# University of Florida/Egm4313/s12.team14.report2

## Problem 1

### Find

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

Consider no excitation:

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

Plot the solution.

### Solution

Characteristic equation:

 ${\displaystyle \displaystyle (\lambda -\lambda _{1})(\lambda -\lambda _{2})=0}$ ${\displaystyle \displaystyle (Eq.3)}$

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

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

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

Non-homogeneous solution:

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

Homogeneous solution:

 ${\displaystyle \displaystyle y_{h}(x)=C_{1}e^{-2x}+C_{2}e^{5x}}$ ${\displaystyle \displaystyle (Eq.7)}$

Overall solution:

 ${\displaystyle \displaystyle y(x)=C_{1}e^{-2x}+C_{2}e^{5x}+y_{p}(x)}$ ${\displaystyle \displaystyle (Eq.8)}$

No excitation:

 ${\displaystyle \displaystyle r(x)=0=>y_{p}(x)=0}$ ${\displaystyle \displaystyle (Eq.9)}$

From intitial conditions:

 ${\displaystyle \displaystyle y(0)=1=C_{1}+C_{2}}$ ${\displaystyle \displaystyle (Eq.10)}$

 ${\displaystyle \displaystyle y'(0)=0=-2C_{1}+5C_{2}}$ ${\displaystyle \displaystyle (Eq.11)}$

Solving for ${\displaystyle \displaystyle C_{1}}$ and ${\displaystyle \displaystyle C_{2}}$:

 ${\displaystyle \displaystyle C_{1}={\frac {5}{7}}}$ $\displaystyle \displaystyle$

and

 ${\displaystyle \displaystyle C_{2}={\frac {2}{7}}}$ $\displaystyle \displaystyle$

So, the final solution is:

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


## Problem 2

### Find

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

### Solution

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

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

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

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

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

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

 ${\displaystyle \displaystyle \lambda _{2}=\lambda _{1}=\lambda =5}$ $\displaystyle \displaystyle$

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

 ${\displaystyle \displaystyle y=C_{1}e^{-5x}+C_{2}xe^{-5x}}$ ${\displaystyle \displaystyle (Eq.7)}$

From intitial conditions:

 ${\displaystyle \displaystyle y(0)=1=C_{1}}$ ${\displaystyle \displaystyle (Eq.8)}$

 ${\displaystyle \displaystyle y'(0)=0=5C_{1}+C_{2}}$ ${\displaystyle \displaystyle (Eq.9)}$

Solving for ${\displaystyle \displaystyle C_{1}}$ and ${\displaystyle \displaystyle C_{2}}$:

 ${\displaystyle \displaystyle C_{1}=1}$ $\displaystyle \displaystyle$

and

 ${\displaystyle \displaystyle C_{2}=-5}$ $\displaystyle \displaystyle$

So, the final solution is:

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


## Problem 3

### Given

(a)

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

(b)

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

### Find

General solution to the differential equation

### Solution

(a)

Characteristic equation:

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

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

where:

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

${\displaystyle \displaystyle {x={\frac {-6\pm {\sqrt {6^{2}-4*1*8.96}}}{2*1}}\,}}$
${\displaystyle \displaystyle {x={\frac {-6\pm {\sqrt {36-35.84}}}{2}}\,}}$
${\displaystyle \displaystyle {x={\frac {-6\pm 0.4}{2}}\,}}$
${\displaystyle \displaystyle {x={\frac {-6.4}{2}}\,,{\frac {-5.6}{2}}}}$
${\displaystyle \displaystyle {x=-3.2,-2.8}}$
${\displaystyle \displaystyle {y=c_{1}e^{-3.2x}+c_{2}e^{-2.8x}}}$


(b)

Characteristic equation:

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

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

where:

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

${\displaystyle \displaystyle {\lambda ={\frac {-4\pm {\sqrt {4^{2}-4*1*(\pi ^{2}+4)}}}{2*1}}\,}}$
${\displaystyle \displaystyle {\lambda ={\frac {-4\pm {\sqrt {16-4\pi ^{2}-16)}}}{2}}\,}}$
${\displaystyle \displaystyle {\lambda ={\frac {-4\pm 2\pi i}{2}}\,}}$
${\displaystyle \displaystyle {\lambda =-2\pm \pi i}}$
${\displaystyle \displaystyle {y=e^{-2x}(C_{1}\cos \pi x+C_{2}\sin \pi x)}}$


