k}=f(t)}$$
 * }
 * }
 * }

Find
Spring-dashpot-mass system in series. Find the values for the parameters k,c,m with a double real root of $$\lambda=-3$$

Solution
Consider the double real root
 * {| style="width:100%" border="0" align="left"

Characteristic equation is
 * $$\displaystyle{\lambda=-3}$$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Recall
 * $$\displaystyle{(\lambda+3)^{2}=\lambda^{2}+6\lambda+9=0}$$
 *
$$\displaystyle (Eq. 1) $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Thus
 * $$\displaystyle{my{_{k}}''+m\frac{k}{c}y{_{k}}'+ky_{k}=f(t)}$$
 *
$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle{m=1}$$
 * }
 * {| style="width:100%" border="0" align="left"
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle{m\frac{k}{c}=6}$$
 * }
 * {| style="width:100%" border="0" align="left"
 * }
 * {| style="width:100%" border="0" align="left"

Solve for c
 * $$\displaystyle{k=9}$$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Therefore $$\displaystyle{c=\frac{3}{2}\;and\;k=9\;and\;m=1}$$
 * $$\displaystyle{c=\frac{9}{6}=\frac{3}{2}}$$
 * }
 * }
 * }

Given
Taylor Series at t=0

Find
Develop the MacLaurin Series for $$ e^t,\cos t,\sin t\! $$

Solution
Taylor series is defined as
 * {| style="width:100%" border="0" align="left"

The MacLaurin series occurs when t=0
 * $$ \sum_{n=0} ^ {\infin } \frac {f^{(n)}(t)}{n!} \, (x-t)^{n}$$
 *
$$\displaystyle (Eq. 1) $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Development of MacLaurin series for $$e^t$$
 * $$ \sum_{n=0} ^ {\infin } \frac {f^{(n)}(0)}{n!} \, (x)^{n}$$
 *
$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Final Answer $$e^t=1+x+\frac{1}{2}t^2+\frac{1}{6}t^3...$$ Explicit form can be written as $$ e^t=\sum_{n=0} ^ {\infin } \frac {1}{n!} \, (t)^{n}$$
 * $$e^t=\frac{1}{0!}+\frac{1}{1!}t+\frac{1}{2!}t^2+\frac{1}{3!}t^3...$$
 *
$$\displaystyle (Eq. 3) $$
 * }
 * }
 * }

Development of MacLaurin series for $$cos(t)$$
 * {| style="width:100%" border="0" align="left"

Final Answer $$cos(t)=1-\frac{1}{2}t^2+\frac{1}{24}t^4...$$ Explicit form can be written as $$ cos(t)=\sum_{n=0} ^ {\infin } \frac {-1}{2n!} \, (t)^{2n}$$
 * $$cos(t)=\frac{1}{0!}+\frac{0}{1!}t-\frac{1}{2!}t^2+\frac{0}{3!}t^3+\frac{1}{4!}t^4...$$
 *
$$\displaystyle (Eq. 4) $$
 * }
 * }
 * }

Development of MacLaurin series for $$sin(t)$$
 * {| style="width:100%" border="0" align="left"

Final Answer $$sin(t)=1t-\frac{1}{6}t^3+\frac{1}{120}t^5...$$ Explicit form can be written as $$ sin(t)=\sum_{n=0} ^ {\infin } \frac {-1}{(2n+1)!} \, (t)^{(2n+1)}$$
 * $$sin(t)=\frac{0}{0!}+\frac{1}{1!}t-\frac{0}{2!}t^2-\frac{1}{3!}t^3+\frac{0}{4!}t^4+\frac{1}{5!}t^5...$$
 *
$$\displaystyle (Eq. 5) $$
 * }
 * }
 * }

Given
(8)
 * {| style="width:100%" border="0" align="left"

{y}'' + y' + 3.25y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1a)
 * }
 * }

(15)
 * {| style="width:100%" border="0" align="left"

{y}'' + (0.54)y' + (0.0729 + \pi)y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1b)
 * }
 * }

Solution
(8)

Assume that the solution is of the form
 * {| style="width:100%" border="0" align="left"

y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

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

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

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

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1a yields

$$\displaystyle \lambda^2 e^{\lambda x} + \lambda e^{\lambda x} + 3.25e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + \lambda + 3.25) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle \therefore (\lambda^2 + 4\lambda + 3.25) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(3.25)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{1 - 13)}}{2}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{-12}}{2}$$ $$\displaystyle \lambda = -0.5 \pm j{\sqrt{3}}$$ Where $$\displaystyle j = \sqrt{-1}$$

$$\displaystyle y_h = {C_1}e^{(-0.5 + j\sqrt{3})x} + {C_2}e^{(-0.5 - j\sqrt{3})x} $$ In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation Let $$\displaystyle a = -0.5, b = \sqrt{3}$$ $$\displaystyle y_h = {C_1}e^{(a + jb)x} + {C_2}e^{(a - jb)x} $$ $$\displaystyle y_h = {C_1}e^{ax}e^{jbx} + {C_2}e^{ax}e^{-jbx} $$ $$\displaystyle {y_h}' = {C_1}[e^{ax}({jb}e^{jbx}) + e^{jbx}({a}e^{ax})]+ {C_2}[e^{ax}(-{jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle {y_h}'' = {C_1}[e^{ax}((jb)^2e^{jbx}) + ({jb}e^{jbx})({a}e^{ax}) + e^{jbx}((a)^2e^{ax}) + ({a}e^{ax})({jb}e^{jbx})]+ {C_2}[e^{ax}((jb)^2e^{-jbx}) + (-{jb}e^{-jbx})({a}e^{ax}) + e^{-jbx}((a)^2e^{ax}) + ({a}e^{ax})(-{jb}e^{-jbx})] $$

Plugging $$\displaystyle y_h, {y_h}' , {y_h}'' $$ into Eq. 1 and collecting terms gives $$\displaystyle 0 = {C_1}e^{ax}e^{jbx}[-b^2 + jab +a^2 + jab + jb + a + 3.25] + {C_2}e^{ax}e^{-jbx}[-b^2 - jab +a^2 - jab - jb + a + 3.25] $$ $$\displaystyle 0 = {C_1}e^{-0.5x}e^{j{\sqrt{3} x}}[-3 - j{\sqrt{3}}/2 + 1/4 - j{\sqrt{3}}/2 + j{\sqrt{3}} - 1/2 + 3.25] + {C_2}e^{-0.5x}e^{-j{\sqrt{3}}x}[-3 + j{\sqrt{3}}/2 + 1/4 + j{\sqrt{3}}/2 - j{\sqrt{3}} - 1/2 + 3.25] $$ $$\displaystyle 0 = {C_1}e^{-0.5x}e^{j{\sqrt{3}}x}(0) + {C_2}e^{-0.5x}e^{-j{\sqrt{3}}x}(0) $$

Therefore the solution holds for any values of $$\displaystyle C_1, C_2 , x $$

(15)

Assume that the solution is of the form
 * {| style="width:100%" border="0" align="left"

y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

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

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

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

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1b yields

$$\displaystyle \lambda^2 e^{\lambda x} + (0.54)\lambda e^{\lambda x} + (0.0729 + \pi)e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + (0.54)\lambda + (0.0729 + \pi)) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle \therefore (\lambda^2 + (0.54)\lambda + (0.0729 + \pi)) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-(0.54) \pm \sqrt{(0.54)^2 - 4(1)(0.0729 + \pi)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-0.54 \pm j{3.545}}{2}$$ $$\displaystyle \lambda = -0.27 \pm j(1.772)$$ Where $$\displaystyle j = \sqrt{-1}$$

$$\displaystyle y_h = {C_1}e^{(-0.27 + j(1.772))x} + {C_2}e^{(-0.27 - j(1.772))x} $$ In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation Let $$\displaystyle a = -0.27, b = 1.772$$ $$\displaystyle y_h = {C_1}e^{(a + jb)x} + {C_2}e^{(a - jb)x} $$ $$\displaystyle y_h = {C_1}e^{ax}e^{jbx} + {C_2}e^{ax}e^{-jbx} $$ $$\displaystyle {y_h}' = {C_1}[e^{ax}({jb}e^{jbx}) + e^{jbx}({a}e^{ax})]+ {C_2}[e^{ax}(-{jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle {y_h}'' = {C_1}[e^{ax}((jb)^2e^{jbx}) + ({jb}e^{jbx})({a}e^{ax}) + e^{jbx}((a)^2e^{ax}) + ({a}e^{ax})({jb}e^{jbx})]+ {C_2}[e^{ax}((jb)^2e^{-jbx}) + (-{jb}e^{-jbx})({a}e^{ax}) + e^{-jbx}((a)^2e^{ax}) + ({a}e^{ax})(-{jb}e^{-jbx})] $$

