Question

Verify that $y=e^{m{\cos }^{-1}x}$ satisfies the differential equation $\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-m^{2}y=0$

Answer 1

We have,
$y=e^{m{\cos }^{-1}x}$ ...(1)
Differentiating both sides of (1) with respect to $x$, we get
$$\begin{gathered} \frac{dy}{dx}=m{e}^{m{\cos }^{-1}x}\left(\frac{-1}{\sqrt{1-x^2}}\right) \\ \Rightarrow \frac{dy}{dx}=-\frac{m{e}^{m{\cos }^{-1}x}}{\sqrt{1-x^2}}...\left(2\right) \end{gathered}$$
Differentiating both sides of (2) with respect to $x$, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(-\frac{m{e}^{m{\cos }^{-1}x}}{\sqrt{1-x^2}}\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\left(-m\right)\left[\frac{\sqrt{1-x^2}m{e}^{m{\cos }^{-1}x}\left(-{\displaystyle \frac{1}{\sqrt{1-x^2}}}\right)-e^{m{\cos }^{-1}x}{\displaystyle \frac{1}{2}}\left(-{\displaystyle \frac{2x}{\sqrt{1-x^2}}}\right)}{\left(1-x^2\right)}\right] \\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}=\left(-m\right)\left[-m{e}^{m{\cos }^{-1}x}+\frac{x{e}^{m{\cos }^{-1}x}}{\sqrt{1-x^2}}\right] \\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}=m^2{e}^{m{\cos }^{-1}x}-mx\frac{e^{m{\cos }^{-1}x}}{\sqrt{1-x^2}} \\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}=m^{2}y+x\frac{dy}{dx}\left[\text{Using}\left(1\right)\text{and}\left(2\right)\right] \\ \Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-m^{2}y=0 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.