User:Egm6321.f09.Team1.andy/HW3

= Problem 9: Finding solution to Homogeneous L2-ODE-VC =

Given
A Homogeneous L2-ODE-VC is given by
 * {| style="width:100%" border="0" align="left"

x^2y_{xx} - 2xy_{x} + 2y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

where


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

\begin{align} & x = e^t \\ \Rightarrow & \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} \\ \Rightarrow & \frac{dy}{dx} = e^{-t}\frac{dy}{dt} \\ \Rightarrow & y_x = e^{-t}y_t \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * $$\displaystyle (Eq. 2)
 * }

and so,


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

\begin{align} \Rightarrow & y_{xx} = e^{-2t}(y_{tt} - y_t) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
To show that,


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

y_{xxx} = e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and


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

y_{xxxx} = e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solution
Part 1:


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

\begin{align} y_{xxx} & = \frac{d}{dx} (y_{xx}) \\ & \Rightarrow \frac{dt}{dx} \frac{d}{dt} (y_{xx}) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

From Equations (2) and (3),


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

\begin{align} y_{xxx} & = \frac{dt}{dx} \frac{d}{dt} \left[e^{-2t}(y_{tt} - y_t) \right] \\ & \Rightarrow e^{-t} \left[-2e^{-2t}(y_{tt} - y_t) + e^{-2t}(y_{ttt} - y_{tt}) \right] \\ & \Rightarrow e^{-3t} \left[-2y_{tt} + 2y_t + y_{ttt} - y_{tt} \right] \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and the solution,


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

$$\displaystyle \Rightarrow y_{xxx} = e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * $$\displaystyle (Eq. 4)
 * }
 * }

Part 2:


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

\begin{align} y_{xxxx} & = \frac{d}{dx} (y_{xxx}) \\ & \Rightarrow \frac{dt}{dx} \frac{d}{dt} (y_{xxx}) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

From Equation (4),


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

\begin{align} y_{xxxx} & = \frac{dt}{dx} \frac{d}{dt} \left[e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) \right] \\ & \Rightarrow e^{-t} \left[-3e^{-3t}(y_{ttt} - 3y_{tt} + 2y_t) + e^{-3t}(y_{tttt} - 3y_{ttt} + 2y_{tt}) \right] \\ & \Rightarrow e^{-4t} \left[-3y_{ttt} + 9y_{tt} - 6y_t + y_{tttt} - 3y_{ttt} + 2y_{tt} \right]  \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and the solution,


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

$$\displaystyle y_{xxxx} = e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * $$\displaystyle
 * }
 * }

= Problem 10: Finding solution to Homogeneous L2-ODE-VC =

Given
A Homogeneous L2-ODE-VC is given by
 * {| style="width:100%" border="0" align="left"

x^2y'' - 2xy' + 2y = 0 $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y' = \frac{dy}{dx} $$ y'' = \frac{d^2y}{dx^2} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
Find the solution for equation (1) directly by the method of trial solution,


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

y = e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Compare the solution with,


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

y = C_1 x + C_2 x^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

with boundary conditions,
 * {| style="width:100%" border="0" align="left"

y(1) = 3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y(2) = 4 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

Solution
= Problem 12: Reduction of Order Method =

Given
To solve the Homogeneous L2-ODE-VC,


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

y'' + a_1(x)y' + a_0(x)y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

which has one solution as


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

y = u_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

where the complete solution is given by,
 * {| style="width:100%" border="0" align="left"

y = k_1u_1(x) + k_2u_2(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Find
To develop reduction of order method using the following algebraic expressions,

Part 1: $$\displaystyle y(x)=U(x)\pm u_1 (x)$$ Part 2: $$y(x)=\frac{U(x)}{u_1 (x)}$$ Part 3: $$y(x)=\frac{u_1 (x)}{U(x)}$$

Solution
Part 1:

Using the first algebraic expression,


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

y(x)=U(x)\pm u_1 (x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

On differentiating the Equation (3), we get the following first derivative,


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

y'(x)=U'(x)\pm u_1'(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Second derivative,


