Note: y2 represents second order derivative i.e.\(\frac{d^2y}{dx^2}\) and y1 = dy/dx
Given,
x = sin(\(\frac{1}{a}\)log y)
\((logy)=asin^{-1}x\)
y = \(e^{asin^{-1}x}\)...............equation 1
to prove : (1 - x2)y2 -xy1 - a2=0
We notice a second–order derivative in the expression to be proved so first take the step to find the second order derivative.
Let’s find \(\frac{d^2y}{dx^2}\)
As, \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)
So, lets first find dy/dx
\(\because\) y = \(e^{asin^{-1}x}\)
And y = et
Again differentiating with respect to x applying product rule:
Using chain rule and equation 2:
Using equation 1 and equation 2 :
∴ (1–x2)y2–xy1–a2y = 0……proved