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If $y=e^{a \cos ^{-1} x},-1 \leq x \leq 1$, show that
$\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0$
$\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0$
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Verified Answer
We have $y=e^{a \cos ^{-1} x}$
Differentiate w.r.t. $x$,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_1=\mathrm{e}^{\mathrm{a}^{\cos ^{-1} \mathrm{x}}} \frac{-\mathrm{a}}{\sqrt{1-\mathrm{x}^2}}$
$\Rightarrow \sqrt{1-\mathrm{x}^2} \mathrm{y}_1=-\mathrm{ae}^{\mathrm{a} \cos ^{-1} \mathrm{x}}$
Again differentiate w.r.t. $\mathrm{x}$, we get
$\Rightarrow \sqrt{1-\mathrm{x}^2} \mathrm{y}_2+\mathrm{y}_1 \cdot \frac{1}{2} \frac{1}{\sqrt{1-\mathrm{x}^2}}(-2 \mathrm{x})$
$=\frac{a^2 e^{a \cos ^{-1} x}}{\sqrt{1-x^2}}$
$\Rightarrow\left(1-x^2\right) y_2-x y_1-a^2 y=0$
Differentiate w.r.t. $x$,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_1=\mathrm{e}^{\mathrm{a}^{\cos ^{-1} \mathrm{x}}} \frac{-\mathrm{a}}{\sqrt{1-\mathrm{x}^2}}$
$\Rightarrow \sqrt{1-\mathrm{x}^2} \mathrm{y}_1=-\mathrm{ae}^{\mathrm{a} \cos ^{-1} \mathrm{x}}$
Again differentiate w.r.t. $\mathrm{x}$, we get
$\Rightarrow \sqrt{1-\mathrm{x}^2} \mathrm{y}_2+\mathrm{y}_1 \cdot \frac{1}{2} \frac{1}{\sqrt{1-\mathrm{x}^2}}(-2 \mathrm{x})$
$=\frac{a^2 e^{a \cos ^{-1} x}}{\sqrt{1-x^2}}$
$\Rightarrow\left(1-x^2\right) y_2-x y_1-a^2 y=0$
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