User:Egm6321.f10.team03/Hwk5

=Problem 1: Higher Order Derivatives =

Given
Given a fourth order differential equation in the form
 * {| style="width:100%" border="0"

$$ {y}_{xxxx}= exp \left(-4t \right) \left( {y}_{tttt}-6{y}_{ttt}+11{y}_{tt}-6{y}_{t} \right) $$
 * $$\displaystyle (Eq. 1.1) $$
 * }
 * }

Find
Find the fifth order derivative, $${y}_{xxxxx}$$

Solution
Equation 1.1 was found by assuming $$x=exp\left(t\right) $$, therefore the derivative $$ \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$, where $$ \frac{dx}{dt}= exp\left(t\right) $$. Rearranging gives us
 * {| style="width:100%" border="0"

$$ \frac{dy}{dx}=\frac{dt}{dx}\frac{dy}{dt}$$ and
 * $$\displaystyle (Eq. 1.2) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{d}{dx} {y}_{xxxx}=\frac{dt}{dx}\frac{d}{dt} {y}_{xxxx} $$ Substituting $$ \frac{dx}{dt}= exp\left(t\right) $$ into equation 1.3 yields
 * $$\displaystyle (Eq. 1.3) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{d}{dx} {y}_{xxxx}=exp \left(-t \right)\frac{d}{dt} {y}_{xxxx} $$ Taking the derivative with respect to $$t $$ of equation 1.1 gives
 * $$\displaystyle (Eq. 1.4) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{d}{dt} {y}_{xxxx}=exp \left(-4t \right) \left({y}_{ttttt}-10{y}_{tttt}+35 {y}_{ttt}-50 {y}_{tt}+24 {y}_{t} \right) $$ Plugging equation 1.5 into equation 1.4 gives us the solution:
 * $$\displaystyle (Eq. 1.5) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$  {y}_{xxxxx}=exp \left(-5t \right) \left({y}_{ttttt}-10{y}_{tttt}+35 {y}_{ttt}-50 {y}_{tt}+24 {y}_{t} \right) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\,(Eq. 1.6) $$
 * style= |
 * }

Contributing Members

 * Solved and posted by James Roark 16:51, 27 October 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 15:34, 2 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 16:14, 2 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.franklin 01:57, 3 November 2010 (UTC)

=Problem 2: Solving Euler Equation using method of trial solutions=

Given
Given an ODE of the form
 * {| style="width:100%" border="0"

$$ {x}^{2} \ddot{y}-2x \dot{y}+2y=0 $$ With boundary conditions $$y\left( 1\right)=-2 $$ and $$y \left(2 \right)=5 $$
 * $$\displaystyle (Eq. 2.1) $$
 * }
 * }

Find
Solve equation 2.1 using the method of trial solution.

Solution
Assume a solution of the form $$\,y= {x}^{r}$$, where $$r$$ is the roots of the characteristic equation. Therefore, $$\dot{y}=r{x}^{r-1}$$ and $$\ddot{y}=r(r-1) {x}^{r-2}$$. Plugging these equations into equation 2.1, we obtain the characteristic equation


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

$$\, {r}^{2}-3r+2=0 $$
 * $$\displaystyle (Eq. 2.2) $$
 * }
 * }

Which yields roots of unity and 2. The solution can then be written as a linear combination of each
 * {| style="width:100%" border="0"

$$ y \left(x \right)= {c}_{1}x+ {c}_{2}{x}^{2} $$ Substituting the boundary conditions,
 * $$\displaystyle (Eq. 2.3) $$
 * }
 * }


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

$$\displaystyle {{c}_{1}}+{{c}_{2}}=-2$$
 * }
 * }


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

$$\displaystyle 2{{c}_{1}}+4{{c}_{2}}=5$$
 * }
 * }

Solving the above equations give our constants.


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

$$\displaystyle {{c}_{1}}=-\frac{13}{2}$$
 * }
 * }


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

$$\displaystyle {{c}_{2}}=\frac{9}{2}$$
 * }
 * }

Hence the solution becomes

Plot of solution

 * [[Image:Sudheesh hw5 PLOT1.jpg|center|650px]]

Contributing Members

 * Solved and posted by James Roark 16:51, 27 October 2010 (UTC)
 * Proofread byEgm6321.f10.team03.sigillo 16:01, 2 November 2010 (UTC)
 * Plotted the graph and Proofread solution byEgm6321.f10.team3.Sudheesh 16:37, 2 November 2010 (UTC)

= Problem 3 - Solution of Euler L2-ODE - with Varying Coefficients and Constant Coefficients =

Given

 * Euler $$\displaystyle L2-ODE-VC $$,


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


 * $$\displaystyle {{a}_{2}}{{x}^{2}}{{y}^{''}}+{{a}_{1}}x{{y}^{'}}+{{a}_{0}}y=0$$
 * $$\displaystyle (Eq. 3.1)$$
 * }
 * }
 * }


 * Consider characteristic equation,


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


 * $${{\left( r-\lambda \right)}^{2}}=0$$
 * $$\displaystyle (Eq. 3.2)$$
 * }
 * }
 * }


 * Where $$\displaystyle \lambda =5$$

Find

 * (1) Find $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$ such $$\displaystyle (Eq. 3.2)$$ is a characteristic equation of $$\displaystyle (Eq. 3.1)$$


 * (1.2-1.4) Solve $$\displaystyle (Eq. 3.1)$$


 * (2) Solve $$\displaystyle {{b}_{2}}{{y}^{''}}+{{b}_{1}}{{y}^{'}}+{{b}_{0}}y=0$$

(1.1) Find $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$

 * Assume, $$\displaystyle y={{x}^{r}}$$
 * Taking derivative,


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


 * $$\displaystyle {{y}^{'}}=r{{x}^{r-1}}$$
 * }
 * }
 * }


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


 * $${{y}^{''}}=r\left( r-1 \right){{x}^{r-2}}$$
 * }
 * }
 * }


 * Now plugg in $$\displaystyle y$$ and its derivatives in $$\displaystyle (Eq. 3.1)$$,to obtain characteristic equation:


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


 * $${{a}_{2}}{{x}^{2}}r\left( r-1 \right){{x}^{r-2}}+{{a}_{1}}xr{{x}^{r-1}}+{{a}_{0}}{{x}^{r}}=0$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{a}_{2}}r\left( r-1 \right)+{{a}_{1}}r+{{a}_{0}}=0$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{a}_{2}}{{r}^{2}}+\left( {{a}_{1}}-{{a}_{2}} \right)r+{{a}_{0}}=0$$
 * $$\displaystyle (Eq. 3.3)$$
 * }
 * }
 * }