## Problem 4

### Given

(5)

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

(6)

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

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

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

This yields:

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

Therefore, we have:

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

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

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

We can see that

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

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

${\displaystyle \displaystyle \triangle =a^{2}-4b}$
${\displaystyle \displaystyle \triangle =(2\pi ^{2})^{2}-4(\pi ^{2})}$
${\displaystyle \displaystyle \triangle =0}$
Therefore, the equation is critically-damped.


The general solution is therefore represented by:

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

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

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

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

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

Therefore, the general solution to the equation is:

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


Verification by substitution:

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

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

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

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

(6)

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

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

So then, we have

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

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

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

Examining the discriminant, we find that

${\displaystyle \displaystyle \triangle =a^{2}-4b=(-3.2)^{2}-4(2.56)=0}$
Therefore, the equation is critically-damped.


The general solution is represented by:

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

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

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

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

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

Therefore, the general solution to the equation is:

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


Verification by substitution:

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

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

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

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

## Problem 5

### Given

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

Given the basis:

(a)

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

(b)

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

### Find

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

### Solution

(a)The general solution can be written as:

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

The characteristic equation can be written in the following way:

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

Giving the ODE:

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

(b) The general solution can be written as:

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

The characteristic equation can be written in the following way:

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

Giving the ODE:

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

## Problem 6

### Given

Mass Spring Dashpot system FBD's

Spring-dashpot equation of motion from sec 1-5

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

### Find

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

### Solution

Consider the double real root

 ${\displaystyle \displaystyle {\lambda =-3}}$

Characteristic equation is

 ${\displaystyle \displaystyle {(\lambda +3)^{2}=\lambda ^{2}+6\lambda +9=0}}$ ${\displaystyle \displaystyle (Eq.1)}$

Recall

 ${\displaystyle \displaystyle {my{_{k}}''+m{\frac {k}{c}}y{_{k}}'+ky_{k}=f(t)}}$ ${\displaystyle \displaystyle (Eq.2)}$

Thus

 ${\displaystyle \displaystyle {m=1}}$
 ${\displaystyle \displaystyle {m{\frac {k}{c}}=6}}$
 ${\displaystyle \displaystyle {k=9}}$

Solve for c

 ${\displaystyle \displaystyle {c={\frac {9}{6}}={\frac {3}{2}}}}$

Therefore

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


## Problem 7

### Given

Taylor Series at t=0

### Find

Develop the MacLaurin Series for ${\displaystyle e^{t},\cos t,\sin t\!}$

### Solution

Taylor series is defined as

 ${\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(t)}{n!}}\,(x-t)^{n}}$ ${\displaystyle \displaystyle (Eq.1)}$

The MacLaurin series occurs when t=0

 ${\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}\,(x)^{n}}$ ${\displaystyle \displaystyle (Eq.2)}$

Development of MacLaurin series for ${\displaystyle e^{t}}$

 ${\displaystyle e^{t}={\frac {1}{0!}}+{\frac {1}{1!}}t+{\frac {1}{2!}}t^{2}+{\frac {1}{3!}}t^{3}...}$ ${\displaystyle \displaystyle (Eq.3)}$

${\displaystyle e^{t}=1+x+{\frac {1}{2}}t^{2}+{\frac {1}{6}}t^{3}...}$

Explicit form can be written as

${\displaystyle e^{t}=\sum _{n=0}^{\infty }{\frac {1}{n!}}\,(t)^{n}}$


Development of MacLaurin series for ${\displaystyle cos(t)}$

 ${\displaystyle cos(t)={\frac {1}{0!}}+{\frac {0}{1!}}t-{\frac {1}{2!}}t^{2}+{\frac {0}{3!}}t^{3}+{\frac {1}{4!}}t^{4}...}$ ${\displaystyle \displaystyle (Eq.4)}$