Plugging $$\displaystyle y_h, {y_h}' , {y_h}'' $$ into Eq. 1 and collecting terms gives $$\displaystyle 0 = {C_1}e^{ax}e^{jbx}[-b^2 + jab +a^2 + jab + (0.54)(jb + a) + 0.0729 + \pi] + {C_2}e^{ax}e^{-jbx}[-b^2 - jab +a^2 - jab + (0.54)(-jb + a) + 0.0729 + \pi] $$ $$\displaystyle 0 = {C_1}e^{-0.27x}e^{j{1.772x}}[-(1.772)^2 + j(-0.27)(1.772) + (0.27)^2 + j(-0.27)(1.772) + (0.54)(j(1.772) - 0.27) + 0.0729 + \pi] + {C_2}e^{-0.27x}e^{-j1.772x}[-(1.772)^2 - j(-0.27)(1.772) + (0.27)^2 - j(-0.27)(1.772) + (0.54)(-j(1.772) - 0.27) + 0.0729 + \pi] $$ $$\displaystyle 0 = {C_1}e^{-0.27x}e^{j1.772x}(0) + {C_2}e^{-0.27x}e^{-j1.772x}(0) $$

Therefore the solution holds for any values of $$\displaystyle C_1, C_2 , x $$

Given
Differential equation, initial conditions, and forcing function as shown:
 * {| style="width:100%" border="0" align="left"

y'' -10 y' + 25y = r(x), $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

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

y(0) = 4, y'(0) = -5, r(x) = 7e^(5x)-2x^2. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
The solution to Eq. 1

Solution

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

y_p = C_{1}x^2e^{5x} + C_{2}x^2 + C_{3}x + C_4 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

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

y_p' = 5C_{1}x^2e^{5x} + 2C_{1}xe^{5x} + 2C_{2}x + C_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

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

y_p'' = 25C_{1}x^2e^{5x} + 20C_{1}xe^{5x} + 2C_{1}e^{5x} + 2C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Substituting Equations 2, 3, and 4 into Equation 1 yields
 * {| style="width:100%" border="0" align="left"

7e^{5x} - 2x^2 = 2C_{1}e^{5x} + 25C_{2}x^{2} + 25C_{3}x - 20C_{2}x + 2C_2 - 10C_3 + 25C_4 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

To solve for the constants:
 * {| style="width:100%" border="0" align="left"

7 = 2C_1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

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

-2 = 25C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

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

0 = 25C_3 - 20C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }

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

0 = 2C_2 - 10C_3 + 25C_4 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }

Solving Equations 6-9 yields
 * {| style="width:100%" border="0" align="left"

C_1 = \frac{7}{2}, C_2 = -\frac{2}{25}, C_3 = -\frac{8}{125}, C_4 = -\frac{12}{625} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

The homogeneous solution:
 * {| style="width:100%" border="0" align="left"

y_h = C_{5}e^{5x} + C_{6}xe^{5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * }

The total solution:
 * {| style="width:100%" border="0" align="left"

y = C_{5}e^{5x} + C_{6}xe^{5x} + 2\frac{7}{2}x^2e^{5x} - \frac{2}{25}x^{2} - \frac{8}{125}x - \frac{12}{625} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * }

Setting $$\displaystyle x = 0 $$ yields
 * {| style="width:100%" border="0" align="left"

y(0) = C_{5} - \frac{12}{625} = 4 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * }

Solving for $$\displaystyle C_5 $$ yields
 * {| style="width:100%" border="0" align="left"

C_{5} = \frac{2512}{625} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 13)
 * }
 * }

Finding $$\displaystyle y' $$ yields
 * {| style="width:100%" border="0" align="left"

y' = \frac{35}{2}x^2e^{5x} + 5C_{6}xe^{5x} + 7xe^{5x} + \frac{2512}{125}e^{5x} + C_{6}e^{5x} - \frac{4}{25}x^2 - \frac{8}{125} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * }

Setting $$\displaystyle x = 0 $$ yields
 * {| style="width:100%" border="0" align="left"

y'(0) = \frac{2512}{125} + C_6 - \frac{8}{125} = -5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * }

Solving for $$\displaystyle C_6 $$ yields
 * {| style="width:100%" border="0" align="left"

C_{6} = -\frac{3129}{125} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 16)
 * }
 * }

Therefore, the total solution is
 * {| style="width:100%" border="0" align="left"

y = \frac{2512}{625}e^{5x} - \frac{3129}{125}xe^{5x} + 2\frac{7}{2}x^2e^{5x} - \frac{2}{25}x^{2} - \frac{8}{125}x - \frac{12}{625} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (ANS Eq. 17)
 * }
 * }

Team Member Tasks
All team members contributed to the coding of this page.

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


title: File:Egm4313.s12.team9.1-4.jpg


text:
User:Egm4313.s12.team3.R2

Davy

title: User:Egm4313.s12.team3.R2


title: File:Egm4313.s12.team9.1-5.jpg


title: File:Egm4313.s12.team9.1-6.jpg


title: File:Egm4313.s12.team9.2-2.jpg


text:
User:Egm4313.s12.team14.kimiagarov/HW 1

Problem R1.2

Given:

]]

FBD:

]]

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

 * $$\displaystyle{F_k = ky}$$
 * $$\displaystyle (Eq. 1) $$
 * }
 * }
 * }

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

 * $$\displaystyle{F_c = cy'}$$
 * $$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

From the FBD:

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

 * $$\displaystyle{ma = -ky - cy' + r(t)}$$
 * $$\displaystyle (Eq. 3) $$
 * }
 * }
 * }

Plug (Eq. 1) and (Eq. 2) into (Eq 3)

$${my'' + ky + cy' = r(t)}$$

Problem R1.6

water level $${h}$$

$${h' = -k\sqrt{h}}$$

Homogeneous:

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

 * $$\displaystyle{h'_h + k\sqrt{h_h} = 0}$$
 * $$\displaystyle (Eq. 1) $$
 * }
 * }
 * }

Particular:

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

 * $$\displaystyle{h'_p + k\sqrt{h_p} = 0}$$
 * $$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

Add (Eq. 1) and (Eq. 2):

$${h'_h + h'_p +k(\sqrt{h_h} +\sqrt{h_p}) = 0}$$

It could be said that:   $${h'_h + h'_p = h'}$$

But, $${\sqrt{h_h} + \sqrt{h_p} \ne \sqrt{h_h + h_p} = \sqrt{h}}$$

Therefore, superposition cannot be used.

title: User:Egm4313.s12.team14.kimiagarov/HW 1


text: File:Lecture slide 2-5.jpg

title: File:Lecture slide 2-5.jpg


text:
Physics (A-level)/Practical skills in physics

Overview
For A-level physics, you will have to partake in certain practical experiments of three types: quantitative, qualitative and evaluative. You will be graded for one of each in both you AS and A2 parts of the course, adding up to six graded experiments in the two years. You will usually have two or three attempts at each category.

Evaluative
You will be asked to evaluate data from an experiment you have done using the examiners results. You will need to comment on an experiments precision, reliability and accuracy. For this you will need to understand the difference between the three. Precision - this refers to the distance of the points from the line of best fit on the graph. If the majority of the points lye on or very near to the line of best fit, the experiment is said to be precise. Reliability - this refers to the difference in results between repeats. If a repeat has a very similar value in the experiment, the experiment is said to be reliable. Accuracy - this refers to the comparison of a calculated value to the accepted value. If a value that you calculate is very close to the accepted value, it is said to be accurate. An example of this would be if you calculated the earths gravitational field strength (g) to be 9.75ms-2, it is very close to the accepted value of 9.81ms-2. Therefore that experiment would be accurate. Another evaluation you must make is working out or comparing gradients and y-intercepts. For this you must know the equation  y=mx+C  where y and x are variables, m is the gradient and C is the y-intercept. All linear equations can be rearranged into this form, in order to work out certain values. For example the equation, m=PM(1/l)+M The equation in this example compares m over 1/l. Comparing this to y=mx+C, m is the y-variable, 1/l is the x-variable, PM is the gradient and M is the y-intercept. You will be given or will have to work out the graph's gradient and y-intercept. With these you can work out values for P and M.

Links
Back to Physics (A-level)


text:
User:Krista jones88

I am a 23 year old female who is persuing a career in the music insdustry as an artist. I have taken a fascination of learninf and becoming a autodidact.... I am very interested in teaching myself different languages like spanish & japanese, I also want to learn the ins and outs of the music business, laws and all, all of I which I intend to educate myself on.. I'm just hungry for knowledge and the ability one has to obtain it on their own.

title: User:Krista jones88


text:
User:Egm4313.s12.team13.rlh

Report 1

Report 2

title: User:Egm4313.s12.team13.rlh


text:
User:Egm4313.s12.team3.landers

=Problem 1=

Problem Statement
For a spring-dashpot system in parallel with mass and an applied force f(t), derive the equation of motion. Figure of a spring-dashpot system in parallel:

