Question

If $x+y={\tan }^{-1}y$ and $\frac{d^{2}y}{d{x}^2}=f(y)\frac{dy}{dx}$ then $f(y)=$ ______

• A. $\frac{-2}{y^3}$
• B. $\frac{2}{y^3}$
• C. $\frac{1}{y}$
• D. $\frac{-1}{y}$

Answer 1

Answer: (B)

Given, $x+y={\tan }^{-1}y$
On differentiating w.r.t. $x$, we get
$1+\frac{dy}{dx}=\frac{1}{1+y^2}\cdot \frac{dy}{dx}$
$\Rightarrow \left(1-\frac{1}{1+y^2}\right)\frac{dy}{dx}=-1$
$\Rightarrow \frac{y^2}{1+y^2}\frac{dy}{dx}=-1$
$\Rightarrow \frac{dy}{dx}=-\left(\frac{1+y^2}{y^2}\right)=-1-\frac{1}{y^2}$
Again, differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{d{x}^2}=0+\frac{2}{y^3}\frac{dy}{dx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{2}{y^3}\cdot \frac{dy}{dx}$ but given,
$\frac{d^{2}y}{d{x}^2}=f(y)\frac{dy}{dx}$
On comparing, we get
$f(y)=\frac{2}{y^3}$