User:EGM6321.f11.team1.pan/HW2

= R2.1 Verify the solution of the specific Legendre differential equation=

Given
The Legendre differential equation is
 * $$\displaystyle L_2(y):=(1-x^2)y''-2xy'+n(n+1)y=0$$ .

For the specific case $$\displaystyle n=1$$ , the Legendre differential equation becomes
 * {| style="width:100%" border="0" align="left"

$$ The two homogeneous solutions are given:
 * $$\displaystyle (1-x^2)y''-2xy'+2y=0$$
 * $$\displaystyle (Eq.2.1.1)
 * $$\displaystyle (Eq.2.1.1)
 * }
 * }
 * $$\displaystyle y_H^1(x)=x$$ ;
 * $$\displaystyle y_H^2(x)=\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1$$ .

Find
Verify that 
 * $$L_2(y_H^1)=L_2(y_H^2)=0$$ .

Solution
For $$y_H^1$$ ,  its first derivative is
 * $$\displaystyle \frac{dy_H^!}{dx}=1$$ ;

and its second derivative is 
 * $$\displaystyle \frac{d^2y_H^1}{dx^2}=0$$ .<br\>

Take them into $$\displaystyle (Eq.2.1.1)$$ ,<br\>
 * $$\displaystyle L_2(y_H^1)=(1-x^2)\cdot 0-2x\cdot 1+2\cdot x=0$$ .<br\>

For $$y_H^2$$ ,<br\> <br\> its first derivative is<br\>
 * $$\displaystyle \frac{dy_H^2}{dx}=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{2}\cdot\frac{1-x}{1+x}\cdot\left(\frac{1+x}{1-x}\right)'$$ <br\>


 * $$\displaystyle =\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}$$ ;<br\>

<br\> and its second derivative is<br\>
 * $$\displaystyle \frac{d^2y_H^2}{dx^2}=\frac{1}{2}\cdot\frac{1-x}{1+x}\cdot\frac{2}{(1-x)^2}+\frac{1-x^2+2x^2}{(1-x^2)^2}$$ <br\>
 * $$\displaystyle =\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}$$ .<br\>

Take them into $$\displaystyle (Eq.2.1.1)$$ ,<br\>
 * $$\displaystyle L_2(y_H^2)=(1-x^2)\cdot \left[\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}\right]-2x\cdot \left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}\right]+2\cdot \left[\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1\right]$$ <br\>
 * $$\displaystyle =1+\frac{1+x^2}{1-x^2}-x\cdot log\left(\frac{1+x}{1-x}\right)-\frac{2x^2}{1-x^2}+x\cdot log\left(\frac{1+x}{1-x}\right)-2$$<br\>
 * $$\displaystyle =0$$ .<br\>

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by EGM6321.f11.team1.pan

Reference for R2.1
Pea1.f11.mtg7.djvu

= R2.2 Verify the solution of an equation=

Find
Verify that<br\>
 * $$\displaystyle p(x)=k_1e^{-x}+x-1$$ <br\>

is indeed the solution for <br\>
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle p'+p=x$$.
 * <p style="text-align:right;">$$\displaystyle (Eq.2.2.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.2.1)
 * }
 * }

Solution
Derivate of $$\displaystyle p(x)=k_1e^{-x}+x-1$$ is <br\>
 * $$\displaystyle p'(x)=-k_1e^{-x}+1$$ .<br\>

Substituting its into $$\displaystyle (Eq.2.2.1)$$ ,<br\>
 * $$\displaystyle p'(x)+p(x)=-k_1e^{-x}+1+k_1e^{-x}+x-1=x$$ .<br\>

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by EGM6321.f11.team1.pan

Reference for R2.2
Pea1.f11.mtg7.djvu

= R2.7 Find G(y',y,x)=0, and verify it is an N1-ODE=

Given
Function $$\displaystyle \phi(x,y)=x^2y^{\frac{3}{2}}+log(x^3t^2)=k $$ .<br\>

Find
Find $$\displaystyle G(y',y,x)=\frac{d}{dx}\phi(x,y)=0$$ .<br\> And show that this is an N1-ODE.

Solution
According to the Chain Rule, <br\>
 * $$\displaystyle \frac{d\phi(x,y)}{dx}=\frac{\partial\phi(x,y)}{\partial x}+\frac{\partial\phi(x,y)}{\partial y}\frac{dy}{dx}$$ <br\>
 * $$\displaystyle =2xy^{\frac{3}{2}}+3x^2y^2\cdot\frac{1}{x^3y^2}+y'\left(\frac{3}{2}x^2y^{\frac{1}{2}}+2x^3y\frac{1}{x^3y^2}\right)$$<br\>
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle =2xy^{\frac{3}{2}}+\frac{3}{x}+\left(\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}\right)y'$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.7.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.7.1)
 * }
 * }

Let <br\>
 * $$\displaystyle G(y',y,x)=\frac{d}{dx}\phi(x,y)=0$$ .<br\>

Let <br\>
 * $$\displaystyle M(x,y)=2xy^{\frac{3}{2}}+\frac{3}{x}$$ ;<br\>
 * $$\displaystyle N(x,y)=\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y} $$ .<br\>

Therefore, $$\displaystyle (Eq.2.7.1)$$ can be shown as <br\>
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$ ,

which is the particular class of N1-ODE (linear in $$\displaystyle y'$$). <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by EGM6321.f11.team1.pan

Reference for R2.7
Pea1.f11.mtg9.djvu