Solution
The kinetics for this problem:
 * {| style="width:100%" border="0"

$$\displaystyle y=y_k=y_c $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle m\ddot{y}+y_I=f(t) $$     (2) $$y_I=$$ forces opposing f(t)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_I=y_k+y_c $$     (3) For a spring:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_k=kx_k $$     (4) For a dashpot:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_c=c\dot{x}_c $$     (5) Taking the derivative of equation 1, you recieve:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \dot{y}=\dot{y_k}=\dot{y_c} $$     (6) Plug equations 4 and 5 into equation 3:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_I=kx_k+c\dot{x_c} $$     (7) Taking equation 7 and plugging it back into equation 2 yields:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$ f(t)=m\ddot{x}+c\dot{x_c}+kx_k $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Problem Statement
Find and plot the solution to the L2 ODE with no excitation.
 * {| style="width:100%" border="0"

$$\displaystyle {y}''-10{y}'+25y=r(x) $$     (1) Given initial conditions:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y(0)=1 $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle {y}'(0)=0 $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * due to there being no excitation:
 * {| style="width:100%" border="0"

$$\displaystyle r(x)=0 $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Problem Solution
The Characteristic Equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+a\lambda+b=r(x)=0 $$     (2) From the original equation (1):
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle a=-10 $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle b=25 $$ Which yields:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2-10\lambda+25=0 $$    (3) Equation 3 can be simplified to:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle (\lambda-5)^2 $$ It has a double root;
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle \lambda=5 $$ Also can be shown by discriminant:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle (-10)^2-(4)(25)=0 $$ The General Solution
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y=(c_1+c_2x)e^{5x} $$     (4) Take the Derivative in order to apply initial conditions
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {y}'=c_2e^{5x}+5(c_1+c_2x)e^{5x} $$    (5) Apply initial conditions into equations 4 and 5:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle y(0)=1=c_1 $$    (6)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle {y}'(0)=0=c_2+5c_1 $$    (7) Solving equations 6 and 7 yields:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle c_1=1 $$    (8)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$\displaystyle c_2=-5 $$    (9) Plugging the these values back into the general solution (4) yields:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ y=(1-5x)e^{5x} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

title: User:Egm4313.s12.team3.landers


text:
File:Spring fbd.png

Licensing
for school project

title: File:Spring fbd.png


text:
File:Dashpot fbd.png

Licensing
for school project

title: File:Dashpot fbd.png


text:
File:Mass fbd.png

Licensing
for school project

title: File:Mass fbd.png


text:
User:RTennantwood

Unit convenor: Dr Robin Tennant-Wood
I am only on campus, in person, Tuesdays, Wednesdays and Thursdays, on other days I am contactable by email.
 * Room: 6D38
 * Phone: 6201 2700
 * Email: [mailto:robin.tennant-wood@canberra.edu.au robin.tennant-wood@canberra.edu.au]

title: User:RTennantwood


text:
User:Egm4313.s12.team8.koester

/R1.5.2/

/R2.9/

/R3.5/

/R4.4.2/

/R5.5/

/R6.3/

/R7.4/

title: User:Egm4313.s12.team8.koester


text:
User:Egm4313.s12.team8.koester/R1.5.2

Problem 5
Problem found on page 59, problem number 5

$$ \displaystyle 5. \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

The characteristic equation that corresponds to this kind of problem is:

$$ \displaystyle \; {y}''+a {y}'+b y=0 $$

So for this problem:

$$ \displaystyle \;  a=2\pi  $$  and  $$ \displaystyle  \;   b=\pi ^{2} $$

Substiting into the equation we have a term $$ \displaystyle a^{2}-4b $$ that tells us the case number we are dealing with

$$ \displaystyle \;  a^{2}-4b= (2\pi)^{2}-4(\pi ^{2})= 4\pi ^{2}- 4\pi ^{2} =0 $$

Since this term equals 0, we have Case II and therefore a real double root.

Therefore:

$$ \displaystyle \;  \lambda_1= -a/2= - 2\pi/2 = -\pi $$

The first solution is:

$$ \displaystyle \;  y_1= e ^{\lambda_1x}= e ^{-\pi x} $$

You can obtain the second solution by using the method of reduction of order

$$ \displaystyle \; y_2= uy_1  $$

$$ \displaystyle \; {y_2}'= {u}'y_1 + u{y_1}'  $$

$$ \displaystyle \; {y_2}= {u}y_1 +2{u}'{y_1}'+ u{y_1}''  $$

Substituting these derivatives into the original characteristic equation we get:

$$ \displaystyle \; ({u}y_1 +2{u}'{y_1}'+ u{y_1}) + a({u}'y_1 + u{y_1}') + buy_1 =0  $$

Collecting $$ \displaystyle u, {u}', {u}''  $$ terms together you obtain:

$$ \displaystyle \; {u}y_1 + {u}'(2{y_1}'+ay_1) + u({y_1}+a{y_1}'+by_1)=0  $$

The second and third terms will disappear to leave you with:

$$ \displaystyle \; {u}y_1=0 $$ and therefore $$ \displaystyle  \; {u}=0  $$

Integrate twice to get the expressions:

$$ \displaystyle \; {u}'=c_1  $$

$$ \displaystyle \; u=c_1x+c_2  $$

We can now simply choose for $$ \displaystyle \; c_1=1 $$ and $$ \displaystyle  \; c_2=0 $$

So now:

$$ \displaystyle \; u=x  $$ and therefore $$ \displaystyle  \;y_2=xy_1   $$

Therefore:

$$ \displaystyle \; y_2= xe^ {-\pi x}  $$

And the General Solution becomes:

$$ \displaystyle \; y=(c_1+c_2x)e^ {-\pi x}   $$

Solved and typed by: Egm4313.s12.team8.koester 02:32, 31 January 2012 (UTC)

Checked by:

title: User:Egm4313.s12.team8.koester/R1.5.2


text:
Caregiving and dementia

Topics

 * Allied health
 * Aboriginal and Torres Strait Islander (ATSI)
 * Behavioural and psychological symptoms
 * Culturally and Linguistically Diverse (CALD)
 * Diagnosing dementia
 * Environmental design and assistive technology
 * GPs and medical specialists
 * Medication management
 * Nursing
 * Palliative
 * Person centred care
 * Sexualities and dementia
 * Younger onset dementia

Featured content
Prize winners from the essay contest:

title: Caregiving and dementia


text: File:Problem2 2.png

title: File:Problem2 2.png


text:
Talk:Caregiving and dementia

Comments
This is where we can discuss the project. Jtneill - Talk - c 01:52, 31 January 2012 (UTC)

title: Talk:Caregiving and dementia


text:
Category:User ru-N

Users in this category indicate they have skill level N for language Russian.

title: Category:User ru-N


text:
User:Egm4313.s12.team16.r2

Report 2
By: Team 16

R2.1
Given:

$$ {\lambda _1=-2 }$$

$$ {\lambda _2=5 }$$

$$ {y'(0)=0 }$$

work:

$$ (-_1)(-_2))=0 $$

$$ ^2-_1-_2+_1_2=0 $$

plug in given values

$$ ^2-3-10=0 $$

Non homogeneous

$${ r(x)=^2-3-10 }$$

Homogeneous solution

$$ {y_n(x)=C_1e^{-2x}+C_2e^{5x}} $$

overall solution

$${ y(x)=C_1e^{-2x}+C_2e^{5x}+y_p(x) }$$

its given that this is a case of no excitation therefore

{ r(x)=0}

$$ { y_p(x)=0 } $$

$${ y(0)=1=C_1+C_2}$$

$$ { y'(0)=0=-2C_1+5C_2} $$

therefore C1&C2 are

$$ C_1=5/7 $$

$$C_2=2/7 $$

part B

$$ {\lambda _1=-2 }$$

$$ {\lambda _2=5 }$$

$$ {y'(0)=0 }$$

$$ m=2 $$

$$ m(-_1)(-_2))=0 $$

$$ m(^2-_1-_2+_1_2)=0 $$

plug in given values

$$ 2^2-6-20=0 $$

Non homogeneous

$$ {\lambda _1=-2 }$$

$$ {\lambda _2=5 }$$

$$ {y'(0)=0 }$$

$$ m=3 $$

$$ m(-_1)(-_2))=0 $$

$$ m(^2-_1-_2+_1_2)=0 $$

plug in given values

$$ 3^2-9-30=0 $$

Non homogeneous

Author: Nyal Jennings

Reviewed by: Brandon Wright

R2.2
For this problem the initial conditions for this equation:

$$ y''+10y'+25=r(x)$$

were given as:

y(0)=1 and y'(0)=0 converting into terms of $$ \lambda $$ this becomes:

$$ \lambda^2+10\lambda+25=0 $$

Factoring this equation becomes $$ (\lambda+5)^2 $$

This means that the solution $$ \lambda = -5 $$ is a real repeated solution.

Because the solution is a double root the general equation is

$$ y_g= C_1e^{-5x}+C_2xe^{-5x} $$

Using the initial conditions y(0)=1 and y'(0)=0 we can solve for $$ C_1, C_2 $$

Plugging x=0,y=1 into the general solution gives

$$ C_1e^{-5(0)}+C_2(0)e^{-5(0)}= 1 $$

The second term in the equation equals 0 making $$ C_1= 1 $$

Next we have to take the derivative of the general equation to solve for $$C_2$$

The derivative of the first term in the equation is simply: $$ -5C_1e^{-5x}$$

Where as the second term derivative is found using the rule of derivatives of two multiplied terms

Using this rule the second term becomes: $$ -5C_2xe^{-5x}+C_2e^{-5x}$$

Making the whole derivative: $$ y'= -5C_1e^{-5x}-5C_2xe^{-5x}+C_2e^{-5x} =0 $$

Now plugging x=0, y=0 and $$ C_1=1 $$ into the derivative we get: $$ -5(1)e^{-5(0)}-5C_2(0)e^{-5(0)}+C_2e^{-5(0)} =0 $$

Solving this equation for $$C_2$$ it is found that $$ C_2=5$$

The final solution to this ODE is

Author: Brandon Wright

Reviewed by: Ryan McDaniel

R2.3
Find the general solution of the ODEs.

