Let $x=x(y)$ be the solution of the differential equation $y^2\,dx+\left(x-\frac1y\right)dy=0$. If $x(1)=1$, then $x\left(\frac12\right)$ is:

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\frac{dx}{dy}+\frac{x}{y^2}=\frac1{y^3}$; the integrating factor is $e^{-1/y}$, giving $x=1+\frac1y+Ce^{1/y}$. $x(1)=1\Rightarrow C=-e^{-1}$, so $x(	frac12)=1+2-e=3-e$. Answer (3).

Q. Let $x=x(y)$ be the solution of the differential equation $y^2\,dx+\left(x-\frac1y\right)dy=0$. If $x(1)=1$, then $x\left(\frac12\right)$ is:

Explanation

Reviewed by Dr. Abubacker

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