 * Expanding the given characteristic equation $$\displaystyle (Eq. 3.2)$$:


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


 * $${{\left( r-5 \right)}^{2}}=0$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{r}^{2}}-10r+25=0$$
 * $$\displaystyle (Eq. 3.4)$$
 * }
 * }
 * }


 * Equating $$\displaystyle (Eq. 3.3)$$and $$\displaystyle (Eq. 3.4)$$ give coefficients as:


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

\begin{array}{*{20}l} {{a}_{2}}=1 & {{a}_{1}}=-9 & {{a}_{0}}=25\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Therefore $$\displaystyle Eq. 3.1$$ becomes:


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


 * $$\displaystyle {{x}^{2}}{{y}^{''}}-9x{{y}^{'}}+25y=0$$
 * $$\displaystyle (Eq. 3.5)$$
 * }
 * }
 * }

(1.2) First Homogenous solution

 * First homogenous solution is


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

$$\displaystyle {{y}_{1}}={{x}^{\lambda }}={{x}^{5}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 3.6)$$
 * }
 * }

(1.3) Second Homogenous solution

 * Second homogenous solution is


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


 * $${{y}_{2}}=U\left( x \right){{y}_{1}}=U\left( x \right){{x}^{5}}$$
 * $$\displaystyle (Eq. 3.7)$$
 * }
 * }
 * }


 * Taking derivatives:


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


 * $${{y}_{2}}^{'}={{x}^{5}}{{U}^{'}}\left( x \right)+5{{x}^{4}}U\left( x \right)$$
 * }
 * }
 * }


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


 * $${{y}_{2}}^{}={{U}^{}}\left( x \right){{x}^{5}}+10{{U}^{'}}\left( x \right){{x}^{4}}+20U\left( x \right){{x}^{3}}$$
 * }
 * }
 * }


 * Substitute above three equations in $$\displaystyle Eq. 3.5$$:


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


 * $${{x}^{2}}\left( {{U}^{''}}\left( x \right){{x}^{5}}+10{{U}^{'}}\left( x \right){{x}^{4}}+20U\left( x \right){{x}^{3}} \right)-9x\left( {{x}^{5}}{{U}^{'}}\left( x \right)+5{{x}^{4}}U\left( x \right) \right)+25U\left( x \right){{x}^{5}}=0$$
 * }
 * }
 * }


 * From lecture and definition, all $$\displaystyle U\left( x \right){{x}^{5}}$$ terms drop out because combined equation equal to zero


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


 * $$\Rightarrow {{U}^{''}}{{x}^{7}}+{{U}^{'}}{{x}^{6}}=0$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{U}^{''}}+\frac{x}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.8)$$
 * }
 * }
 * }


 * Let $$\displaystyle {{U}^{'}}=S$$ and then $$\displaystyle (Eq. 3.8)$$ becomes an equation with order reduced by 1:


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


 * $${{S}^{'}}+\frac{S}{x}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.9)$$
 * }
 * }
 * }


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


 * $$\Rightarrow \frac{\partial S}{\partial x}=-\frac{S}{x}$$
 * }
 * }
 * }


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


 * $$\Rightarrow \frac{\partial S}{S}=-\frac{\partial x}{x}$$
 * }
 * }
 * }


 * Integrating the above equation implies:


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


 * $$\displaystyle \ln S=-\ln x+c$$
 * }
 * }
 * }


 * Taking exponetial on both sides results in:


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


 * $$S={{C}_{1}}\frac{1}{x}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.10)$$
 * }
 * }
 * }


 * But,


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


 * $$\displaystyle {{U}^{'}}=S$$
 * }
 * }
 * }


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


 * $$\Rightarrow U=\int{Sdx}+{{C}_{2}}$$
 * }
 * }
 * }


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


 * $$\Rightarrow U=\int{\left( {{C}_{1}}\frac{1}{x} \right)dx}+{{C}_{2}}$$
 * }
 * }
 * }


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


 * $$\Rightarrow U={{C}_{1}}\ln x+{{C}_{2}}$$
 * }
 * }
 * }


 * Substituting the above equation in $$\displaystyle (Eq. 3.7)$$ yields,


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

$${{y}_{2}}=\left( {{C}_{1}}\ln x+{{C}_{2}} \right){{x}^{5}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.11)$$
 * }
 * }

(1.4) General Homogenous solution

 * General homogenous solution is given by


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


 * $$\displaystyle y={{k}_{1}}{{y}_{1}}+{{k}_{2}}{{y}_{2}}$$
 * }
 * }
 * }

Substituting the $$\displaystyle Eq. 3.6$$ and $$\displaystyle Eq. 3.11$$ yields:


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


 * $$y={{k}_{1}}{{x}^{5}}+{{k}_{2}}\left( {{C}_{1}}\ln x+{{C}_{2}} \right){{x}^{5}}$$
 * }
 * }
 * }


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


 * $$y=\left( {{k}_{1}}+{{C}_{2}} \right){{x}^{5}}+{{k}_{2}}\left( {{C}_{1}}\ln x \right){{x}^{5}}$$
 * }
 * }
 * }


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

$$y={{K}_{1}}{{x}^{5}}+{{K}_{2}}\ln x\cdot {{x}^{5}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.12)$$
 * }
 * }

(2) Solve Euler L2-ODE-CC

 * Euler L2-ODE-CC:


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


 * $$\displaystyle {{b}_{2}}{{y}^{''}}+{{b}_{1}}{{y}^{'}}+{{b}_{0}}y=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.13)$$
 * }
 * }
 * }


 * Assume,


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


 * $$\displaystyle y={{e}^{rx}}$$
 * }
 * }
 * }


 * Taking derivative:


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


 * $$\displaystyle {{y}^{'}}=r{{e}^{rx}}$$
 * }
 * }
 * }


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


 * $$\displaystyle {{y}^{''}}={{r}^{2}}{{e}^{rx}}$$
 * }
 * }
 * }

Substituting the above three equations in $$\displaystyle Eq. 3.13$$:


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


 * $$\displaystyle {{b}_{2}}{{r}^{2}}+{{b}_{1}}r+{{b}_{0}}=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.14)$$
 * }
 * }
 * }


 * Comparing $$\displaystyle Eq. 3.14$$ with $$\displaystyle Eq. 3.4$$ give coefficients:


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

\begin{array}{*{20}l} {{b}_{2}}=1 & {{b}_{1}}=-10 & {{b}_{0}}=25\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Now $$\displaystyle Eq. 3.13$$ becomes:


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


 * $$\displaystyle {{y}^{''}}-10{{y}^{'}}+25y=0$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.15)$$
 * }
 * }
 * }