R2.3a
Given:

$$ y''+6y'+8.96y=0 \! $$, "Which has the general form of $ y''+ay'+by=0 \! $" Solve for $$ a^2-4b \! $$ where $$ a= 6 \! $$ and $$ b= 8.96 \! $$: $$ 6^2-4(8.96) = 0.16 > 0 \! $$ "Therefore, this is a Case I - 2 distinct roots: $\lambda _1 \! $ and $\lambda _2 \! $"

"with a general solution of $ y=C_1e^{\lambda _1x}+ C_2e^{\lambda _2x} $"

$$ \lambda _1 = \frac{1}{2}\left [ -a+\sqrt{a^2-4b} \right ] $$ "$= \frac{1}{2}\left [ -6+\sqrt{0.16} \right ] = -2.8 $" $$ \lambda _2 = \frac{1}{2}\left [ -a-\sqrt{a^2-4b} \right ]$$ "$= \frac{1}{2}\left [ -6-\sqrt{0.16} \right ] = -3.2 $" Substituting into the general solution where $$ \lambda_1= -2.8 \! $$ and $$ \lambda_2= -3.2 \! $$:

R2.3b
Given:

$$ y''+4y'+(\pi ^2+4)y=0 \! $$,

"Which has a general form of $ y''+ay'+by=0 \! $" Solve for $$ a^2-4b \! $$ where $$ a= 4\! $$ and $$ b=\pi^2+4 \! $$: $$ 4^2-4(\pi^2+4) = -39.4784 < 0 \! $$ "Therefore, this is a Case III - complex conjugate roots: $ \lambda= -\frac{1}{2}a \pm i\omega $"

"with a general solution of $ y=e^{-ax/2}\left ( A\cos(\omega x) + B\sin (\omega x) \right )\! $"

Also, $$ \omega^2=b-\frac{1}{4}a^2 \! $$

"$ =(\pi^2+4)-\frac{1}{4}(4^2) \! $"

"$ \omega ^2= \pi^2 \! $"

"$ \omega = \pi \! $"

Substituting into the general solution where $$ a= 4 \! $$ and $$ \omega = \pi \! $$:

Author: Erin Standish Reviewed by:Brandon Wright

R2.4
Find the general solution of the ODE.

R2.4a
Given:

$$ y''+2 \pi y'+\pi^2y=0 \! $$

"which has the general form of $ y''+ay'+by=0 \! $." Solve for $$ a^2-4b \! $$ where $$a=2 \pi \! $$ and $$ b=\pi \! $$: $$ (2\pi)^2-4(\pi)^2 = 0 = 0 \! $$ "Since 0 = 0, this is a Case II - real double root: $\lambda = -a/2 \!$"

"with a general solution of $ y=(C_1+C_2x)e^{-ax/2} \!$." Substituting into the general solution where $$ a= 2 \pi \!$$ :

R2.4b
Given:

$$ 10y''-32y'+25.6y=0 \! $$

"We must get into the form $y''+ay'+by=0 \! $ by diving through by 10:" $$ y''-3.2y'+2.56y=0 \! $$ Solve for $$ a^2-4b \! $$ where $$ a= 3.2 \! $$ and $$ b= 2.56 \! $$: $$ 3.2^2-4(2.56) = 0 = 0 \! $$ "Since 0 = 0, this is a Case II - real double root: $ \lambda = -a/2 \! $"

"with a general solution of $ y=(C_1+C_2x)e^{-ax/2} \! $." Substituting into the general solution where $$ a = 3.2 \! $$:

Author: Erin Standish Reviewed by: Nyal Jennings

R2.5
Find an ODE with form $$ y''+ay'+by=0 \! $$ for the given basis.

R2.5a
Given: $$ e^{2.6x} \! $$, $$ e^{-4.3x} \! $$

This basis implies that we are dealing with an ODE of Case I - 2 distinct roots: "$ \lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b})=2.6$ and    $\lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b})=-4.3$" "Also, $ a^2-4b>0 \! $."

The given basis of an ODE of Case I is: $$e^{\lambda_1x}$$ and $$e^{\lambda_2x}$$.

Substituting $$ c = \sqrt{a^2-4b} $$ yields: "$ \lambda_1=\frac{1}{2}(-a+c)=2.6$ and    $\lambda_2=\frac{1}{2}(-a-c)=-4.3 $"

This yields two equations & two unknowns:

$$ \lambda_1=\frac{1}{2}(-a+c)=2.6$$......(1)

$$\lambda_2=\frac{1}{2}(-a-c)=-4.3$$.....(2)

Solving for $$ a \! $$ by adding (1) to (2) yields: "$ (5.2=-a+c) \! $" "$+(-8.6=-a-c) \! $" "$ -3.4=-2a \! $" "$ 1.7=a \! $"

Substituting $$ a \! $$ into (1) and solving for $$ c \! $$ yields: "$ \frac{1}{2}(-1.7+c)=2.6 $" "$ 5.2=-1.7+c \! $" "$ 6.9=c \! $"

Substituting back into $$ c = \sqrt{a^2-4b} $$ and solving for $$ b \! $$ yields: "$6.9=\sqrt{1.7^2-4b}$" "$47.61=2.89-4b \! $" "$ -11.18=b \! $"

Substituting into the general form:

R2.5b
Given: $$e^{ \sqrt{5} x} $$, $$ xe^{ -\sqrt{5} x} $$

This basis implies that we are dealing with an ODE of Case II - real double root: "$ \lambda =-\frac{1}{2}a $." "Also, $ a^2-4b=0 \! $."

The given basis of an ODE of Case II is: $$e^{\frac{-ax}{2}}$$.

Solving for a: "$e^{\frac{-ax}{2}}=e^{-\sqrt{5}x} $|undefined" "$+\frac{a}{2}=+ \sqrt{5} $" "$a=2\sqrt{5}$"

Solving for b using $$ a^2-4b=0 \! $$: "$(2\sqrt{5})^2-4b=0$" "$-4b=-20 \! $" "$b=5 \! $"

Substituting into the general form:

Author: Erin Standish Reviewed by: Martin Dolgiej

R2.6
Given

From (3) p.1-5 the equation of motion of the spring-mass-dashpot system

$$ my_k'' + \frac{mk}{c}y_k' + ky =f(t)...(1a) $$

Solution

Using the double real root λ=-3 the characteristic equation can be developed

$$ (\lambda+3)^2=\lambda^2 + 6\lambda + 9 = 0 \! $$

Relating this to the equation of motion in (1a)

$$ m=1 \! $$

$$ k=9 \! $$

$$ c=\frac{3}{2} $$

Author:James Welch

Reviewed by: Erin Standish

R2.7
Equation for Maclaurin Series

$$ f(x) = \sum_{n=0}^\infty a_0 (x - x_0)^n $$

where

$$ a_0 = \frac{1}{n!}f^{(n)}(x_0) \! $$

and

$$ x_0 = 0 \! $$

PART 1

$$ f(t) = e^t = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$

$$ f(t) = f(0) + (1)f^{(1)}(0)t + \frac{1}{2}f^{(2)}(0)t^{(2)} + \frac{1}{6}f^{(3)}(0)t^{(3)} \! $$

PART 2

$$ f(t) = cos(t) = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$

$$ f(t) = f(0) + (1)f^{(1)}(0)t + \frac{1}{2}f^{(2)}(0)t^{(2)} + \frac{1}{6}f^{(3)}(0)t^{(3)} + \frac{1}{24}f^{(4)}(0)t^{(4)} \! $$

PART 3

$$ f(t) = sin(t) = \sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)(t)^n \! $$

$$ f(t) = f(0) + (1)f^{(1)}(0)t + \frac{1}{2}f^{(2)}(0)t^{(2)} + \frac{1}{6}f^{(3)}(0)t^{(3)} + \frac{1}{24}f^{(4)}(0)t^{(4)} \! $$

Author:Ryan McDaniel

Reviewed By: Martin Dolgiej

R2.8
Page 59

8. Put the equation in general form $$y''+y'+3.25y=0 \! $$ We can start with $$y = e^{st} \! $$ and from that derive $$y' = se^{st \! }$$ and $$y'' = s^{2}e^{st} \! $$ where $$s = i\omega \! $$. Replacing into the original equation, $$s^{2}e^{st}+se^{st}+3.25e^{st}=0 \! $$ and factoring out the exponential term, $$e^{st}(s^{2}+s+3.25)=0 \! $$ Since $$e^{st}\neq 0 $$, we have our characteristic equation as $$s^2+s+3.25=0 \! $$ using the quadratic formula to solve, the roots are found at $$ \frac{1}{2}\pm i\sqrt{3} \! $$ and substituting into standard form, the equation becomes $$y = e^{\frac{-x}{2}}(Acos\sqrt3 x+Bsin\sqrt3 x)\! $$ Page 59 15. Put the equation in general form $$y''+0.54y'+(0.0729+\pi)y=0 \! $$ We can start with $$y = e^{st} \! $$ and from that derive $$y' = se^{st} \! $$ and $$y'' = s^{2}e^{st} \! $$ where $$s = i\omega \! $$. Replacing into the original equation, $$s^{2}e^{st}+0.54se^{st}+(0.0729+\pi)e^{st}=0 \! $$ and factoring out the exponential term, $$e^{st}(s^{2}+0.54s+(0.0729+\pi))=0 \! $$ Since $$e^{st}\neq 0$$, we have our characteristic equation as $$s^2+0.54s+(0.0729+\pi)=0 \! $$ using the quadratic formula to solve, the roots are found at $$-0.27\pm \sqrt{\pi}i \! $$ and substituting into standard form, the equation becomes $$y = e^{-0.27x}(Acos\sqrt{\pi} x+Bsin\sqrt{\pi} x)\! $$ Author: Matt Hilsenrath Reviewed by: Martin Dolgiej

