User:Egm6321.f12.team6.Hu.M/report6

==Problem *6.4:Euler L2-ODE-VC and Euler L2-OED-CC ==

Given
Characteristic equation:

where $$ \displaystyle \lambda=5 $$

Euler L2-ODE-VC:

Euler L2-ODE-CC:

Problem Statement
1.1: Find $$ \displaystyle a_2,a_1,a_0 $$ such that $$is characteristic equation of $$.

1.2: Find 1st homogeneous solution: $$ \displaystyle y_1(x)=x^\lambda $$.

1.3: Complete solution: Find $$ \displaystyle c(x) $$ such that $$ \displaystyle y(x)=c(x)y_1(x) $$.

1.4: Find the 2nd homogeneous solution $$ \displaystyle y_2(x) $$.

2.1: Find $$ \displaystyle b_2,b_1,b_0 $$ such that $$is characteristic equation of $$.

2.2: Find 1st homogeneous solution: $$ \displaystyle y_1(x)=e^{x\lambda} $$.

2.3: Complete solution: Find $$ \displaystyle c(x) $$ such that $$ \displaystyle y(x)=c(x)y_1(x) $$.

2.4: Find the 2nd homogeneous solution $$ \displaystyle y_2(x) $$.

Nomenclature
L2-ODE-VC stands for Linear 2nd order differential equation with various coefficient.

L2-ODE-CC stands for Linear 2nd order differential equation with constant coefficient.

Problem 1.1: Finding Charateristic Equation
Assume that $$ \displaystyle y=x^r $$ where $$ \displaystyle r $$ is a constant.

Then we have:
 * $$ \displaystyle y'=rx^{r-1} $$
 * $$ \displaystyle y''=r(r-1)x^{r-2} $$

Substitute these into $$we have:


 * $$ \displaystyle a_2x^2r(r-1)x^{r-2}+a_1xrx^{r-1}+a_0x^r=0 $$
 * $$ \displaystyle a_2r(r-1)x^r+a_1rx^r+a_0x^r=0 $$

Compare $$with $$, we can conclude that:


 * $$ \displaystyle a_2=1 $$
 * $$ \displaystyle a_1-a_2=-2\lambda $$
 * $$ \displaystyle a_0=\lambda^2 $$

With $$ \displaystyle \lambda=5 $$ we have:


 * $$ \displaystyle a_0=25, a_1=-9, a_2=1$$.

Problem 1.2: First Homogeneous Equation
In problem 1.1 we assume that $$ \displaystyle y=x^r $$. With $$we have $$ \displaystyle r=\lambda $$.

Thus the 1st homogeneous solution is:
 * $$ \displaystyle y_1(x)=x^\lambda=x^5 $$

Problem 1.3: Finding Constant
Given that $$ \displaystyle y(x)=c(x)y_1(x) $$

Using the chain rules we have:
 * $$ \displaystyle y'(x)=c(x)y_1'(x)+c'(x)y_1(x) $$
 * $$ \displaystyle y(x)=c(x)y_1(x)+c'(x)y_1'(x)+c'(x)y_1'(x)+c(x)y_1(x)=c(x)y_1(x)+2c'(x)y_1'(x)+c''(x)y_1(x) $$

Substitute above equations into $$and yields, we have:

Group $$by $$ \displaystyle c(x) $$:

The section $$ \displaystyle c(x) $$ will be zero because of $$. Then substitute $$ \displaystyle y_1(x)=x^5 $$ into $$and using the answer in Problem 1.1 we have:
 * $$ \displaystyle c'(x)[10a_2x^6+a_1x^6]+c''(x)a_2x^7=0 $$
 * $$ \displaystyle \Rightarrow c'(x)+c''(x)x=0 $$

Then we have:
 * $$ \displaystyle c'(x)+c''(x)x=[c'(x)x]'=0 $$
 * $$ \displaystyle c'(x)x=k_1 $$ where $$ \displaystyle k_1 $$ is a constant.
 * $$ \displaystyle c(x)=k_1 ln(x)+k_2 $$ where $$ \displaystyle k_1,k_2 $$ are constant

Therefore:
 * $$ \displaystyle y(x)=[k_1 ln(x)+k_2]x^5 $$

Problem 1.4: Second Homogeneous Equation
From Problem 1.3 we have the general homogeneous solution as $$ \displaystyle y(x)=[k_1 ln(x)+k_2]x^5 $$.

Since $$ \displaystyle y_1(x)=x^5 $$

Then $$ \displaystyle y_2(x)=ln(x)x^5 $$.

