Question

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

• A. $-2$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

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