${\displaystyle cos(t)=1-{\frac {1}{2}}t^{2}+{\frac {1}{24}}t^{4}...}$

Explicit form can be written as

${\displaystyle cos(t)=\sum _{n=0}^{\infty }{\frac {-1}{2n!}}\,(t)^{2n}}$


Development of MacLaurin series for ${\displaystyle sin(t)}$

 ${\displaystyle sin(t)={\frac {0}{0!}}+{\frac {1}{1!}}t-{\frac {0}{2!}}t^{2}-{\frac {1}{3!}}t^{3}+{\frac {0}{4!}}t^{4}+{\frac {1}{5!}}t^{5}...}$ ${\displaystyle \displaystyle (Eq.5)}$

${\displaystyle sin(t)=1t-{\frac {1}{6}}t^{3}+{\frac {1}{120}}t^{5}...}$

Explicit form can be written as

${\displaystyle sin(t)=\sum _{n=0}^{\infty }{\frac {-1}{(2n+1)!}}\,(t)^{(2n+1)}}$


## Problem 8

### Given

(8)

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

(15)

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

### Solution

(8)

Assume that the solution is of the form

 ${\displaystyle \displaystyle y_{h}=e^{\lambda x}}$ ${\displaystyle \displaystyle (Eq.2)}$

 ${\displaystyle \displaystyle \therefore y'_{h}=\lambda e^{\lambda x}}$ ${\displaystyle \displaystyle (Eq.3)}$

And

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

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

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

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

${\displaystyle \displaystyle y_{h}={C_{1}}e^{(-0.5+j{\sqrt {3}})x}+{C_{2}}e^{(-0.5-j{\sqrt {3}})x}}$

In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let ${\displaystyle \displaystyle a=-0.5,b={\sqrt {3}}}$
${\displaystyle \displaystyle y_{h}={C_{1}}e^{(a+jb)x}+{C_{2}}e^{(a-jb)x}}$
${\displaystyle \displaystyle y_{h}={C_{1}}e^{ax}e^{jbx}+{C_{2}}e^{ax}e^{-jbx}}$
${\displaystyle \displaystyle {y_{h}}'={C_{1}}[e^{ax}({jb}e^{jbx})+e^{jbx}({a}e^{ax})]+{C_{2}}[e^{ax}(-{jb}e^{jbx})+e^{jbx}({a}e^{ax})]}$
${\displaystyle \displaystyle {y_{h}}''={C_{1}}[e^{ax}((jb)^{2}e^{jbx})+({jb}e^{jbx})({a}e^{ax})+e^{jbx}((a)^{2}e^{ax})+({a}e^{ax})({jb}e^{jbx})]+{C_{2}}[e^{ax}((jb)^{2}e^{-jbx})+(-{jb}e^{-jbx})({a}e^{ax})+e^{-jbx}((a)^{2}e^{ax})+({a}e^{ax})(-{jb}e^{-jbx})]}$

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

Therefore the solution holds for any values of ${\displaystyle \displaystyle C_{1},C_{2},x}$

(15)

Assume that the solution is of the form

 ${\displaystyle \displaystyle y_{h}=e^{\lambda x}}$ ${\displaystyle \displaystyle (Eq.2)}$

 ${\displaystyle \displaystyle \therefore y'_{h}=\lambda e^{\lambda x}}$ ${\displaystyle \displaystyle (Eq.3)}$

And

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

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

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

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

${\displaystyle \displaystyle y_{h}={C_{1}}e^{(-0.27+j(1.772))x}+{C_{2}}e^{(-0.27-j(1.772))x}}$

In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation
Let ${\displaystyle \displaystyle a=-0.27,b=1.772}$
${\displaystyle \displaystyle y_{h}={C_{1}}e^{(a+jb)x}+{C_{2}}e^{(a-jb)x}}$
${\displaystyle \displaystyle y_{h}={C_{1}}e^{ax}e^{jbx}+{C_{2}}e^{ax}e^{-jbx}}$
${\displaystyle \displaystyle {y_{h}}'={C_{1}}[e^{ax}({jb}e^{jbx})+e^{jbx}({a}e^{ax})]+{C_{2}}[e^{ax}(-{jb}e^{jbx})+e^{jbx}({a}e^{ax})]}$
${\displaystyle \displaystyle {y_{h}}''={C_{1}}[e^{ax}((jb)^{2}e^{jbx})+({jb}e^{jbx})({a}e^{ax})+e^{jbx}((a)^{2}e^{ax})+({a}e^{ax})({jb}e^{jbx})]+{C_{2}}[e^{ax}((jb)^{2}e^{-jbx})+(-{jb}e^{-jbx})({a}e^{ax})+e^{-jbx}((a)^{2}e^{ax})+({a}e^{ax})(-{jb}e^{-jbx})]}$