 * First homogenous solution is:


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


 * $$\displaystyle {{y}_{1}}={{e}^{5x}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.16)$$
 * }
 * }
 * }


 * Second homogenous solution is:


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


 * $$\displaystyle {{y}_{2}}=U\left( x \right){{e}^{5x}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.17)$$
 * }
 * }
 * }


 * taking derivative of $$\displaystyle Eq. 3.17$$:


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


 * $$\displaystyle {{y}_{2}}^{'}={{U}^{'}}\left( x \right){{e}^{5x}}+5U\left( x \right){{e}^{5x}}$$
 * }
 * }
 * }


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


 * $$\displaystyle {{y}_{2}}^{}={{U}^{}}\left( x \right){{e}^{5x}}+10{{U}^{'}}\left( x \right){{e}^{5x}}+25U\left( x \right){{e}^{5x}}$$
 * }
 * }
 * }

Plugging the above three equations into $$\displaystyle Eq. 3.15$$ and dropping the terms with $$\displaystyle U\left( x \right)$$ as in the first case, results in:


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


 * $$\displaystyle {{U}^{''}}\left( x \right){{e}^{5x}}=0$$
 * }
 * }
 * }


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


 * $$\displaystyle \Rightarrow {{U}^{''}}\left( x \right)=0$$
 * }
 * }
 * }


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


 * $$\displaystyle \Rightarrow {{U}^{'}}\left( x \right)={{C}_{1}}$$
 * }
 * }
 * }


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


 * $$\displaystyle \Rightarrow U\left( x \right)={{C}_{1}}x+{{C}_{2}}$$
 * }
 * }
 * }


 * Therefore second homogenous solution is;


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


 * $$\displaystyle {{y}_{2}}=\left( {{C}_{1}}x+{{C}_{2}} \right){{e}^{5x}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.18)$$
 * }
 * }
 * }


 * General Homogenous solution


 * General homogenous solution is given by


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


 * $$\displaystyle y={{k}_{1}}{{y}_{1}}+{{k}_{2}}{{y}_{2}}$$
 * }
 * }
 * }


 * Substituting the $$\displaystyle Eq. 3.16$$ and $$\displaystyle Eq. 3.18$$ yields:


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


 * $$y={{k}_{1}}{{e}^{5x}}+{{k}_{2}}\left( {{C}_{1}}x+{{C}_{2}} \right){{e}^{5x}}$$
 * }
 * }
 * }


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


 * $$y=\left( {{k}_{1}}+{{C}_{2}} \right){{e}^{5x}}+{{k}_{2}}{{C}_{1}}x\cdot {{e}^{5x}}$$
 * }
 * }
 * }


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

$$\displaystyle y={{K}_{1}}{{e}^{5x}}+{{K}_{2}}x{{e}^{5x}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.19)$$
 * }
 * }

Contributing members

 * Solved by Egm6321.f10.team03.schulze (Kurt)
 * Posted and rechecked the solution by Egm6321.f10.team3.Sudheesh 04:25, 2 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 17:36, 2 November 2010 (UTC)

=Problem 4: Find a particular solution using the method "Variation of Constants"=

Given
The following equation is given from page 11-2 out of the class notes:
 * {| style="width:100%" border="0"

$$\,{y}'+xy=2x+3$$
 * <p style="text-align:right;">$$\,(Eq. 4.1) $$
 * style= |
 * }

Find
The particular solution $${{y}_{p}}$$ has to be found using the method of variation of constants after having found the homogeneous solution $${{y}_{h}}$$.

Solution
First we find the homogeneous solution to the homogeneous equation out of Eq. 4.1, that is:
 * {| style="width:100%" border="0"

$$\,{y}'+xy=0$$ With the separation of variables we get:
 * <p style="text-align:right;">$$\,(Eq. 4.1) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{1}{y}dy=-x\,dx$$ The result of the integration on both sides of the equation yields to:
 * <p style="text-align:right;">$$\,(Eq. 4.2) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{\frac{1}{y}dy}=\int{-x\,dx+C\Leftrightarrow \log \left( y \right)=\frac{-{{x}^{2}}}{2}+C}$$ hence the homogeneous solution $${{y}_{h}}$$ is: The result of the integration on both sides of the equation yields to:
 * <p style="text-align:right;">$$\,(Eq. 4.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{y}_{h}}={{e}^{\frac{-{{x}^{2}}}{2}}}\cdot A$$ with
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\,(Eq. 4.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,A={{e}^{C}}$$ The next Step is the variation of the constant $$\, A$$ that now becomes a function of the variable $$\, x$$. The solution $$\, y$$ and its derviation can thus be written as:
 * <p style="text-align:right;">$$\,(Eq. 4.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$y=A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}$$
 * <p style="text-align:right;">$$\,(Eq. 4.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'={A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}-A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x$$ Inserting Eq. 4.5 and Eq. 4.6 into Eq. 4.1 the given equation can be written as:
 * <p style="text-align:right;">$$\,(Eq. 4.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}-A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x+A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x=2x+3$$ The second and third term of the left-hand-side cancel out and Eq. 4.7 can be written as:
 * <p style="text-align:right;">$$\,(Eq. 4.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}=2x+3$$ To find the particular solution for $$y$$ the second last step is to find an expression for $$A(x)$$. This can be obtained by integrating $${A}'(x)$$ of Eq. 4.9:
 * <p style="text-align:right;">$$\,(Eq. 4.9) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=\int{\left( 2x+3 \right)}\cdot {{e}^{\frac{2}}}\,dx+k$$
 * <p style="text-align:right;">$$\,(Eq. 4.10) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=3\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,ds}+2\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,\cdot s\,ds}+k$$
 * <p style="text-align:right;">$$\,(Eq. 4.11) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=3\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,ds}+2\cdot {{e}^{\frac{2}}}+k$$
 * <p style="text-align:right;">$$\,(Eq. 4.12) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=3\sqrt{\frac{\pi }{2}}\cdot erfi\left( \frac{x}{\sqrt{2}} \right)+2{{e}^{\frac{2}}}+k$$ Inserting Eq. 4.13 into Eq. 4.6 gives the particular soluton $${{y}_{p}}$$:
 * <p style="text-align:right;">$$\,(Eq. 4.13) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{y}_{p}}=2+3\sqrt{\frac{\pi }{2}}\cdot erfi\left( \frac{x}{\sqrt{2}} \right)+{{e}^{\frac{-{{x}^{2}}}{2}}}+k$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\,(Eq. 4.14) $$
 * style= |
 * }

Contributing Members
Solved and posted byEgm6321.f10.team03.sigillo 01:33, 31 October 2010 (UTC)

= Problem 5 - Solution of Single Degree of Freedom Dynamic System =

Statement

 * Consider the Forced Single Degree of Freedom System shown below:


 * [[Image:Sudheesh hw5.jpg|center|650px]]


 * Equation of Motion of the system is given by


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

$$
 * $${{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y=f\left( t \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.1)
 * }
 * }


 * Where, $$\displaystyle{{a}_{2}}=m$$, $$\displaystyle{{a}_{1}}=c$$ and $$\displaystyle {{a}_{0}}=k$$


 * $$\displaystyle (Eq. 5.1)$$ is a $$\displaystyle Nonhomogenous-L2-ODE-CC$$

(1.1)Find PDEs for Integrating Factor $$\displaystyle h\left( t,y \right)$$

 * It could be proved that the $$\displaystyle Eq.5.1$$ in the given form is not exact. But it can be made exact using Integrating Factor. For that reason multiply $$\displaystyle Eq.5.1$$ by integrating factor $$\displaystyle h\left( t,y \right)$$.