Given:
This is the characteristic equation from the ODE: Given the following initial conditions: $$ y(0) = 1, y'(0) = 0 \! $$ $$ r(x) = 0 \! $$

Solution:
Solving the characteristic equation (Eq. 1) for $$ \lambda \! $$: $$ \lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \! $$ $$ \lambda = \frac{-4 \pm \sqrt{4^2-4(1)(13)}}{2(1)} \! $$ $$ \lambda = \frac{-4 \pm 6i}{2} \! $$ $$ \lambda = -2 \pm 3i \! $$ Therefore, the solution will take the form of: $$ y(x) = e^{-ax}(C_1\cos({\omega x}) + C_2\sin({\omega x})) \! $$ Where the a and $$ \omega $$ are: $$ a = -2, \omega = 3 \! $$ Substituting the values and taking the derivative of $$ y(x) $$: $$ y(x) = e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ $$ y(x)' = e^{-2x}(-3C_1\sin({3x}) + 3C_2\cos({3x})) - 2e^{-2x}(C_1\cos({3x}) + C_2\sin({3x})) \! $$ Using the initial condition states above and solving for $$ C_1, C_2 $$: $$ y(0)= 1 = e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$ y(0)' = 0 = e^{-2(0)}(-3C_1\sin({3(0)}) + 3C_2\cos({3(0)})) - 2e^{-2(0)}(C_1\cos({3(0)}) + C_2\sin({3(0)})) \! $$ $$ C_1 = 1, C_2 = \frac{2}{3} $$ Final solution: Equations for Graph: Author: Martin Dolgiej Reviewed by: Ryan McDaniel

title: User:Egm4313.s12.team16.r2


text:
User:Egm4313.s12.team8.jmurillo

Problem Statement
'''Kreyszig 2011 Pg. 59 Problem 4''' Find a general solution for the following ODE.

Given: $$\displaystyle y''+4y'+(\pi^2+4)y=0 $$

Solution
The characteristic equation for the given ODE is  $$\displaystyle \lambda^2+\lambda+(\pi^2+4)=0 $$

The roots of the characteristic equation are complex roots in the form $$\displaystyle \lambda=-\frac{1}{2}a\pm i \omega $$

The roots of the characteristic equation are

$$\displaystyle \lambda=\frac{1}{2}(-4\pm 2 \pi i) $$

Therefore this ODE falls under Case III which has a general solution in the form

$$\displaystyle y=e^\frac{-ax}{2} (Acos \omega x+Bsin \omega x) $$

The general solution to the ODE is

$$\displaystyle y= e^{-2x} (Acos\pi x+Bsin\pi x) $$

title: User:Egm4313.s12.team8.jmurillo


text:
User:Egm4313.s12.team18.burgher.wb

Swoop!

title: User:Egm4313.s12.team18.burgher.wb


text:
User:Egm4313.s12.team18.gibson.mg

=Michelle Gibson=

Useful Links
Team 18 Course Wiki Course Website Report Table

Problem
Derive the equation of motion of the spring-mass-dashpot in fig. 5.3 in K 2011 p 85 with an applied force r(t) on the ball.

Solution
Find: The equation of motion with the applied force, r(t).

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

$$  \displaystyle {F_c = cy'} $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {F_k = ky} $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Using Newton's Second Law:
 * {| style="width:100%" border="0"

$$  \displaystyle {\Sigma\vec{F}=m\vec{a}} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {\vec{a}=y''} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From the Free Body Diagram above:

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

$$  \displaystyle -ma=F_c+F_k-r(t) $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Or

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

$$  \displaystyle r(t)=F_c+F_k+ma $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Plug Equation 1 and Equation 2 into Equation 3.

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

$$  \displaystyle {r(t)=cy'+ky+my} $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Or

 * {| style="width:100%" border="2"

$$  \displaystyle {r(t)=my''+cy'+ky} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Problem
K 2011 p. 60 problems 8 and 15. These problems give you an equation and ask for the general solution. Once the general solution has been found, use substitution to check the answer.

Problem 8) $$\displaystyle y^{''}+y^{'}+3.25y=0 $$

Problem 15) $$\displaystyle y^{''}+0.54y^{'}+(0.0729+\pi)y=0 $$

Solution
Problem 8)
 * $$ \displaystyle y^{''}+y^{'}+3.25y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+\lambda+3.25 =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:

 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=1^{2}-4(3.25)=-12 $$

Or
 * $$\displaystyle a^{2}-4b=-12 $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:

 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{3.25-\frac{1}{4}(1)^{2}}) \rightarrow \omega=\sqrt{3} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{3}$$
 * $$ \displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

Checking the solution by using substitution:

 * $$\displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

 * $$\displaystyle y^{'}=\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{-2}+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{'}=-\frac{1}{2}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$

 * $$\displaystyle y^{''}=-\frac{e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))}{4}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}-\frac{e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))}{2}+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Or
 * $$\displaystyle y^{''}=-\frac{1}{4}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$

 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$

 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-x/2}Bsin(\sqrt{3}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.

Problem 15)
 * $$ \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y =0 $$

Convert the given equation to the characteristic equation
 * $$ \displaystyle \lambda^{2}+0.54\lambda+(0.0729+\pi ) =0 $$

The equation becomes:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Using the quadric formula, solve for the roots:

 * $$ \displaystyle \lambda = \frac{1}{2}(-a+\sqrt{a^{2}-4b}) $$
 * $$ \displaystyle \lambda = \frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

Since the quadratic formula involves a ± function, $$ \displaystyle \lambda$$ will be equal to 2 values.

With the values given in the problem statement, the discriminant value will be a negative number. Thus, the roots will be complex conjugates.
 * $$\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi)=-4\pi $$

Or
 * $$\displaystyle a^{2}-4b=-4\pi $$

The solution form must be:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Here, $$ \displaystyle \omega$$=:

 * $$ \displaystyle \omega = (\sqrt{b-\frac{1}{4}{a^{2}}}) $$
 * $$ \displaystyle \omega = (\sqrt{(0.0729-\pi)-\frac{1}{4}(0.54)^{2}}) =\sqrt{\pi} \rightarrow \omega=\sqrt{\pi} $$

This leads us to the solution.
 * $$ \displaystyle y=e^{-x/2}(Acos(\omega x)+Bsin(\omega x)) $$

Substituting in $$ \displaystyle \omega=\sqrt{\pi}$$ $$ \displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

Checking the solution by using substitution:

 * $$\displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

 * $$\displaystyle y^{'}=-0.27{e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))}+e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)) $$

 * $$\displaystyle y^{''}=0.0729e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))+e^{-0.27x}(-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)) $$

Using the original values from the equation given in the problem statement and doing all possible term combination:

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$

 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$

 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$

 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

You are then left with:
 * $$\displaystyle [0]e^{-0.27}Bsin(\sqrt{\pi}x) $$

Since the term inside the bracket is zero, we know we the equation equals zero and that we solved the problem correctly.

title: User:Egm4313.s12.team18.gibson.mg


text:
University of Florida/Egm4313/s12.team11.perez.gp/R1.4

Problem: Derive (3) and (4) from (2).

Given:

$$ V = LC\frac{dV^2_c}{dt^2} + RC\frac{dV_c}{dt} + V_c $$ (2)

$$ V' = LI'' + RI' + \frac{1}{C}I $$                     (3)

$$ V = LQ'' + RQ' + \frac{1}{C}Q $$                      (4)

Solution:

First, let's solve for (3).

Recall that:

$$ Q = CV_c = \int Idt $$,

and

$$ I = C\frac{dV_c}{dt} = \frac{dQ}{dt} $$.

We can use this information to replace the differentiating terms accordingly.

After doing so, we get:

$$ V = L{I}' + RI + V_c \! $$

but knowing that $$ I = \frac{dQ}{dt} $$, we can rearrange the terms to get $$ V_c = \frac{1}{C}\int Idt $$.

Using this information in the previously derived equation, we find that:

$$ V = LI' + RI + \frac{1}{C}\int Idt $$.

Finally, after differentiating $$ V $$ with respect to $$ t $$, we get:

$$ V' = LI'' + RI' + \frac{1}{C}I $$.

Now, let's solve for (4).

Once again considering that $$ Q = CV_c \! $$, we can solve for (4) by differentiating

twice and then plugging it into (2).

Deriving twice, we find that:

$$ Q = CV_c \! $$

$$ Q' = C\frac{dV_c}{dt} $$

$$ Q'' = C\frac{dV^2_c}{dt^2} $$

After plugging this into (2), we see that:

$$ V = LQ'' + RQ' + V_c \! $$.

