Let $x=x(y)$ be the solution of the differential equation $y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0$. If $x(1)=1$, then $x\left(\frac{1}{2}\right)$ is :

Question

Aicharya · Accepted Answer

$\begin{aligned}
& y^2 d x+\left(x-\frac{1}{y}\right) d y=0 \\ & y^2 d x=\left(\frac{1}{y}-x\right) d y \\ & \Rightarrow y^2 \frac{d x}{d y}=\frac{1}{y}-x \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{y^2}=\frac{1}{y^3} \\ & \text { I.F. }=e^{\int \frac{1}{y^2} d y}=e^{\frac{-1}{y}}
\end{aligned}$
$\therefore$ Solution is
$x e^{\frac{-1}{y}}=\int e^{-\frac{1}{y}} \times \frac{1}{y^3} d y+C$
Let $\frac{-1}{y}=t$
$\begin{aligned}
& \Rightarrow \frac{1}{y^2} d y=d t \\ & \Rightarrow x e^{-\frac{1}{y}}=-\int e^t t d t+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^t(t-1)+C \\ & \Rightarrow x e^{-\frac{1}{y}}=-e^{\frac{-1}{y}}\left(\frac{-1}{y}-1\right)+C
\end{aligned}$
$x(1)=1$
$\Rightarrow e^{-1}=-e^{-1}(-2)+C$
$\Rightarrow C=-e^{-1}$
$\Rightarrow x=\frac{1}{y}+1-e^{-1+\frac{1}{y}}$
$x\left(\frac{1}{2}\right)=3-e$

Let $x=x(y)$ be the solution of the differential equation $y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right)…

Solution