Differential Equations Was abe to show $y'=-4x$ $y''=-4$ and when substitute $=0$

Question

$(1-x^2)y''-xy'+4y=0$ Show that $1-2x^2$ is a solution try to find another using Wronskian method.

Was abe to show $y'=-4x$ $y''=-4$ and when substitute $=0$

Answer 1

Hint

There are two ways to calculate the Wronskian

$$\begin{align} &(1)~W(y_1,y_2)(t)~=~\begin{vmatrix}y_1(t) & y_2(t)\\y_1'(t) & y_2'(t)\end{vmatrix}~=~y_1(t)y_2'(t)-y_2(t)y_1'(t)\\ &(2)~W(y_1,y_2)(t)~=~e^{-\int p(x)\mathrm{d}x} \end{align}$$

You already know one solution, $y_1$, of the differential equation.

Answer 2

Hint: With $p(x)=\dfrac{-x}{1-x^2}$ we write $$W(x)=\exp\left(-\int\dfrac{-x}{1-x^2}\right)dx=\dfrac{1}{\sqrt{1-x^2}}$$ for $|x|<1$. Then solve the integral $$y_2=(1-2x^2)\int\dfrac{W(x)}{(1-2x^2)^2}dx$$