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

Question

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

Accepted Answer

$\begin{gathered} {y}^{2}dx+(x-\frac{1}{y})dy=0 \ {y}^{2}dx=(\frac{1}{y}-x)dy \ \Rightarrow {y}^{2}\frac{dx}{dy}=\frac{1}{y}-x \ \Rightarrow \frac{dx}{dy}+\frac{x}{{y}^{2}}=\frac{1}{{y}^{3}} \ ={e}^{\int \frac{1}{{y}^{2}}dy}={e}^{\frac{-1}{y}} \end{gathered}$

$	herefore $ Solution is

$x{e}^{\frac{-1}{y}}=\int {e}^{-\frac{1}{y}}	imes \frac{1}{{y}^{3}}dy+C$

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

$\Rightarrow \frac{1}{{y}^{2}}dy=dt$

$\begin{gathered} \Rightarrow x{e}^{-\frac{1}{y}}=-\int {e}^{t}tdt+C \ \Rightarrow x{e}^{-\frac{1}{y}}=-{e}^{t}(t-1)+C \ \Rightarrow x{e}^{-\frac{1}{y}}=-{e}^{\frac{-1}{y}}(\frac{-1}{y}-1)+C \end{gathered}$

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

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