Value of $\lambda$

Question

If $y=e^{2\cos^{-1}x}$ also $$(1-x^2)y_2-xy_1-\lambda y=0$$ then the value of $\lambda$ is. I see that the question is incomplete but answer is given as $2$. Am I missing on anything.

Answer 1

On differentiating;

• $y'=$$y\frac{-2}{√(1-x^2)}$ which on squaring gives, $(1-x^2)y^2=4y^2$. Differentiate again and after cancellation of $2y'$ on both sides, you get $(1-x^2)y''-xy'-4y=0$

Answer 2

I assume that $y_1$ means $y'$ and $y_2$ means $y''$.

So differentiating we get $y'=-2y\frac{1}{\sqrt{1-x^2}}$. Differentiating again we get $y''=-2xy\frac{1}{(1-x^2)^{3/2}}+4y\frac{1}{1-x^2}$. Hence $$(1-x^2)y''-xy'=-2xy\frac{1}{\sqrt{1-x^2}}+4y+2xy\frac{1}{\sqrt{1-x^2}}=4y$$

So the equation holds with $\lambda=4$.

Note that this does not agree with the answer given in the Question.