Question

If $x+y={\tan }^{-1}y$ and $y^{\prime \prime }=f\left(y\right)y^{\prime }$ then $f\left(y\right)=$

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

Answer 1

The correct option is D $\frac{-2}{y(1+y^2)}$

$x+y=ta{n}^{-1}y$

differentiating w.r.t. x

$1+y^{\prime }=\frac{1}{1+y^2}{y}^{\prime }$

diffentiating once more,

$y^{\prime \prime }=\frac{-2y}{(1+y^2{)}^2}{y}^{\prime }+\frac{1}{1+y^2}{y}^{\prime \prime }$

$y^{\prime \prime }(1-\frac{1}{1+y^2})=\frac{-2y}{(1+y^2{)}^2}{y}^{\prime }$

$y^{\prime \prime }\left(\frac{y^2}{1+y^2}\right)=\frac{-2y}{(1+y^2{)}^2}{y}^{\prime }$

$y^{\prime \prime }=\frac{-2}{y(1+y^2)}{y}^{\prime }$

$f\left(y\right)=\frac{-2}{y(1+y^2)}$