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


 * $$h\left( t,y \right)\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=h\left( t,y \right)f\left( t \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.2.1)$$
 * }
 * }
 * }


 * Consider the LHS of the above equation and which is in the form as shown below:


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


 * $$h\left( t,y \right){{a}_{2}}{{y}^{}}+h\left( t,y \right)\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=f\left( t,y,p \right){{y}^{}}+g\left( t,y,p \right)$$
 * }
 * }
 * }


 * Where,


 * $$\displaystyle f\left( t,y,p \right)=h\left( t,y \right){{a}_{2}}$$
 * $$\displaystyle g\left( t,y,p \right)=h\left( t,y \right)\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)$$
 * $$\displaystyle p={{y}^{'}}$$


 * Now taking partial derivatives,


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

\begin{array}{*{20}l} {{f}_{tt}}={{h}_{tt}}{{a}_{2}} & {{f}_{ty}}={{h}_{ty}}{{a}_{2}}& {{f}_{y}}={{h}_{y}}{{a}_{2}}\\ {{f}_{yy}}={{h}_{yy}}{{a}_{2}} & {{f}_{tp}}=0 & {{f}_{yp}}=0 \\ {{g}_{tp}}={{h}_{t}}{{a}_{1}}& {{g}_{y}}={{h}_{y}}\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)+h{{a}_{0}} \\ {{g}_{yp}}={{h}_{y}}{{a}_{1}} & {{f}_{tp}}=0 & {{g}_{pp}}=0 \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Now consider the two relations of second conditions of exactness [[media:2010_09_23_14_52_54.djvu|Eq.1&2,page3,Mtg 15 (b)]]


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


 * $$\displaystyle {{f}_{tt}}+2p{{f}_{ty}}+{{p}^{2}}{{f}_{yy}}={{g}_{tp}}+p{{g}_{yp}}-{{g}_{y}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.2.2)$$
 * }
 * }
 * }


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


 * $$\displaystyle {{f}_{tp}}+p{{f}_{yp}}+2{{f}_{y}}={{g}_{pp}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.2.3)$$
 * }
 * }
 * }


 * First substitute partial derivatives in $$\displaystyle Eq. 5.2.3$$


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


 * $$\displaystyle 0+0+2{{h}_{y}}{{a}_{2}}=0$$
 * }
 * }
 * }


 * The above condiiton is valid only if $$\displaystyle {{h}_{y}}=0$$ implies $$\displaystyle h$$ is independent of $$\displaystyle y$$. that is


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


 * $$h\left( t,y \right)=h\left( t \right)$$
 * }
 * }
 * }


 * Now substitute partial derivatives in $$\displaystyle Eq. 5.2.2$$


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


 * $${{h}_{tt}}{{a}_{2}}+\underbrace{2p{{h}_{ty}}{{a}_{2}}}_{0}+\underbrace_{0}={{h}_{t}}{{a}_{1}}+\underbrace{p{{h}_{y}}{{a}_{1}}}_{0}-\underbrace{{{h}_{y}}\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)-}_{0}h{{a}_{0}}$$
 * }
 * }
 * }


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

$$\displaystyle {{a}_{2}}{{h}_{tt}}-{{a}_{1}}{{h}_{t}}+{{a}_{0}}h=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.2.4)
 * }
 * }


 * The above equation is the PDE for $$\displaystyle h$$

Given

 * Consider trial solution for Integrating Factor,


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

$$
 * $$\displaystyle h\left( t \right)={{e}^{\alpha t}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.5.3.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.5.3.1)
 * }
 * where, $$\displaystyle \alpha $$, is unknown, which has to be determined.
 * where, $$\displaystyle \alpha $$, is unknown, which has to be determined.


 * Multiplying $$\displaystyle Eq.5.1 $$ by $$\displaystyle Eq.5.3.1 $$ implies:


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


 * $${{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)={{e}^{\alpha t}}f\left( t \right)$$
 * }
 * }
 * }


 * Integrating both sides gives:


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

.
 * $$\int{{{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)dt}=\int{{{e}^{\alpha t}}f\left( t \right)dt}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.2)$$
 * }
 * }
 * }


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

\parallel \\ 0 \end{smallmatrix}}{{y}^{''}}+{{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right)$$ $$
 * $$\int{{{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)dt}={{e}^{\alpha t}}\left( \underbrace_{\begin{smallmatrix}
 * $$\int{{{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)dt}={{e}^{\alpha t}}\left( \underbrace_{\begin{smallmatrix}
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.3)
 * }
 * }

====(2.1)Find $$\displaystyle{{\overset{-}{\mathop{a}}\,}_{1}}$$, $$\displaystyle{{\overset{-}{\mathop{a}}\,}_{0}}$$ in terms of $$\displaystyle{{a}_{2}}$$, $$\displaystyle{{a}_{1}}$$, $$\displaystyle{{a}_{0}}$$====


 * Differentiating both sides of $$\displaystyle Eq.5.3.3 $$


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


 * $$\displaystyle {{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=\frac{d}{dt}\left[ {{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right) \right]$$
 * }
 * }
 * }


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


 * $${{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=\alpha {{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right)+{{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{}}+{{\overset{-}{\mathop{a}}\,}_{0}}{{y}^{'}} \right)$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{a}_{2}}{{y}^{}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y={{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{}}+\left( \alpha {{\overset{-}{\mathop{a}}\,}_{1}}+{{\overset{-}{\mathop{a}}\,}_{0}} \right){{y}^{'}}+\alpha {{\overset{-}{\mathop{a}}\,}_{0}}y$$
 * }
 * }
 * }


 * Now comparing the coefficents on both sides,


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

$${{\overset{-}{\mathop{a}}\,}_{1}}={{a}_{2}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.4)
 * }
 * }