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

y(x)=U(x)\pm u_1''(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

Using Equations (3), (4) and (5) in Equation (1),


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

y + a_1(x)y' + a_0(x)y = \left[U(x)\pm u_1''(x) \right] + \left[U'(x)\pm u_1'(x) \right] a_1(x) +  \left[U(x)\pm u_1(x) \right] a_0(x)  = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\begin{align} \Rightarrow y + a_1(x)y' + a_0(x)y = & \left[U(x) + U'(x)a_1(x) + U(x)a_0(x)\right] \pm \underbrace{\left[u_1''(x) + u_1'(x)a_1(x) + u_1(x)a_0(x) \right]}_{=0\ as\ per\ Eq(2)} \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\begin{align} \Rightarrow y + a_1(x)y' + a_0(x)y = & \left[U(x) + U'(x)a_1(x) + U(x)a_0(x)\right] = 0 \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.


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



So, assumption that $$\displaystyle y(x)=U(x)\pm u_1 (x)$$ as full homogeneous solution for Equation (1) is incorrect.
 * style="width:70%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Part 2: 

Using the second algebraic expression,


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

y(x)=\frac{U(x)}{u_1 (x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

On differentiating the Equation (6), we get the following first derivative,


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

y'=\frac{U'u_1 - Uu_1'}{u_1^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Second derivative,


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

y =\frac{1}{u_1^2}(U'u_1' + Uu_1 - Uu_1'' - U'u_1') - \frac{u_1'}{u_1^3}(U'u_1 - Uu_1') $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)
 * }
 * }

Using Equations (6), (7) and (8) in Equation (1),


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

y + a_1(x)y' + a_0(x)y = \frac{U}{u_1}a_0 + \left[\frac{U'}{u_1} - \frac{U'u_1'}{u_1^2}\right] a_1+ \left[\frac{U}{u_1} - \frac{Uu_1''}{u_1^2} - \frac{U'u_1'}{u_1^2} + \frac{Uu_1'^2}{u_1^3}\right]  = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.


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



So, assumption that $$\displaystyle y(x)=\frac{U(x)}{u_1 (x)}$$ as full homogeneous solution for Equation (1) is incorrect.
 * style="width:70%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

Part 3: 

Using the third algebraic expression,


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

y(x)=\frac{u_1(x)}{U(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

It is obvious from the above Part 2 section, if we let $$U(x) = u_1(x)$$ and $$ u1(x) = U(x)$$, then the same argument holds that Equation (9) cannot be used as a full homogeneous solution for Equation (1). Following is the derivation, anyways.

On differentiating the Equation (9), we get the following first derivative,


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

y'=\frac{u_1'U - u_1U'}{U^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }

Second derivative,


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

y =\frac{1}{U^2}(u_1'U' + u_1U - u_1U'' - u_1'U') - \frac{U'}{U^3}(u_1'U - u_1U') $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

using Equations (9), (10) and (11) in Equation (1),


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

y + a_1(x)y' + a_0(x)y = \frac{u_1}{U}a_0 + \left[\frac{u_1'}{U} - \frac{u_1'U'}{U^2}\right] a_1+ \left[\frac{u_1}{U} - \frac{u_1U''}{U^2} - \frac{u_1'U'}{U^2} + \frac{u_1U'^2}{U^3}\right]  = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.


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



So, assumption that $$\displaystyle y(x)=\frac{u_1(x)}{U (x)}$$ as full homogeneous solution for Equation (1) is incorrect.
 * style="width:70%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= Problem 13: Using given trial solutions to solve function =

Given
Given the following homegenious L2_ODE_VC
 * {| style="width:100%" border="0" align="left"

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

The two test trial solutions are
 * {| style="width:100%" border="0" align="left"

y=ax^b $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y=e^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Find
Find $$u_1(x)$$ and $$u_2(x)$$ of Eq. 1 using two given trial solutions. Compare it with the solution, given by reduction of order method 2,


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

u_1=x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

and


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

u_2=\frac{x}{2}log_e\left(\frac{1+x}{1-x}\right) - 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Solution
Part 1:

With the trial solution,


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

y=ax^b $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The derivatives are,


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

y'=abx^{(b-1)} $$  $$  y''=ab(b-1)x^{(b-2)} $$  $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)


 * }

Substituting Equations (2), (6), and (7) in Equation (1),


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

\Rightarrow (1-x^2)ab(b-1)x^{(b-2)} - 2xabx^{(b-1)} + 2ax^b = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\Rightarrow ab(b-1)x^{(b-2)}-ab(b-1)x^b - 2abx^b + 2ax^b = 0 $$  \Rightarrow ab(b-1)x^{(b-2)}-\left[ab(b-1) + 2ab - 2a \right]x^b = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 8)


 * }

In Equation (8), equating co-efficients of $$ x^{(b-2)}$$ ,


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

\Rightarrow ab(b-1) = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since a and b both cannot be zero, one value of $$ b=1$$.

In Equation (8), equating co-efficients of $$ x^b$$ ,


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

\Rightarrow ab(b-1) + 2ab - 2a = 0 $$  \Rightarrow a(b^2+b-2) = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }

Again $$ a$$ cannot be zero, so $$ b =1,-2$$.

Therefore,


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

\Rightarrow u_1(x) = ax $$ \Rightarrow u_2(x) = \frac{a}{x^2} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }

So, the trial solution,


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

\Rightarrow y = k_1u_1(x) + k_2u_2(x) $$  \Rightarrow y = K_1x + K_2\frac{1}{x^2} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Using the boundary conditions,


 * {| style="width:30%" border="0" align="left"

y(\frac{\pi}{2\sqrt{2}}) = 1 $$  $$\displaystyle and $$ $$\displaystyle y(\frac{\pi}{\sqrt{2}}) = 2 $$  We arrive at,
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * {| style="width:30%" border="0" align="left"

K_1 = \frac{2\sqrt{2}}{\pi} $$  $$\displaystyle and $$ $$\displaystyle K_2 = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

\Rightarrow y = \frac{2\sqrt{2}}{\pi}x $$  $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9)
 * }
 * }

Therefore,


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

u_1 = \frac{2\sqrt{2}}{\pi}x;\ and\ u_2 = 0 $$
 * $$\displaystyle
 * $$\displaystyle


 * }

Part 2:

With the trial solution,


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

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

The derivatives are,


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

y'=re^{rx} $$  $$  y''=r^2e^{rx} $$  $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)


 * }

Substituting Equations (3), (8), and (9) in Equation (1),


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

\Rightarrow \left[(1-x^2)r^2 - 2xr + 2\right]e^{rx} = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

$$ e^{rx}$$ cannot be zero.

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

\Rightarrow -x^2r^2-2xr + r^2 +2 = 0 $$  \Rightarrow -x^2[r^2]-x[2r] + r^2 +2 = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }

We do not have any terms on the right hand side to equate with co-efficients of $$ x^2$$ and $$ x$$, so equating the constant co-efficients,

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

\Rightarrow r^2 +2 = 0 $$  \Rightarrow r = \pm \sqrt2i $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }

The trial solution becomes,


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

\Rightarrow y = e^{\sqrt2ix} $$ \Rightarrow y = K_1 cos(\sqrt2x) + K_2 sin(\sqrt2x) $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Using the boundary conditions,


 * {| style="width:30%" border="0" align="left"

y(\frac{\pi}{2\sqrt{2}}) = 1 $$  $$\displaystyle and $$ $$\displaystyle y(\frac{\pi}{\sqrt{2}}) = 2 $$  We arrive at,
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * {| style="width:30%" border="0" align="left"

K_1 = -2 $$  $$\displaystyle and $$ $$\displaystyle K_2 = 1 $$  The trial solution then becomes,
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


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

\Rightarrow y = -2 cos(\sqrt2x) + sin(\sqrt2x) $$ $$ and so,
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 12)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

u_1 = -2 cos(\sqrt2x);\ and\ u_2 = sin(\sqrt2x) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Plotting solutions y1,y2,y3 respectively, from Equations (9), (12) and solution from equations (4),(5).