# If the solve the problem

Question:

If $y=e^{a \cos ^{-1} x}$, prove that $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$

Solution:

Formula: -

(i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$

(ii) $\frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}=\frac{1}{\mathrm{x}}$

(ii) $\frac{\mathrm{d}\left(\cos ^{-1} \mathrm{x}\right)}{\mathrm{dx}}=\frac{-1}{\sqrt{1+\mathrm{x}^{2}}}$

(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{nx}^{\mathrm{n}-1}$

(iv) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d} \text { (wou) }}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

(v) logarithms differentiation $\frac{d y}{d x}=y\left[\frac{v(x)}{u(x)} \cdot u^{\prime}(x)+v^{\prime}(x) \cdot \log [u(x)]\right]$

Given: -

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

Taking logarithm on both sides we obtain

$\frac{1}{y} \frac{d y}{d x}=a \frac{-1}{\sqrt{1-x^{2}}}$

$\frac{d y}{d x}=\frac{-a y}{\sqrt{1-x^{2}}}$

By squaring both sides, wee obtain

$\left(\frac{d y}{d x}\right)^{2}=\frac{a^{2} y^{2}}{1-x^{2}}$

$\Rightarrow\left(1-x^{2}\right) \cdot\left(\frac{d y}{d x}\right)^{2}=a^{2} y^{2}$

$\Rightarrow\left(1-x^{2}\right)\left(\frac{d y}{d x}\right)^{2}=a^{2} y^{2}$

Again differentiating both sides with respect to $x$, we obtain

$\left(\frac{d y}{d x}\right)^{2} \cdot \frac{d\left(1-x^{2}\right)+\left(1-x^{2}\right)}{d x} \cdot \frac{d}{d x}\left[\left(\frac{d y}{d x}\right)^{2}\right]=a^{2} \frac{d\left(y^{2}\right)}{d x}$

$\Rightarrow\left(\frac{d y}{d x}\right)^{2}\left[(-2 x)+\left(1-x^{2}\right)\right] 2 \cdot \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow-x \frac{d y}{d x}+\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} y$

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

Hence proved