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
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

From intitial conditions:


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

y(0)=1=C_1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }


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

y'(0)=0=5C_1+C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

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


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

C_1=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

and


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

C_2=-5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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"


 * $$\displaystyle {y}'' + 2\pi {y}' + {\pi}^2y = 0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1) $$
 * }
 * }
 * }

(6)


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


 * $$\displaystyle {10{y}'' - 32{y}' + 25.6y = 0}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

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
 * $$\displaystyle{(\lambda+3)^{2}=\lambda^{2}+6\lambda+9=0}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1) $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Thus
 * $$\displaystyle{my{_{k}}''+m\frac{k}{c}y{_{k}}'+ky_{k}=f(t)}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }
 * {| 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
 * $$ \sum_{n=0} ^ {\infin } \frac {f^{(n)}(t)}{n!} \, (x-t)^{n}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 1) $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

Development of MacLaurin series for $$e^t$$
 * $$ \sum_{n=0} ^ {\infin } \frac {f^{(n)}(0)}{n!} \, (x)^{n}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }
 * {| 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 $$ e^t=\sum_{n=0} ^ {\infin } \frac {1}{n!} \, (t)^{n}$$
 * $$e^t=\frac{1}{0!}+\frac{1}{1!}t+\frac{1}{2!}t^2+\frac{1}{3!}t^3...$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3) $$
 * }
 * }
 * }

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 $$ cos(t)=\sum_{n=0} ^ {\infin } \frac {-1}{2n!} \, (t)^{2n}$$
 * $$cos(t)=\frac{1}{0!}+\frac{0}{1!}t-\frac{1}{2!}t^2+\frac{0}{3!}t^3+\frac{1}{4!}t^4...$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4) $$
 * }
 * }
 * }

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 $$ sin(t)=\sum_{n=0} ^ {\infin } \frac {-1}{(2n+1)!} \, (t)^{(2n+1)}$$
 * $$sin(t)=\frac{0}{0!}+\frac{1}{1!}t-\frac{0}{2!}t^2-\frac{1}{3!}t^3+\frac{0}{4!}t^4+\frac{1}{5!}t^5...$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5) $$
 * }
 * }
 * }

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

{y}'' + y' + 3.25y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1a)
 * }
 * }

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

{y}'' + (0.54)y' + (0.0729 + \pi)y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }


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

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }


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

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }


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

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }


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

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }


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

y(0) = 4, y'(0) = -5, r(x) = 7e^(5x)-2x^2. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

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

7 = 2C_1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }


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

-2 = 25C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }


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

0 = 25C_3 - 20C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }


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

0 = 2C_2 - 10C_3 + 25C_4 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

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

y_h = C_{5}e^{5x} + C_{6}xe^{5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }

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

C_{5} = \frac{2512}{625} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (Eq. 15)
 * }
 * }

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

C_{6} = -\frac{3129}{125} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\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
 * <p style="text-align:right;">$$\displaystyle (ANS Eq. 17)
 * }
 * }

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