Once we rearrange $$ Q = CV_c \! $$, we find that $$ V_c = \frac{Q}{C} $$.

We can plug this in to the above equation to get the solution:

$$ V = LQ'' + RQ' + \frac{1}{C}Q $$.


text:
User:Egm4313.s12.team8/R1

=R1.1 - Equation of Motion for a Parallel Spring-Dashpot System=

Problem Statement

Derive the equation of motion of a spring-dashpot system in parallel, with a mass and applied force $$f(t)$$.

Solution
Solved on my own

Kinematics

 * $$\displaystyle\begin{align}

y = y_k = y_c \end{align}$$

Kinetics

 * $$\displaystyle m y'' + f_k + f_c = f(t)$$

Constitutive Relations

 * $$\displaystyle\begin{align}

f_k &= k y_k \\ f_c &= c y_c' \end{align}$$

Therefore
$$\displaystyle m y''(t) + c y'(t) + k y(t) = f(t)$$

The general solution of the above ODE is:

 * $$\displaystyle\begin{align}

y_g (t) &= C_1 \; e^{\frac{-c - \sqrt{c^2 - 4 m k}}{2 m} t} \\ &+ C_2 \; e^{\frac{-c + \sqrt{c^2 - 4 m k}}{2 m} t} \\ &- \frac{\int e^{\frac{\left( c + \sqrt{c^2 + 4 m k} \right) t}{2 m}} f(t) \,dt}{\sqrt{c^2 - 4 m k}} \; e^{\frac{-c - \sqrt{c^2 - 4 m k}}{2 m} t} \\ &+ \frac{\int e^{\frac{\left( c - \sqrt{c^2 + 4 m k} \right) t}{2 m}} f(t) \,dt}{\sqrt{c^2 - 4 m k}} \; e^{\frac{-c + \sqrt{c^2 - 4 m k}}{2 m} t} \end{align}$$

=R1.2 - Equation of Motion for a Series Spring-Dashpot System=

Problem Statement

Derive the equation of the spring-mass-dashpot in Fig.53 (shown above) in K 2011 p.85, with an applied force $$\displaystyle r(t) $$ on the ball.

Kinematics
$$\displaystyle y=y_{c}=y_{k} $$ (2-0)

Kinetics
$$\displaystyle r(t)=m y''+f_{c}+f_{k} $$ (2-1)

Where
$$\displaystyle f_{k}=k y_{k}$$ (2-2) and $$\displaystyle f_{c}=c y_{c}$$ (2-3) Plugging into (2-1) gives us: $$\displaystyle r(t)=my''+cy_{c}+ky_{k}$$ (2-4) Using (2-0) and substituting into (2-4), we find that the final equation of motion for the spring-mash-dashpot is: $$\displaystyle r(t)=my''+cy'+ky $$ (2-5)

=R1.3 - Draw FBD of System on p.1-4=

Problem Statement:
For the spring-dashpot-mass system on p.1-4, draw the FBDs and derive the equation of motion (2) on p.1-4.

Solution:
Break the system into its three components: spring, dashpot, and mass.

Derivation of Equation of Motion
Looking at the FBD of the mass: $$f_I$$ represents the internal forces in the system from both the spring and dashpot. $$my''$$ represents the inertial force of the mass. $$f(t)$$ represents the excitation on the spring-dashpot-mass system. Sum the forces on this FBD and set equal to 0: $$ \sum f_y=-my''-f_{I}+f(t)=0$$ Rearrange to get the desired equation of motion: $$my''+f_{I}=f(t)$$

=R1.4 - Derive RLC Circuit Equations from p.2-2=

Problem Statement
Derive (i) Equation 3 ($$\displaystyle LI+RI'+\frac{1}{C}I=V'$$) and (ii) Equation 4 ($$\displaystyle LQ+RQ'+\frac{1}{C}Q=V$$) using Equation 2 ($$\displaystyle V=LC\frac{\mathrm{d^2v_{c}} }{\mathrm{d} t^2}+RC\frac{\mathrm{d} v_{c}}{\mathrm{d} t}+v_c$$) on page 2-2 in the lecture notes.

Part i
Equation 2:

Equation 1:

Equation 1 can be rewritten as Eqn 1-b:

Differentiating Eqn 1-b yields:

Differentiating Eqn 1-c gives:

Then we differentiate Equation 2 to get:

Then we subsitute Eqn. 1-b, Eqn. 1-c, and Eqn. 1-d into Equation 2-b to get:

By cancelling like terms we derive Equation 3:

Which can be rewritten as:

Part ii
To derive Equation 4, we again start with Equation 2:

Then we will use another version of Equation 1 on p. 2-2 in the lecture notes.

Then rearranging the variables, we get:

Differentiating this yields:

And differentiating again yields:

Substituting these derivatives into Equation 2 gives:

Cancelling like terms gives Equation 4:

=R1.5.1 - Find the General Solution to Kreyszig Pg. 59 Problem 4=

Problem Statement
'''Kreyszig 2011 Pg. 59 Problem 4''' Find a general solution for the following ODE.

Given: $$\displaystyle y''+4y'+(\pi^2+4)y=0 $$

Solution
The characteristic equation for the given ODE is  $$\displaystyle \lambda^2+\lambda+(\pi^2+4)=0 $$

The roots of the characteristic equation are complex roots in the form $$\displaystyle \lambda=-\frac{1}{2}a\pm i \omega $$

The roots of the characteristic equation are

$$\displaystyle \lambda=\frac{1}{2}(-4\pm 2 \pi i) $$

Therefore this ODE falls under Case III which has a general solution in the form

$$\displaystyle y=e^\frac{-ax}{2} (Acos \omega x+Bsin \omega x) $$

The general solution to the ODE is

$$\displaystyle y= e^{-2x} (Acos\pi x+Bsin\pi x) $$

=R1.5.2 - Find the General Solution to Kreyszig Pg. 59 Problem 5=

Problem Statement
Problem found on page 59, problem number 5

$$ \displaystyle 5. \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

Solution
The characteristic equation that corresponds to this kind of problem is:

$$ \displaystyle \; {y}''+a {y}'+b y=0 $$

So for this problem:

$$ \displaystyle \;  a=2\pi  $$  and  $$ \displaystyle  \;   b=\pi ^{2} $$

Substiting into the equation we have a term $$ \displaystyle a^{2}-4b $$ that tells us the case number we are dealing with

$$ \displaystyle \;  a^{2}-4b= (2\pi)^{2}-4(\pi ^{2})= 4\pi ^{2}- 4\pi ^{2} =0 $$

Since this term equals 0, we have Case II and therefore a real double root.

Therefore:

$$ \displaystyle \;  \lambda_1= -a/2= - 2\pi/2 = -\pi $$

The first solution is:

$$ \displaystyle \;  y_1= e ^{\lambda_1x}= e ^{-\pi x} $$

You can obtain the second solution by using the method of reduction of order

$$ \displaystyle \; y_2= uy_1  $$

$$ \displaystyle \; {y_2}'= {u}'y_1 + u{y_1}'  $$

$$ \displaystyle \; {y_2}= {u}y_1 +2{u}'{y_1}'+ u{y_1}''  $$

Substituting these derivatives into the original characteristic equation we get:

$$ \displaystyle \; ({u}y_1 +2{u}'{y_1}'+ u{y_1}) + a({u}'y_1 + u{y_1}') + buy_1 =0  $$

Collecting $$ \displaystyle u, {u}', {u}''  $$ terms together you obtain:

$$ \displaystyle \; {u}y_1 + {u}'(2{y_1}'+ay_1) + u({y_1}+a{y_1}'+by_1)=0  $$

The second and third terms will disappear to leave you with:

$$ \displaystyle \; {u}y_1=0 $$ and therefore $$ \displaystyle  \; {u}=0  $$

Integrate twice to get the expressions:

$$ \displaystyle \; {u}'=c_1  $$

$$ \displaystyle \; u=c_1x+c_2  $$

We can now simply choose for $$ \displaystyle \; c_1=1 $$ and $$ \displaystyle  \; c_2=0 $$

So now:

$$ \displaystyle \; u=x  $$ and therefore $$ \displaystyle  \;y_2=xy_1   $$

Therefore:

$$ \displaystyle \; y_2= xe^ {-\pi x}  $$

And the General Solution becomes:

$$ \displaystyle \; y=(c_1+c_2x)e^ {-\pi x}   $$

=R1.6 - Determine Properties of ODEs=

Problem Statement
For each ODE in Fig.2 in K2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Falling Stone (Sec. 1.1)
$$ \displaystyle y''=g=const $$

Parachutist (Sec. 1.2)
$$ \displaystyle mv'=mg-bv^2 $$

Out-flowing Water (Sec.1.3)
$$ \displaystyle h'=-k\sqrt{h} $$

Vibrating Mass on a Spring (Sec. 2.4,2.8)
$$ \displaystyle my''+ky=0 $$

Beats of a vibrating system (Sec. 2.8)
$$ \displaystyle y''+w_0^2y= \cos wt$$

Current I in RLC circuit (Sec. 2.9)
$$ \displaystyle LI''+RI'+ \frac 1 c I=E' $$

Deformation of a Beam (Sec. 3.3)
$$ \displaystyle EIy^{iv} = f(x) $$

Pendulum (Sec. 4.5)
$$ \displaystyle L \theta''+g \sin \theta = 0 $$

Falling Stone (Sec. 1.1)
Order: 2nd Linearity: Linear Superposition: Yes The Homogeneous Solution, $$ \displaystyle y_h''=0 $$ plus the Particular Solution, $$ \displaystyle y_p''=g $$ yields $$ \displaystyle y_h+y_p=g $$ which is the same as $$ \displaystyle (y_h+y_p)= \bar y=g $$. Therefore, the Superposition Principle can be applied.