Problem 2.1: Finding Charateristic Equation
Similar to Problem 1.1 we assume that $$ \displaystyle y(x)=e^{xr} $$.

Then we have:
 * $$ \displaystyle y'(x)=re^{rx} $$
 * $$ \displaystyle y''(x)=r^2e^{rx} $$

Substitute these equations into $$we have:
 * $$ \displaystyle b_2r^2e^{rx}+b_1re^{rx}+b_0e^{rx}=0 $$

Compare $$with $$we have: $$ \displaystyle b_2=1 $$ $$ \displaystyle b_1=-2\lambda $$ $$ \displaystyle b_0=\lambda^2 $$ With $$ \displaystyle \lambda-5 $$ we have:
 * $$ \displaystyle b_0=25, b_1=-10, b_2=1 $$

Problem 2.2: First Homogeneous Equation
In problem 2.1 we assume that $$ \displaystyle y=e^{xr} $$. With $$we have $$ \displaystyle r=\lambda $$.

Thus the 1st homogeneous solution is:
 * $$ \displaystyle y_1(x)=e^{x\lambda}=e^{5x} $$

Problem 2.3: Finding Constant
Given that $$ \displaystyle y(x)=c(x)y_1(x) $$

Using the chain rules we have:
 * $$ \displaystyle y'(x)=c(x)y_1'(x)+c'(x)y_1(x) $$
 * $$ \displaystyle y(x)=c(x)y_1(x)+c'(x)y_1'(x)+c'(x)y_1'(x)+c(x)y_1(x)=c(x)y_1(x)+2c'(x)y_1'(x)+c''(x)y_1(x) $$

Substitute above equations into $$and yields, we have:

Group $$by $$ \displaystyle c(x) $$:

The section $$ \displaystyle c(x) $$ will be zero because of $$. Then substitute $$ \displaystyle y_1(x)=e^{5x} $$ into $$and using the answer in Problem 2.1 we have:
 * $$ \displaystyle c'(x)[2b_2y_1'(x)+b_1y_1(x)]+c''(x)b_2y_1(x)=0 $$
 * $$ \displaystyle \Rightarrow c''(x)e^{5x}=0 $$

Then we have:
 * $$ \displaystyle c''(x)=0 $$
 * $$ \displaystyle c'(x)=k_1 $$ where $$ \displaystyle k_2 $$ is a constant.
 * $$ \displaystyle c(x)=k_1x+k_2 $$ where $$ \displaystyle k_1,k_2 $$ are constant

Therefore:
 * $$ \displaystyle y(x)=[k_1x+k_2]e^{5x} $$

Problem 2.4: Second Homogeneous Equation
From Problem 2.3 we have the general homogeneous solution as $$ \displaystyle y(x)=[k_1x+k_2]e^{5x} $$.

Since $$ \displaystyle y_1(x)=e^{5x} $$

Then $$ \displaystyle y_2(x)=xe^{5x} $$.

Given
In the Lecture note it is given as below:

With $$ \displaystyle W:=u_1u_2'-u_2u_1'= \begin{vmatrix} u_1 & u_2 \\ u_1' & u_2' \end{vmatrix} $$ In King's book it is given as below:

With $$ \displaystyle \frac{1}{h(x)}=(\frac{u_2(x)}{u_1(x)})' $$

Problem Statement
Show that $$agrees with $$.

Solution
With $$ \displaystyle \frac{1}{h(x)}=(\frac{u_2(x)}{u_1(x)})'=\frac{u_1(x)u_2'(x)-u_2(x)u_1'(x)}{u_1^2(x)}=\frac{W(x)}{u_1^2(x)} $$.

We can rewrite $$as below:
 * $$ \displaystyle y_P(x)=u_1(x) \int \frac{W(x)}{u_1^2(x)} \left [\int h(x) \frac{f(x)u_1(x)}{W(x)}dx \right ]dx $$

It can be rewrite with dummy integrals as:
 * $$ \displaystyle y_P(x)=u_1(x) \left [ \left ( \int^x \frac{f(s)u_1(s)}{W(s)}ds \right ) \frac{u_2(x)}{u_1(x)} - \int^x \frac{f(s)u_2(s)}{W(s)} ds \right ] $$

Put $$ \displaystyle u_1(x), u_2(x) $$ inside integrals we have:
 * $$ \displaystyle y_P(x)= \int^x \frac{f(s)u_1(s)u_2(x)}{W(s)}ds - \int^x \frac{f(s)u_2(s)u_1(x)}{W(s)} ds=\int^x f(s) \left [ \frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)} \right ]ds $$

Obviously this is $$, thus we have the conclusion that $$agrees with $$.