Let x=x(y) be the solution of the differential equation y^2 ~d x+(x-(1)/(y)) d y=0.

Question

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 :

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 $\frac{-1}{y}=t$

$$ \Rightarrow \frac{1}{y^2} d y=d t $$

$\begin{aligned} & \Rightarrow x e^{-\frac{1}{y}}=-\int e^t t d t+C \ & \Rightarrow x e^{-\frac{1}{y}}=-e^t(t-1)+C \ & \Rightarrow x e^{-\frac{1}{y}}=-e^{\frac{-1}{y}}\left(\frac{-1}{y}-1\right)+C\end{aligned}$

$\begin{aligned} & x(1)=1 \ & \Rightarrow e^{-1}=-e^{-1}(-2)+C \ & \Rightarrow C=-e^{-1} \ & \Rightarrow x=\frac{1}{y}+1-e^{-1+\frac{1}{y}} \ & x\left(\frac{1}{2}\right)=3-e\end{aligned}$

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