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


 * $$\alpha {{\overset{-}{\mathop{a}}\,}_{1}}+{{\overset{-}{\mathop{a}}\,}_{0}}={{a}_{1}}$$
 * }
 * }
 * }


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

$$\Rightarrow {{\overset{-}{\mathop{a}}\,}_{0}}={{a}_{1}}-\alpha {{\overset{-}{\mathop{a}}\,}_{1}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.5)
 * }
 * }


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


 * $$\alpha {{\overset{-}{\mathop{a}}\,}_{0}}={{a}_{0}}$$
 * }
 * }
 * }


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

$$\Rightarrow {{\overset{-}{\mathop{a}}\,}_{0}}=\frac{\alpha }$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.6)
 * }
 * }

(2.2)Find Quadratic equation for $$\displaystyle\alpha $$

 * $$\displaystyle Eq.5.3.5$$ and $$\displaystyle Eq.5.3.6$$ implies:


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


 * $${{a}_{1}}-\alpha {{\overset{-}{\mathop{a}}\,}_{1}}=\frac{\alpha }$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{a}_{1}}\alpha -{{\overset{-}{\mathop{a}}\,}_{1}}{{\alpha }^{2}}={{a}_{0}}$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{\overset{-}{\mathop{a}}\,}_{1}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0$$
 * }
 * }
 * }

.
 * Now substituting $$\displaystyle Eq.5.3.4$$ into the above equation implies:


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

$$\displaystyle {{a}_{2}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.7)
 * }
 * }

.

(2.3)Find Reduced order equation

 * $$\displaystyle Eq.5.3.3$$ and $$\displaystyle Eq.5.3.4$$ implies:


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


 * $${{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right)=\int{{{e}^{\alpha t}}f\left( t \right)dt}$$
 * }
 * }
 * }


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

$$
 * $$\Rightarrow {{y}^{'}}+\fracy=\underbrace{\frac\int{{{e}^{\alpha t}}f\left( t \right)dt}}_{F\left( t \right)}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.8)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.8)
 * }
 * }

(2.4)Use IFM to find $$\displaystyle y$$

 * Since $$\displaystyle Eq.5.3.8 $$ is $$\displaystyle L1-ODE-CC$$,the Integrating Factor is given by,


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

$$
 * $$h\left( t \right)={{e}^{\int{\beta dt}}}={{e}^{\beta t}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.9)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.9)
 * }
 * }


 * where, $$\displaystyle \beta =\frac$$


 * Now $$\displaystyle y\left( t \right)$$ is,


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


 * $$y\left( t \right)=\frac{1}{h\left( t \right)}\int{h\left( t \right)F\left( t \right)dt}$$
 * }
 * }
 * }


 * Substituting $$\displaystyle h\left( t \right)$$ and $$\displaystyle F\left( t \right)$$ in the above equation implies:


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

$$ \displaystyle \begin{align} y &=\frac\int_ – ^{t}{\left( {{e}^{(\beta -\alpha )\tau }}\int_ – ^{\tau }{{{e}^{\alpha x}}f(x)dx} \right)}d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau }\\ & =\frac\left[ \int_ – ^ – {{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+{{k}_{1}}\int_ – ^ – {{{e}^{(\beta -\alpha )t}}dt} \right]\\ & =\frac\left[ \int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+\frac{\beta -\alpha }{{e}^{(\beta -\alpha )t}} \right]\\ & =\frac{(\beta -\alpha ){_{2}}}{{e}^{-\alpha t}}+\frac{{e}^{-\beta t}}+\frac{{{{{a}}}_{2}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \end{align} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Combine constants to form $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$,


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

$$ \displaystyle y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.10) $$
 * }
 * }

(2.5)Show $$\displaystyle\alpha \beta =\frac$$ and $$\displaystyle\alpha +\beta =\frac$$

 * We know $$\displaystyle\beta =\frac$$ and $$\displaystyle\alpha =\frac$$


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

$$\alpha \beta =\frac\cdot \frac=\frac=\frac$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.11)
 * }
 * }


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


 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\alpha +\beta =\frac+\frac=\frac{{{a}_{0}}{{\overset{-}{\mathop{a}}\,}_{1}}+{{\overset{-}{\mathop{a}}\,}_{0}}^{2}}=\frac{{{a}_{0}}{{a}_{2}}+{{\overset{-}{\mathop{a}}\,}_{0}}\left( \frac{\alpha } \right)}{\left( \frac{\alpha } \right){{a}_{2}}}=\frac{\alpha {{a}_{2}}+{{\overset{-}{\mathop{a}}\,}_{0}}}=\frac$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.12)
 * }
 * }

(2.6)Deduce expression for particular solution $$\displaystyle{{y}_{p}}$$ for general excitation $$\displaystyle f\left( t \right)$$

 * We can see from the solution for $$\displaystyle y,(Eq.5.3.10)$$ that it is a linear combination of homogenous solutions and particular solution


 * Where, homogenous solutions are:


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


 * $${{y}_{H}}^{1}={{e}^{-\alpha t}}$$
 * }
 * }
 * }


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


 * $${{y}_{H}}^{2}={{e}^{-\beta t}}$$
 * }
 * }
 * }


 * Particular solution is,


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

$${{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.13)
 * }
 * }

(2.7)Verify particular solution with table [[media:egm6321.f09.mtg20.djvu|Mtg 20, Fall 2009]]for $$\displaystyle f\left( t \right)={{t}^{2}}$$

 * Substitute $$\displaystyle f\left( t \right)={{t}^{2}}$$ in $$\displaystyle (Eq.5.3.13)$$:


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


 * $${{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}{{t}^{2}}dt} \right)dt}$$
 * }
 * }
 * }


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


 * $${{y}_{P}}=\frac{{{\alpha }^{2}}{{\beta }^{2}}{{t}^{2}}-2\left( \alpha {{\beta }^{2}}+{{\alpha }^{2}}\beta \right)t+2\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta  \right)}$$
 * }
 * }
 * }


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


 * $$=\frac-\frac{2\left( \alpha {{\beta }^{2}}+{{\alpha }^{2}}\beta \right)t}+\frac{2\left( {{\left( \alpha +\beta  \right)}^{2}}-\alpha \beta  \right)}$$
 * }
 * }
 * }


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

.
 * $$=\frac{{{a}_{2}}\alpha \beta }-\frac{2\left( \alpha +\beta \right)t}+\frac{2\left( {{\left( \alpha +\beta  \right)}^{2}}-\alpha \beta  \right)}$$
 * }
 * }
 * }


 * Now subtituting $$\displaystyle(Eq.5.3.11)$$ and $$\displaystyle(Eq.5.3.12))$$ for$$\displaystyle\left( \alpha +\beta \right)$$ and $$\displaystyle\left( \alpha \beta  \right)$$ respectively in the above equation, implies:


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

$$
 * $${{y}_{P}}=\frac-\frac{2{{a}_{1}}t}{{{a}_{0}}^{2}}+2\left( \frac{{{a}_{1}}^{2}}{{{a}_{0}}^{3}}-\frac{{{a}_{0}}^{2}} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.14)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.14)
 * }
 * }

Using the table for particular solution in Mtg 20, fall 2009, consider a trial solution,


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

$$
 * $$\displaystyle {{y}_{P}}=A{{t}^{2}}+Bt+C$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.15)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.15)
 * }
 * }


 * where A,B,and C are unknown cofficients to be determined.