Parachutist (Sec. 1.2)
Order: 1st Linearity: Nonlinear Superposition: No

Rewriting the original equation with $$ \displaystyle \mathit{v}$$ on one side yields $$ \displaystyle mv'+bv^2=mg $$

The Homogeneous Solution is $$ \displaystyle mv_h'+bv_h^2=0 $$.

The Particular Solution is $$ \displaystyle mv_p'+bv_p^2=mg $$. Summing both solutions yields $$ \displaystyle mv_h'+bv_h^2 + mv_p'+bv_p^2 = m(v_h+v_p)' + b(v_h^2+v_p^2) = mg $$.

$$ \bar v = v_h+v_p $$, however, $$ v_h^2+v_p^2 \neq (v_h+v_p)^2 $$

Thus, the Superposition Principle CANNOT be applied.

Out-flowing Water (Sec.1.3)
Order: 1st Linearity: Nonlinear Superposition: No

The Homogeneous Solution $$ \displaystyle h_h'=-k\sqrt{h_h} $$

plus the Particular Solution $$ \displaystyle h_p'=-k\sqrt{h_p} $$

yields $$ \displaystyle h_h'+h_p'=-k\sqrt{h_h}-k\sqrt{h_p}$$

or $$ \displaystyle (h_h+h_p)'= \bar h' = -k(h_h^{1/2}+h_p^{1/2})$$.

Yet, $$ \displaystyle -k(h_h^{1/2}+h_p^{1/2}) \neq -k(h_h+h_p)^{1/2}$$.

Therefore, The Superposition Principle CANNOT be applied.

Vibrating Mass on a Spring (Sec. 2.4,2.8)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution $$ \displaystyle my_h''+ky_h=0 $$

plus the Particular Solution $$ \displaystyle my_p''+ky_p=0 $$

yields $$ \displaystyle m(y_h+y_p)''+k(y_h+y_p)=0 $$.

or $$ \displaystyle m \bar y''+k \bar y=0 $$.

Therefore, the Superposition Principle applies.

Beats of a vibrating system (Sec. 2.8)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution is $$ \displaystyle y_h''+w_0^2y_h=0 $$

The Particular Solution is $$ \displaystyle y_p''+w_0^2y_p= \cos wt $$ Summing both solutions yields $$ \displaystyle (y_h+y_p)'' + w_0^2(y_h+y_p) = \cos wt $$

We know $$ \displaystyle \bar y = y_h+y_p $$,

so $$ \displaystyle \bar y'' + w_0^2 \bar y = \cos wt $$

Therefore, the Superposition Principle can be applied.

Current I in RLC circuit (Sec. 2.9)
Order: 2nd Linearity: Linear Superposition: Yes

The Homogeneous Solution is $$ \displaystyle LI_h''+RI_h'+ \frac 1 c I_h=0 $$

The Particular Solution is $$ \displaystyle LI_p''+RI_p'+ \frac 1 c I_p=E' $$ Summing both solutions yields $$ \displaystyle L(I_h+I_p)''+R(I_h+I_p)'+ \frac 1 c (I_h+I_p)=E' $$

We know $$ \displaystyle \bar I = I_h+I_p $$,

so $$ \displaystyle L \bar I''+R \bar I'+ \frac 1 c \bar I=E' $$

Therefore, the Superposition Principle can be applied.

Deformation of a Beam (Sec. 3.3)
Order: 4th Linearity: Linear Superposition: Yes

The Homogeneous Solution $$ \displaystyle EIy_h^{iv} = 0 $$

plus the Particular Solution $$ \displaystyle EIy_p^{iv} = f(x) $$

yields $$ \displaystyle EI(y_h^{iv}+y_p^{iv}) = EI(y_h+y_p)^{iv} = f(x) $$.

Knowing $$\bar y = y_h+y_p$$, we have

$$ \displaystyle EI(\bar y)^{iv} = f(x) $$.

Thus, the Superposition Principle can be applied.

Pendulum (Sec. 4.5)
Order: 2nd Linearity: Nonlinear Superposition: No

The Homogeneous Solution $$ \displaystyle L \theta_h''+g \sin \theta_h = 0 $$

plus the Particular Solution $$ \displaystyle L \theta_p''+g \sin \theta_p = 0 $$

yields $$ \displaystyle L ( \theta_h+ \theta_p)+g{( \sin \theta_h+ \sin \theta_p)} = L ( \theta_h+ \theta_p)''+g{( \sin \theta_h+ \sin \theta_p)} = 0 $$

But, $$ g{( \sin \theta_h+ \sin \theta_p)} \neq g \sin ( \theta_h + \theta_p) $$

Therefore, the Superposition Principle CANNOT be applied.

=Team Contributions=

title: User:Egm4313.s12.team8/R1


text:
User talk:RTennantwood

 Hello RTennantwood, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 05:22, 31 January 2012 (UTC)

title: User talk:RTennantwood


text:
User:Egm4313.s12.team18.windham

=Alissa Windham=

Useful Links
Team 18 Course Wiki Report Table LaTeX Equation Editor

Problem 1.6
For each ODE in Fig. 2 in K2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition applies.

Solution

Tower of Pisa

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

$$ \displaystyle { y{}''=g } $$ (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

Order: Second Linear Principle of Superposition applies:

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

$$ \displaystyle { y_{h}{}''=0} $$ (1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle { y_{p}{}''=g} $$ (1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle {(y_{h}+y_{p})=\bar{y}} $$ (1.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle { (y_{h}+y_{p}){}''=g} $$ (1.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle {\mathbf{\bar{ y}{}''=g}} $$ Is the same as $$ \displaystyle { \mathbf{y{}''=g}} $$ (1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

Parachutist

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

$$ \displaystyle {mv{}'+bv^{2}=mg} $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

Order: First Non-Linear Principle of Superposition does not apply:

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

$$ \displaystyle { mv_{h}{}'+bv_{h}^{2}=0} $$ (2.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle { mv_{p}{}'+bv_{p}^{2}=mg} $$ (2.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle {(v_{h}+v_{p})=\bar{v}} $$ (2.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle { m(v_{h}+v_{p}){}'+b(v_{h}^{2}+v_{p}^{2})=mg} $$ (2.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle {\mathbf{ m\bar{v}{}'+b(v_{h}^{2}+v_{p}^{2})=mg} } $$ Is not the same as $$ \displaystyle { \mathbf{mv{}'+bv^{2}=mg}} $$ (2.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

Out-flowing Water

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

$$ \displaystyle {h{}'=-k\sqrt{h}} $$ (3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

Order: First Non-Linear Principle of Superposition does not apply:

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

$$ \displaystyle { h_{h}{}'+k\sqrt{h_{h}}=0} $$ (3.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle { h_{p}{}'+k\sqrt{h_{p}}=0} $$ (3.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle {(h_{h}+h_{p})=\bar{h}} $$ (3.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$ \displaystyle { (h_{h}+h_{p}){}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0} $$ (3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{\bar{h}{}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0 }} $$ Is not the same as $$ \displaystyle {\mathbf{ h{}'+k\sqrt{h}=0}} $$ (3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Vibrating Mass on a String

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

$$ \displaystyle {my{}''+ky=0} $$ (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Order:2nd Linear Principle of Superposition applies:

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

$$ \displaystyle {my_{h}{}''+ky_{h}=0} $$ (4.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {my_{p}{}''+ky_{p}=0} $$ (4.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (4.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {m(y_{p}+y_{h}){}''+k(y_{p}+y_{h})=0} $$ (4.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$  \displaystyle {\mathbf{m\bar{y}''+k\bar{y}=0 }} $$ is the same as $$ \displaystyle { \mathbf{my{}''+ky=0}} $$     (4.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Beats of a Vibrating Sytem

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

$$ \displaystyle {y{}''+\omega _{0}^{2}y=\cos( \omega t)} $$ (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Order:2nd Linear Principle of Superposition applies:

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

$$ \displaystyle {y_{h}{}''+\omega _{0}^{2}y_{h}=0} $$ (5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {y_{p}{}''+\omega _{0}^{2}y_{p}=\cos( \omega t)} $$ (5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(y_{p}+y_{h}){}''+\omega _{0}^{2}(y_{p}+y_{h})=\cos( \omega t)} $$ (5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{\bar{y}{}''+\omega _{0}^{2}\bar{y}=\cos( \omega t)}} $$ Is the same as $$ \displaystyle {\mathbf{ y{}''+\omega _{0}^{2}y=\cos( \omega t)}} $$ (5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Current I in RLC Circuit

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

$$ \displaystyle {LI{}''+RI{}'+\frac{1}{c}I=E{}'} $$ (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Order:2nd Linear Principle of Superposition applies:

