Question

If $x+y=ta{n}^{-1}yand\frac{d^{2}y}{d{x}^2}=f\left(y\right)\frac{dy}{dx}$, then $f\left(y\right)=$ ______

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

Answer 1

Answer: (B)

$\frac{2}{y^3}$

$x+y=ta{n}^{-1}y$
$1+\frac{dy}{dx}=\frac{1}{1+y^2}.\frac{dy}{dx}$
$\frac{dy}{dx}\left(\frac{y^2}{1+y^2}\right)+1=0$
$\frac{dy}{dx}\left(\frac{y^2}{1+y^2}\right)=-1$
$\frac{dy}{dx}=\frac{-1-y^2}{y^2}$
$\frac{dy}{dx}=\frac{-1}{y^2}-1$
$\frac{d^{2}y}{d{x}^2}=\frac{2}{y^3}\frac{dy}{dx}$
Hence
$f\left(y\right)=\frac{2}{y^3}$