Login/Sign Up
If $y=e^{a c o^{-1} x},-1 \leq x \leq 1$, show that $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$.
If $y=e^{a c o^{-1} x},-1 \leq x \leq 1$, show that $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$.
CBSE | Class 12 | Continuity and Differentiability | Maths | Miscellaneous | NCERT
If $y=e^{a c o^{-1} x},-1 \leq x \leq 1$, show that $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$.

Solution:

The provided expression is $y=e^{a c o s^{-} x}$

$\frac{d y}{d x}=e^{a \cos ^{-1} x} \cdot \frac{d}{d x} a \cos ^{-1} x$

$=e^{a \operatorname{acs}^{-1 x} x} \cdot a\left(\frac{-1}{\sqrt{1-x^{2}}}\right)$

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

Which implies,

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

With respect to $x$ differentiating both the sides, we obtain

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

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

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

As a result, hence proved.

i

More Material