User:Egm6321.f10.team1.allen/HW1

Given
We are given a Legendre differential equatons with n=1 as shown below.

$$L_2(y)=(1-x)^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=0 $$

Both of the following homogeneous solutions satisfy the above differential equation.

$$ y_H^1=x=P_1(x) $$

$$ y_H^2=\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1 = Q_1(x) $$

Find
We are asked to verify that

$${{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$.

Solution
In order to verify these solutions we must first obtain their first and second derivatives.

$$y_H^1(x)=P_n(x)=x$$

$$\frac{dy_H^1(x)}{dx}=1$$

$$\frac{d^2y_H^1(x)}{dx^2}=0$$

These solutions are plugged into the Legendre Differential Equation.

$$L_2(y_H^1)=(1-x)^2(0)-2x(1)+2(x)=0 $$

After simplifying.

$$L_2(y_H^1)=-2x+2x=0 $$

After cancelling terms, we obtain our desired results.

$${{L}_{2}}(y_{H}^{1})=0$$.

$$y_H^2(x)=Q_n(x)=\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1$$

$${{L}_{2}}(x)={{L}_{2}}({\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1})=0$$.

Author
Allen