 * Taking derivative,


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


 * $$\displaystyle {{y}_{P}}^{'}=2At+B$$
 * }
 * }
 * }


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


 * $$\displaystyle {{y}_{P}}^{''}=2A$$
 * }
 * }
 * }


 * Substitute above equations in $$\displaystyle(Eq.5.1)$$


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


 * $$\displaystyle 2A{{a}_{2}}+\left( 2At+B \right){{a}_{1}}+\left( A{{t}^{2}}+Bt+C \right){{a}_{0}}={{t}^{2}}$$
 * }
 * }
 * }


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


 * $$\displaystyle {{a}_{0}}A{{t}^{2}}+\left( 2{{a}_{1}}A+{{a}_{0}}B \right)t+\left( 2{{a}_{2}}A+{{a}_{1}}B+{{a}_{0}}C \right)={{t}^{2}}$$
 * }
 * }
 * }


 * Equating coefficients on both side of equations gives:


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


 * $$\displaystyle A=\frac{1}$$
 * }
 * }
 * }


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


 * $$\displaystyle B=\frac{-2{{a}_{1}}}{{{a}_{0}}^{2}}$$
 * }
 * }
 * }


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


 * $$C=2\left( \frac{{{a}_{1}}^{2}}{{{a}_{0}}^{3}}-\frac{{{a}_{0}}^{2}} \right)$$
 * }
 * }
 * }

Plugg in A,B and C in $$\displaystyle(Eq.5.3.15)$$ gives


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

$$
 * $${{y}_{P}}=\frac-\frac{2{{a}_{1}}t}{{{a}_{0}}^{2}}+2\left( \frac{{{a}_{1}}^{2}}{{{a}_{0}}^{3}}-\frac{{{a}_{0}}^{2}} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.16)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.16)
 * }
 * }


 * Evident from $$\displaystyle(Eq.5.3.14)$$ and $$\displaystyle(Eq.5.3.16)$$ that particular solution obtained using the deduced expression and table are same.

(2.8)Solve L2-ODE-CC with the forcing function $$\displaystyle f\left( t \right)={{e}^{-{{t}^{2}}}}$$ (Gaussian distribution)

 * Homogenous solutions will be same as determined earlier.
 * To determine particular solution, substitute $$\displaystyle f\left( t \right)={{e}^{-{{t}^{2}}}}$$ in $$\displaystyle (Eq.5.3.13)$$:


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


 * $${{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}{{t}^{2}}dt} \right)dt}$$
 * }
 * }
 * }


 * Simplifying,


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

$$
 * $${{y}_{P}}=\frac{\sqrt{\pi }\cdot {{e}^{\frac{4}}}}{2\left( \beta -\alpha \right){{a}_{2}}}\left( {{e}^{-\alpha t}}erf\left( t-\frac{\alpha }{2} \right)-erf\left( t-\frac{\beta }{2} \right) \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.17)
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.17)
 * }
 * }


 * Now $$\displaystyle y$$ is obtained by combining homogenous solutions and particular solution ($$\displaystyle Eq.5.3.17$$)


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

$$y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac{\sqrt{\pi }\cdot {{e}^{\frac{4}}}}{2\left( \beta -\alpha \right){{a}_{2}}}\left( {{e}^{-\alpha t}}erf\left( t-\frac{\alpha }{2} \right)-erf\left( t-\frac{\beta }{2} \right) \right)$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 5.3.18)
 * }
 * }

Find coefficients $$\displaystyle {{a}_{0}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{2}}$$

 * $$\displaystyle(a) Eq.5.1$$ accept a characteristic equation $$\displaystyle\left( r+1 \right)\left( r-2 \right)=0$$


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


 * $$\left( r+1 \right)\left( r-2 \right)=0$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{r}^{2}}-r-2=0$$
 * }
 * }
 * }

.
 * So the $$\displaystyle L2-ODE $$ is given by


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


 * $$\displaystyle {{y}^{''}}-y-2y=0$$
 * }
 * }
 * }

.
 * The coefficients are:


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

\begin{array}{*{20}l} {{a}_{2}}=1 & {{a}_{1}}=-1 & {{a}_{0}}=-2\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

.
 * $$\displaystyle(b) Eq.5.1$$ accept a characteristic equation $$\displaystyle{{\left( r-4 \right)}^{2}}=0$$


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


 * $${{\left( r-4 \right)}^{2}}=0$$
 * }
 * }
 * }


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


 * $$\Rightarrow {{r}^{2}}-8r+16=0$$
 * }
 * }
 * }

.
 * Now the $$\displaystyle L2-ODE $$ is given by


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


 * $$\displaystyle {{y}^{''}}-8y+16y=0$$
 * }
 * }
 * }

.
 * The coefficients are:


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

\begin{array}{*{20}l} {{a}_{2}}=1 & {{a}_{1}}=-8 & {{a}_{0}}=16\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Contributing Members

 * Solved and posted byEgm6321.f10.team3.Sudheesh 00:47, 2 November 2010 (UTC)
 * Solved by Egm6321.f10.team3.franklin 01:55, 3 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 22:47, 2 November 2010 (UTC)

=Problem 6: Equivalence of particular solutions=

Given
The particular solution for a L2-ODE-VC developed in class during Mtg 30 is given as:
 * {| style="width:100%" border="0"

$${{y}_{p}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}$$ Where $$h(x)$$ is equal to:
 * <p style="text-align:right;">$$\,(Eq. 6.1) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$h(x)=\frac{{{u}_{1}}^{2}(x)}{{{u}_{2}}^{\prime }(x)\cdot {{u}_{1}}(x)-{{u}_{2}}(x)\cdot {{u}_{1}}^{\prime }(x)}$$
 * <p style="text-align:right;">$$\,(Eq. 6.2) $$
 * style= |
 * }

