Question

If $y=e^{a{\cos }^{-1}x}$, $(-1\le x\le 1)$ show that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y=0$.

Answer 1

Given curve is $y=e^{a{\cos }^{-1}x}$

Differentiating given curve,

$\Rightarrow y^{\prime }=e^{a{\cos }^{-1}x}.\frac{-a}{\sqrt{(1-x^2)}}$

$\Rightarrow y^{\prime }=\frac{-ay}{\sqrt{(1-x^2)}}$.........(1)

On differentiating above equation again w.r.t $x$, we get

$\Rightarrow y^{\prime \prime }=\frac{-a(-a{e}^{a{\cos }^{-1}x}+\frac{x.e^{a{\cos }^{-1}x}}{\sqrt{(1-x^2)}})}{(1-x^2)}$

$\Rightarrow (1-x^2)y^{\prime \prime }=-a(-a{e}^{a{\cos }^{-1}x}+\frac{x.e^a{\cos }^{-1}x}{\sqrt{(1-x^2)}})$....(from equation (1))

$\Rightarrow (1-x^2)y^{\prime \prime }=a^2{e}^{a{\cos }^{-1}x}+x.y^{\prime })$

$\Rightarrow (1-x^2)y^{\prime \prime }-x.y^{\prime }-a^2.y=0$

$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-a^{2}y=0$

hence proved