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

$$ \displaystyle {LI_{h}{}''+RI_{h}{}'+\frac{1}{c}I_{h}=0} $$ (6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {LI_{p}{}''+RI_{p}{}'+\frac{1}{c}I_{p}=E{}'} $$ (6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(I_{p}+I_{h})=\bar{I}} $$ (6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {L(I_{p}+I_{h}){}''+R(I_{p}+I_{h}){}'+\frac{1}{c}(I_{p}+I_{h})=E{}'} $$ (6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{L\bar{I}{}''+R\bar{I}{}'+\frac{1}{c}\bar{I}=E{}' }} $$ is the same as $$ \displaystyle {\mathbf{ LI{}''+RI{}'+\frac{1}{c}I=E{}'}} $$ (6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Deformation of a Beam

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

$$ \displaystyle {EIy^{iv}=f(x)} $$ (7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Order:Fourth Linear Principle of Superposition applies:

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

$$ \displaystyle {EIy_{h}^{iv}=0} $$ (7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {EIy_{p}^{iv}=f(x)} $$ (7.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (7.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {EI(y_{h}+y_{p})^{iv}=f(x)} $$ (7.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{EI\bar{y}^{iv}=f(x)} } $$ Is the same as $$ \displaystyle {\mathbf{ EIy^{iv}=f(x)}} $$ (7.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Pendulum

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

$$ \displaystyle {L\Theta {}''+gsin(\Theta )=0} $$ (8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Order:Second Non-Linear Principle of Superposition does not apply:

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

$$ \displaystyle {L\Theta_{h} {}''+gsin(\Theta_{h} )=0} $$ (8.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {L\Theta_{p} {}''+gsin(\Theta_{p} )=0} $$ (8.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {(\Theta_{p}+\Theta_{h})=\bar{\Theta}} $$ (8.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {L(\Theta_{p}+\Theta_{h}){}''+gsin(\Theta_{p}+\Theta_{h})=0} $$ (8.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\mathbf{L\bar{\Theta} {}+gsin(\Theta_{p}+\Theta_{h} )=0}} $$ Is not the same as $$ \mathbf{L\Theta {}+gsin(\Theta )=0} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Problem 2.3
Solve problems 3 and 4 on page 59 in the textbook, Kreyszig 2011. "Find a general solution for the ODE and check by substitution."

3.$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$

4.$$ \displaystyle {y{}''+4y{}'+\left ( \pi ^{2}+4 \right )y=0} $$

Solution
Background information: For homogeneous linear ODE's with constant coefficients, a general equation can be described as, $$ \displaystyle {y{}''+ay{}'+by=0.} $$

The function $$ \displaystyle {y=e^{\lambda x}} $$ may be tried as a solution for the general equation above.

After substituting this equation and its derivatives: $$ \displaystyle {y{}'=\lambda e^{\lambda x}} $$ and $$ \displaystyle {y{}''=\lambda ^{2}e^{\lambda x}} $$ into the general equation,the result is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)e^{\lambda x}=0} $$

and the characteristic equation is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)=0} $$ which means that if $$ \displaystyle {\lambda} $$ is a solution of this characteristic equation, then the exponential function is a solution of the general equation. If we solve this characteristic equation for $$ \displaystyle {\lambda} $$, the result is$$ \displaystyle {\lambda _{1,2}=\frac{1}{2}(-a\pm \sqrt{a^{^{2}}-4b})} $$

Therefore there are 3 cases when it comes to finding the solution to Homogeneous Linear ODE's with CC:

Case 1:    Two real roots if  $$ \displaystyle {a^{^{2}}-4b>0} $$

Case 2:    A real double root if $$\displaystyle {a^{^{2}}-4b=0} $$

Case 3: Complex conjugate roots if $$\displaystyle {a^{^{2}}-4b=0} $$

Problem 3 page 59 :
 * {| style="width:100%" border="0"

$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$ (1) The discriminant for this equation (1) is:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle { \sqrt{6^{^{2}}-4(8.96)}=.16} $$ which is >0 and therefore a case 1 (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {\lambda _{1}=-2.8, \lambda _{2}=-3.2} $$ and $$ \displaystyle {y _{1}=e^{-2.8x}, y _{2}=e^{-3.2x}} $$ (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

The general solution to (1) is

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

$$ \displaystyle {y=c_{1}e^{-2.8x}+c_{2}e^{^{-3.2x}}} $$ (4) Check this solution by plugging in
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {y _{1}=e^{-2.8x}} $$ and $$ \displaystyle {y _{2}=e^{-3.2x}} $$ (5) into (1). The results for plugging in the above equations(5) and their derivatives:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {y_{1}=e^{-2.8x},y_{1}{}'=-2.8e^{-2.8x},y_{1}{}''=7.84e^{-2.8x}} $$ and $$ \displaystyle {y_{2}=e^{-3.2x},y_{2}{}'=-3.2e^{-3.2x},y_{2}{}''=10.24e^{-3.2x}} $$ (6) into equation (1) respectively are:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {e^{-2.8x}(7.84-(6)(2.8)+8.96)=0} $$ and $$ \displaystyle {e^{-3.2x}(10.24-(6)(3.2)+8.96)=0} $$
 * style="width:95%" |
 * style="width:95%" |

(7)
 *

 * }
 * }

Both of these reduce to
 * {| style="width:100%" border="0"

$$ \displaystyle {0=0} $$ (8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Problem 4 page 59:
 * {| style="width:100%" border="0"

$$ \displaystyle {y{}''+4y{}'+\left ( \pi ^{2}+4 \right )y=0} $$ (9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

The discriminant for this equation is:

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

$$ \displaystyle {4^{2}-4(\pi ^{2}+4)=-4\pi^{2}} $$ (10) which is a case 3 and solutions to the characteristic equation are:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {\lambda _{1}=-\frac{1}{2}a+i\omega, \lambda _{2}=-\frac{1}{2}a-i\omega} $$ (11) and our particular solutions for characteristic equation for this problem are:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {\lambda _{1}=-\frac{1}{2}(4)+i\pi, \lambda _{2}=-\frac{1}{2}4-i\pi} $$ (12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

The general solution to (9) is

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

$$ \displaystyle {y=e^{-\frac{4}{2}x}(A\cos \pi x+Bsin\pi x)} $$ (13) Check this solution by plugging in
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {y _{1}=e^{-\frac{4}{2}x}\cos \pi x} $$ and $$ \displaystyle {y _{2}=e^{-\frac{4}{2}x}\sin \pi x} $$ (14) into (1). The results for plugging in the equations (14) and their derivatives:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {y_{1}=e^{-2x}\cos \pi x, y_{1}{}'=-\pi e^{-2x}-2e^{-2x}\cos \pi x, y_{1}{}''=e^{-2x}((4-\pi ^{2})\cos \pi x+4\pi \sin \pi x)} $$ (15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {y_{2}=e^{-2x}\sin \pi x, y_{2}{}'=e^{-2x}(\pi \cos \pi x-2\sin \pi x), y_{2}{}''=e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)} $$ (16)  into equation (1) respectively are:
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {e^{-2x}((4-\pi ^{2})\cos \pi x+4\sin \pi x+4(-\pi \sin \pi x-2\cos \pi x)+(\pi ^{2}+4\cos \pi x)=0} $$ (17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$ \displaystyle {e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)+4e^{-2x}(\pi \cos \pi x-2\sin \pi x)+(\pi ^{2}+4)e^{-2x}\sin \pi x=0} $$ (18) Both of these reduce to
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {0=0} $$ (19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Solution
Background information:

Basic Rule- if r(x) in $$ \displaystyle {y{}''+ay{}'+by=r(x)} $$ is one of the functions in the first column in table 2.1, choose $$ \displaystyle {y_{p}} $$ in the same line and determine its undetermined coefficients by substituting $$ \displaystyle {y_{p}} $$ and its derivatives into the above equation.

title: User:Egm4313.s12.team18.windham


text: File:Spring-mass-dashpot.jpg

title: File:Spring-mass-dashpot.jpg


text: Category:Caregiving and dementia

title: Category:Caregiving and dementia


title: File:CORDIC.Background.4.A.20120131.pdf


text:
Talk:Technical writing/UI Design

Hi,

I am an MA Student studying technical writing and elearning. I am very interested in the User Interface Design section. I think the next big challenge facing interfacing is to design/ redesign for the iphone and ipad type product. The screens are so small it will be difficult to rewrite existing material to suit. Aslo older people with poor sight (like myself) may feel left behind. --Angelaryan23 14:35, 31 January 2012 (UTC)

title: Talk:Technical writing/UI Design


text:
File:BallFBD.png

Licensing
for school project

title: File:BallFBD.png


text:
File:SpringFBD.png

Licensing
for school project

title: File:SpringFBD.png


text:
File:DashpotFBD.png

Licensing
for school project

title: File:DashpotFBD.png


text:
User talk:Egm4313.s12.Team15.j.shugart

Gerald (Jerry) Shugart

Important Links

______________________

Lecture Notes

Almost WYSIWYG Manual

Google Course Site

Report Grading Criteria

Good Model Reports

Report Table

Writing Tools

Team 15

title: User talk:Egm4313.s12.Team15.j.shugart