Find
Show that the developed particular solution is equal to the particular solution in King, Billingham and Otto page 8 Eq. 1.6 as:
 * {| style="width:100%" border="0"

$${{y}_{p}}=\oint{f\left( s \right)\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{W(s)} \right)}\,ds$$ Where $$W(s)$$ is the Wronskian and given as:
 * <p style="text-align:right;">$$\,(Eq. 6.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$W(s)={{u}_{1}}(s)\cdot {{u}_{2}}^{\prime }(s)-{{u}_{1}}^{\prime }(s)\cdot {{u}_{2}}(s)=\left| \begin{matrix} {{u}_{1}}(x) & {{u}_{2}}(x) \\ {{u}_{1}}^{\prime }(x) & {{u}_{2}}^{\prime }(x) \\ \end{matrix} \right|$$
 * <p style="text-align:right;">$$\,(Eq. 6.4) $$
 * style= |
 * }

Solution
The right hand side of Eq. 6.1 can be integrated using the partial integration method:
 * {| style="width:100%" border="0"

$$\int(t)\cdot r(t)\,dt=r(t)\cdot v(t)-\int{{r}'(t)\cdot v(t)}\,dt$$ Where in the specific case of Eq. 6.1, $$r(t)$$ and $$v(t)$$ can be found as:
 * <p style="text-align:right;">$$\,(Eq. 6.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{u}_{1}}\left( x \right)\int{\underbrace{\frac{1}{h\left( x \right)}}_\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{r}}\,dx$$ Applying the partial integration method to Eq. 6.1 the right hand side can be rewritten as:
 * <p style="text-align:right;">$$\,(Eq. 6.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{u}_{1}}(x)\cdot \left[ \underbrace{\int{h(x)f(x)\,dx}}_{r}\cdot \underbrace{\frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}}_{v}-\int{\underbrace{h(x)\cdot f(x)}_\cdot }\underbrace{\frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}}_{v}\,dx \right]$$ Applying Eq. 6.2, $$h(x)$$ can be rewritten as:
 * <p style="text-align:right;">$$\,(Eq. 6.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{u}_{1}}(x)\cdot \left[ \int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds\cdot \frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}-\int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{2}}(s)}{{{u}_{1}}(s)}\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds \right]$$ Multiplying $${{u}_{1}}$$ into the squared brackets we get:
 * <p style="text-align:right;">$$\,(Eq. 6.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\left[ \int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds\cdot \frac{{{u}_{2}}(x)}{1}-\int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{2}}(s)}{1}\cdot \frac{{{u}_{1}}(x)\cdot {{u}_{1}}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds \right]$$ Which is equal to the contour integral:
 * <p style="text-align:right;">$$\,(Eq. 6.9) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\oint\limits_{x}{f(s)\cdot }\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)} \right)\,ds=\oint\limits_{x}{f\left( s \right)\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{W(s)} \right)\,ds}$$ '''Eq. 6.10, thus agrees with Eq. 1.6 in King, Billingham and Otto page 8.'''
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\,(Eq. 6.10) $$
 * style= |
 * }

Contributing Members

 * Solved and posted byEgm6321.f10.team03.sigillo 19:42, 31 October 2010 (UTC)
 * Proofread by James Roark 18:40, 3 November 2010 (UTC)

=Problem 7: Find $$ {u}_{1},  {u}_{2} $$ using trial solution and variation of parameters=

Given
Give the reverse engineered equation on [|Mtg 31-1] of the lecture notes


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

$$ \left(x+1 \right) {y}^{''} - \left(2x+3 \right) {y}^{'} + 2y=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.1)
 * }

Find
Find $$ {u}_{1},  {u}_{2} $$ 2 Homogeneous solutions of $$\displaystyle (Eq 7.1) $$ using the trial solution:


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

y= {e}^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.2)
 * style= |
 * }

Solution
To solve we first find the roots to the characteristic equation by plugging $$\displaystyle (Eq 7.2) $$ into $$\displaystyle (Eq 7.1) $$ to obtain


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

\left(x+1 \right) {r}^{2} - \left(2x+3 \right) r + 2=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.3)
 * style= |
 * }

Find the roots by using the quadratic formula
 * {| style="width:100%" border="0"

r= \frac{\left(2x+3 \right) \pm \sqrt{{\left(2x+3 \right)}^{2}-8\left(x+1 \right)}} {2 \left( x+1\right)}= \frac{\left(2x+3 \right) \pm \left(2x+1 \right) } {2 \left( x+1\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.4)
 * style= |
 * }

Solving we get the two roots to the characteristic equation:


 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{r}_{1} = 2 ; {r}_{2}=  \frac{1}{ \left( x+1\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.5)
 * style= |
 * }

Now we use the trial solution $$\displaystyle y= {e}^{rx} $$ and the constant root to obtain:


 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{u}_{1} = exp(2x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.6)
 * style= |
 * }

To find $$\displaystyle {u}_{2}(x) $$ we will use the variation of parameters method.


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

y(x)=U(x) {u}_{1}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.7)
 * style= |
 * }

Plugging $$\displaystyle (Eq. 7.7)$$ into $$\displaystyle (Eq. 7.1)$$ gives us


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

{U}^{'}(x) \left[ -\left(2x+3 \right) {u}_{1}(x)+ \left( x+1\right) {u}^{'}_{1}(x) \right] + \left( x+1\right) {u}_{1}(x) {U}^{''}(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.8)
 * style= |
 * }

Now use the reduction of order method 0 to find the expression for $$\displaystyle {u}_{2}$$. Using equation 1 from [|Mtg 31-1] we can now solve for $$\displaystyle {u}_{2}(x)$$


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

{u}_{2}(x)= {u}_{1}(x) \int \frac{1}{ {u}^{2}_{1}}exp \left[ - \int {a}_{1}dx\right]dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.9)
 * style= |
 * }

where $$\displaystyle {a}_{1}:= \frac{2x+3}{x+1} $$. Solving the integral for $$\displaystyle {a}_{1}$$ first we have
 * {| style="width:100%" border="0"

\int {a}_{1}dx= \int \frac{2x+3}{x+1} dx = \left[2x+ log(x+1)\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.10)
 * style= |
 * }

Now plug $$\displaystyle (Eq. 7.10)$$ into $$\displaystyle (Eq. 7.9) $$ to obtain


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

{u}_{2}(x)= {u}_{1}(x) \int \frac{x+1}{ {u}^{2}_{1}}exp \left[ -2x\right]dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.11)
 * style= |
 * }

Substituting in our expression for $$\displaystyle {u}_{1}(x)=exp\left(2x\right)$$ we have


