Question

If $y=y\left(x\right)$ is the solution of the differential equation $x^{2}dy+(y-\frac{1}{x})dx=0;x>0$ and $y\left(1\right)=1,$ then the value of $y\left(\frac{1}{2}\right)$ is

• A. $3-e$
• B. $3+\sqrt{e}$
• C. $\frac{3}{2}-\frac{1}{\sqrt{e}}$
• D. $3+e$

Answer 1

The correct option is A $3-e$

$x^{2}dy+(y-\frac{1}{x})dx=0;x>0$
$\Rightarrow \frac{dy}{dx}=\frac{1-xy}{x^3}$
$\Rightarrow \frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}$
$I.F.=e^{\int \frac{1}{x^2}dx}=e^{\frac{-1}{x}}$
$\Rightarrow y\cdot e^{-1/x}=\int e^{-1/x}\cdot \frac{1}{x^3}dx$
Let $-\frac{1}{x}=t\Rightarrow dx=x^{2}dt$
$\Rightarrow y\cdot e^{-1/x}=\int -e^{t}tdt$
$\Rightarrow y\cdot e^{-1/x}=e^{-1/x}(\frac{1}{x}+1)+c$
Given $y\left(1\right)=1$
$\Rightarrow c=-e^{-1}$
$\Rightarrow y\cdot e^{-1/x}=e^{-1/x}(\frac{1}{x}+1)-e^{-1}$
$\Rightarrow y=(\frac{1}{x}+1)-e^{1/x}{e}^{-1}$
$\therefore y\left(\frac{1}{2}\right)=3-e$