Question

If $x + y = \tan^{-1} y$ and $\frac {d^2y}{dx^2}=f(y) \frac {dy}{dx}$ 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{d y}{d x}=\frac{1}{1+y^{2}} \cdot \frac{d y}{d x} $
$\Rightarrow \left(1-\frac{1}{1+y^{2}}\right) \frac{d y}{d x}=-1 $
$\Rightarrow \frac{y^{2}}{1+y^{2}} \frac{d y}{d x}=-1 $
$\Rightarrow \frac{d y}{d x}=-\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{d y}{d x} $
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{2}{y^{3}} \cdot \frac{d y}{d x} $ but given,
$\frac{d^{2} y}{d x^{2}} =f(y) \frac{d y}{d x}$
On comparing, we get
$f(y)=\frac{2}{y^{3}}$