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

{u}_{2}(x)= exp \left(2x \right) \int \left(x+1\right) {exp \left(-2x \right)}^{2}exp \left[ -2x\right]dx= exp \left(2x \right) \int \left(x+1\right) exp \left[ -6x\right]dx $$ $$ Solving the above equation by parts yields the solution for $$\displaystyle {u}_{2}(x)$$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.11)
 * style= |
 * }


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

{u}_{2}(x)= exp \left(2x \right) \left[ - \frac{1}{6}x exp \left(-6x \right)- \frac{1}{36}exp \left(-6x \right)-\frac{1}{6}exp \left(-6x \right)\right]= - \frac{1}{36} \left( 6x+7\right)exp \left( -4x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.12)
 * style= |
 * }

The two homogeneous solutions are:


 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{u}_{1} = exp(2x); {u}_{2}(x)= - \frac{1}{36} \left( 6x+7\right)exp \left( -4x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }

Contributing Members
Solved and posted by--Egm6321.f10.team3.cook 02:12, 2 November 2010 (UTC)

Proof read by Egm6321.f10.team03 21:39, 2 November 2010 (UTC) Schulze

=Problem 8 - Create L2-ODE-VC (Reverse Engineering) =

Given
Trial solution


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

$$
 * $$ y(x) = \frac {e^{rx}}{x^2} $$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

And char.eq.


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

r^{2} + 3 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Find
Find L2-ODE-VC

Solution

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

$$
 * $$ \frac{\partial }{\partial x} \left( \frac {e^{rx} }{x^2}\right) = \left( \frac {r e^{rx} }{x^2} - \frac {2 e^{rx} }{x^3} \right) = \frac {e^{rx} }{x^2} \left( r - \frac {2 }{x}\right)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


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

(r-\frac{2}{x})\right)$$ $$
 * $$ \frac{\partial^2 }{\partial^2 x} \left( \frac {e^{rx} }{x^2}\right) = \left( \frac {2e^{rx} }{x^4} - \frac {2e^{rx}(r-\frac {2} {x}) }{x^3} + \frac {re^{rx}(r-\frac {2}{x}) }{x^2} \right) = \frac {e^{rx} }{x^2} \left(  \frac{2}{x^2} -\frac{2}{x}(r-\frac{2}{x}) +r
 * $$ \frac{\partial^2 }{\partial^2 x} \left( \frac {e^{rx} }{x^2}\right) = \left( \frac {2e^{rx} }{x^4} - \frac {2e^{rx}(r-\frac {2} {x}) }{x^3} + \frac {re^{rx}(r-\frac {2}{x}) }{x^2} \right) = \frac {e^{rx} }{x^2} \left(  \frac{2}{x^2} -\frac{2}{x}(r-\frac{2}{x}) +r
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

To find L2-ODE-VC, multiply (Eqt 5.8.1), (Eqt 5.8.2) and (Eqt 5.8.3) by the fuctions of x
 * {| style="width:100%" border="0"

$$ \left[ \frac{e^{rx}}{x^2} \right ] * a_{0}(x) $$ || <p style="text-align:right;">$$\displaystyle (Eq. 5.8.1) $$
 * }
 * }


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

$$ \left[ \frac{\partial }{\partial x} \left( \frac {e^{rx} }{x^2}\right) =  \frac {e^{rx} }{x^2} \left( r - \frac {2 }{x} \right) \right] * a_{1}(x)  $$ || <p style="text-align:right;">$$\displaystyle (Eq. 5.8.2) $$
 * }
 * }


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

$$ \left[ \frac{\partial^2 }{\partial^2 x} \left( \frac {e^{rx} }{x^2}\right)  =  \frac {e^{rx} }{x^2} \left(  \frac{2}{x^2} -\frac{2}{x}(r-\frac{2}{x}) +r(r-\frac{2}{x})\right)\right]  * a_{2}(x) $$ || <p style="text-align:right;">$$\displaystyle (Eq. 5.8.3) $$
 * }
 * }

Do the sum of (Eqt 5.8.1), (Eqt 5.8.2) and (Eqt 5.8.3)


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

$$ \frac {e^{rx}}{x^2} \left( a_{2}(x)\left [ \frac {2}{x^2} - \frac {2}{x}(r-\frac {2}{x}) + r(r-\frac {2}{x})\right ] + a_{1}(x)\left[ r-\frac{2}{x} \right] + a_{0}(x) \right ) = \frac {e^{rx}}{x} (r^2 + 3) = 0$$ || <p style="text-align:right;">$$\displaystyle $$
 * }
 * }


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

$$ \frac {e^{rx}}{x^2} \left( a_{2}(x) r^2 + \left [ a_{1}(x) - \frac{4}{x}a_{2}(x)\right ]r + \left[\frac{6}{x^2}a_{2}(x) - \frac{2}{x}a_{1}(x) \right ] + a_{0}(x) \right ) = \frac {e^{rx}}{x} (r^2 + 3) = 0$$ || <p style="text-align:right;">$$\displaystyle (Eq. 5.8.4) $$
 * }
 * }

from (Eqt. 5.8.4),


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

$$ \therefore a_{2}(x) = 1$$ || <p style="text-align:right;">$$\displaystyle $$
 * }
 * }


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

$$ \therefore a_{1}(x) = \frac{4}{x}$$ || <p style="text-align:right;">$$\displaystyle $$
 * }
 * }


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

$$ \therefore a_{0}(x) = 3 + \frac{2}{x^2}$$ || <p style="text-align:right;">$$\displaystyle $$
 * }
 * }

Therefore,


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

$$ {{y}'}' + \frac {4}{x}{y}' + \left (3+\frac{2}{x^2}\right )y = 0$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.) $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style= |
 * }

Contributing Members
Solved and posted byEgm6321.f10.team03.Hong SJ 16:25, 01 November 2010 (UTC)

Proof read by Egm6321.f10.team03 21:41, 2 November 2010 (UTC)Schulze

Proofread by Egm6321.f10.team3.Sudheesh 23:46, 2 November 2010 (UTC)

=Contributing members=
 * Egm6321.f10.team3.Sudheesh 04:31, 2 November 2010 (UTC) Solved prob #5, posted prob #3 and #5, rechecked solution of prob #3, Proof read prob #1,2,8
 * Egm6321.f10.team03.sigillo 17:56, 2 November 2010 (UTC) Solved problem 4 and 6. Proofread problem 1,2,3,5
 * Egm6321.f10.team3.Schulze 04:31, 2 November 2010 (UTC) Solved prob 3, proof read 7 and 8.
 * Egm6321.f10.team3.franklin 01:59, 3 November 2010 (UTC) Solved prob 5

=Notes and references=