Question

If y=e^acos^-1x -1≤x≤1, show that (1-x)^2 d^2y/dx^2 -xdy/dx-a^2y=0

Answer 1

$\large{\mathfrak{HELLO\;\; DEAR}}$

it seems there is some typing mistake
(1 - x)² it should be (1 - x²)

y = $\bold{e^{acos^{-1}x}}$

we have to prove, (1 - x²)d²y/dx² - xdy/dx - a²y = 0

taking log both side

we get,

logy = acos^-1x

dy/dx*1/y = -a/√(1 - x²)

$\bold{dy/dx = \frac{-ae^{acos^{-1}x}}{\sqrt{1 - x^2}}}$

differentiating again,

$\large{\bold{d^2y/dx^2 = \frac{\sqrt{1 - x^2}[\frac{a^2*e^{acos^{-1}x}}{\sqrt{1 - x^2}}] - (-ae^{acos^{-1}x})*[\frac{-2x}{2\sqrt{1 - x^2}}]}{(\sqrt{1 - x^2})^2}}}$

$\bold{(1 - x^2)d^2y/dx^2 = a^2e^{acos^{-1}x} + x[\frac{-ae^{cos^{-1}x}}{\sqrt{1 - x^2}}]}$

$\bold{(1 - x^2)*d^2y/dx^2 = a^2y + x*dy/dx}$

$\bold{\underline{HENCE, \;\; (1 - x^2)*d^2y/dx^2 - xdy/dx - a^2y = 0}}$

$\bold{(1 - x^2)d^2y/dx^2 - a^2y - x*dy/dx = 0}$

$\large{\mathit{\underline{I \;\;HOPE\;\; ITS\;\; HELP \;\;YOU\;\; DEAR,\;\; THANKS}}}$