How to derive this differential equation from equation of ellipse??

Question

If $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$

Then show that $$\left(1-x^2\right)y'' -xy' -a^2y=0$$

Can anyone please help me with this problem?

Answer 1

Observe that $b$ is not in the equation you need to derive. So isolate $b^2$ on one side of the equation and then differentiate twice.

Answer 2

$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ differentiate $$\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} =0 \implies y'y=-x\frac {b^2}{a^2} $$ Multiply by $y (y \neq 0)$ $$y'y^2=-xy\frac {b^2}{a^2} $$ $$y'(1-\frac {x^2}{a^2})b^2=-xy\frac {b^2}{a^2} $$ $$y'(1-\frac {x^2}{a^2})=-\frac {xy}{a^2} $$ $$y'(a^2-x^2)=- {xy} $$ Differentiate again... $$(a^2-x^2)y'' -xy' +y=0$$ Are you sure you wrote the correct differential equation ?