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If $x+y=\tan ^{-1} y$ and $\frac{d^{2} y}{d x^{2}}=f(y) \frac{d y}{d x}$, then $f(y)$ is equal to
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Verified Answer
The correct answer is:
$\frac{2}{y^{3}}$
Given, $x+y=\tan ^{-1} y$
On differentiating w.r.t. $x$, we get
$\begin{array}{ll}
& 1+\frac{d y}{d x}=\frac{1}{1+y^{2}} \cdot \frac{d y}{d x} \\
\Rightarrow \quad & \left(1-\frac{1}{1+y^{2}}\right) \frac{d y}{d x}=-1 \\
\Rightarrow \quad & \frac{y^{2}}{1+y^{2}} \frac{d y}{d x}=-1 \\
\Rightarrow \quad & \frac{d y}{d x}=-\left(\frac{1+y^{2}}{y^{2}}\right)=-1-\frac{1}{y^{2}}
\end{array}$
Again, differentiating w.r.t. $x$, we get
$\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=0+\frac{2}{y^{3}} \frac{d y}{d x} \\
\Rightarrow \quad \frac{d^{2} y}{d x^{2}} &=\frac{2}{y^{3}} \cdot \frac{d y}{d x} \text { but given, } \\
\frac{d^{2} y}{d x^{2}} &=f(y) \frac{d y}{d x}
\end{aligned}$
On comparing, we get
$f(y)=\frac{2}{y^{3}}$
On differentiating w.r.t. $x$, we get
$\begin{array}{ll}
& 1+\frac{d y}{d x}=\frac{1}{1+y^{2}} \cdot \frac{d y}{d x} \\
\Rightarrow \quad & \left(1-\frac{1}{1+y^{2}}\right) \frac{d y}{d x}=-1 \\
\Rightarrow \quad & \frac{y^{2}}{1+y^{2}} \frac{d y}{d x}=-1 \\
\Rightarrow \quad & \frac{d y}{d x}=-\left(\frac{1+y^{2}}{y^{2}}\right)=-1-\frac{1}{y^{2}}
\end{array}$
Again, differentiating w.r.t. $x$, we get
$\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=0+\frac{2}{y^{3}} \frac{d y}{d x} \\
\Rightarrow \quad \frac{d^{2} y}{d x^{2}} &=\frac{2}{y^{3}} \cdot \frac{d y}{d x} \text { but given, } \\
\frac{d^{2} y}{d x^{2}} &=f(y) \frac{d y}{d x}
\end{aligned}$
On comparing, we get
$f(y)=\frac{2}{y^{3}}$
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