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

Therefore the solution holds for any values of ${\displaystyle \displaystyle C_{1},C_{2},x}$

## Problem 10

### Given

Differential equation, initial conditions, and forcing function as shown:

 ${\displaystyle \displaystyle y''-10y'+25y=r(x),}$ ${\displaystyle \displaystyle (Eq.1)}$
 ${\displaystyle \displaystyle y(0)=4,y'(0)=-5,r(x)=7e^{(}5x)-2x^{2}.}$ ${\displaystyle \displaystyle (Eq.1)}$

### Find

The solution to Eq. 1

### Solution

 ${\displaystyle \displaystyle y_{p}=C_{1}x^{2}e^{5x}+C_{2}x^{2}+C_{3}x+C_{4}}$ ${\displaystyle \displaystyle (Eq.2)}$
 ${\displaystyle \displaystyle y_{p}'=5C_{1}x^{2}e^{5x}+2C_{1}xe^{5x}+2C_{2}x+C_{3}}$ ${\displaystyle \displaystyle (Eq.3)}$
 ${\displaystyle \displaystyle y_{p}''=25C_{1}x^{2}e^{5x}+20C_{1}xe^{5x}+2C_{1}e^{5x}+2C_{2}}$ ${\displaystyle \displaystyle (Eq.4)}$

Substituting Equations 2, 3, and 4 into Equation 1 yields

 ${\displaystyle \displaystyle 7e^{5x}-2x^{2}=2C_{1}e^{5x}+25C_{2}x^{2}+25C_{3}x-20C_{2}x+2C_{2}-10C_{3}+25C_{4}}$ ${\displaystyle \displaystyle (Eq.5)}$

To solve for the constants:

 ${\displaystyle \displaystyle 7=2C_{1}}$ ${\displaystyle \displaystyle (Eq.6)}$
 ${\displaystyle \displaystyle -2=25C_{2}}$ ${\displaystyle \displaystyle (Eq.7)}$
 ${\displaystyle \displaystyle 0=25C_{3}-20C_{2}}$ ${\displaystyle \displaystyle (Eq.8)}$
 ${\displaystyle \displaystyle 0=2C_{2}-10C_{3}+25C_{4}}$ ${\displaystyle \displaystyle (Eq.9)}$

Solving Equations 6-9 yields

 ${\displaystyle \displaystyle C_{1}={\frac {7}{2}},C_{2}=-{\frac {2}{25}},C_{3}=-{\frac {8}{125}},C_{4}=-{\frac {12}{625}}}$ $\displaystyle \displaystyle$

The homogeneous solution:

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

The total solution:

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

Setting ${\displaystyle \displaystyle x=0}$ yields

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

Solving for ${\displaystyle \displaystyle C_{5}}$ yields

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

Finding ${\displaystyle \displaystyle y'}$ yields

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

Setting ${\displaystyle \displaystyle x=0}$ yields

 ${\displaystyle \displaystyle y'(0)={\frac {2512}{125}}+C_{6}-{\frac {8}{125}}=-5}$ ${\displaystyle \displaystyle (Eq.15)}$

Solving for ${\displaystyle \displaystyle C_{6}}$ yields

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

Therefore, the total solution is

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

Name Responsibilities Checked by
Bo Turano Problem R2. --
David Parsons Problem R2.4 --
Dean Pickett Problem R2.8 --
Giacomo Savardi Problem R2. --
Isaac Kimiagarov Problem R2.3 --
Kyle Steiner Problem R2. --
Tony Han Problem R2.6, R2.7 --