Consider the differential equation $y^2 d x+\left(x-\frac{1}{y}\right) d y=0$. If $y(1)=1$, then $x$ is given by
Answer & explanation
Correct answer: option 2
We have,
$y^2 d x+\left(x-\frac{1}{y}\right) d y=0$
$\Rightarrow \frac{d x}{d y}+\frac{1}{y^2} x=\frac{1}{y^3}$
It is a linear differential equation with integrating factor $=e^{-1 / y}$. So, its solution is given by
$x e^{-1 / y}=\int \frac{1}{y^3} e^{-1 / y} d y+C$
$\Rightarrow x e^{-1 / y}=-e^{-1 / y}\left(\frac{1}{y}-1\right)+C$
It is given that $y(1)=1$.
∴ $-\frac{1}{e}=C$
Hence, $x e^{-1 / y}=-e^{-1 / y}\left(-\frac{1}{y}-1\right)-\frac{1}{e} \Rightarrow x=1+\frac{1}{y}-\frac{e^{1 / y}}{e}$