Question

If the substitution $x={\tan }^{-1}(t)$ transforms the differential equation $\frac{d^{2}y}{d{x}^2}+x$ $y\frac{dy}{dx}+{\sec }^{2}x=0$ into a differential equation $1+t^2\frac{d^{2}y}{d{t}^2}+$ $2t+y{\tan }^{-1}(t)\frac{dy}{dt}+k=0$, then $k$ is equal to

• A. $-2$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\left(1+t^2\right)\frac{dy}{dt}...\left(1\right)$
$\frac{d^{2}y}{d{x}^2}=\frac{d\left(\frac{dy}{dx}\right)}{dx}=\left[\frac{d\left(\frac{dy}{dx}\right)}{dt}\right]\cdot \frac{dt}{dx}=\left[\left(1+t^2\right)\frac{d^{2}y}{d{t}^2}+(2t)\frac{dy}{dt}\right]\cdot \left(1+t^2\right)...\left(2\right)$
Also, $t=\tan x\Rightarrow {\left(sec\right)}^{2}x=1+t^2...\left(3\right)$
Putting these values, we get
$\left[\left(1+t^2\right)\frac{d^{2}y}{d{t}^2}+\left(2t\right)\frac{dy}{dt}\right]\cdot \left(1+t^2\right)+{\left(\tan \right)}^{-1}\left(t\right)\cdot y\cdot \left(1+t^2\right)\frac{dy}{dt}+\left(1+t^2\right)=0$
$\Rightarrow \left(1+t^2\right)\frac{d^{2}y}{d{t}^2}+(2t+{\left(\tan \right)}^{-1}\left(t\right)\cdot y)\frac{dy}{dt}+1=0$
Comparing, we get $k=1$ .