Question

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

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

Answer 1

The correct option is A $3-e$

$x^{2}dy=(\frac{1}{x}-y)dx$
$x^2\frac{dy}{dx}=\frac{1}{x}-y$
$x^2\frac{dy}{dx}+y=\frac{1}{x}$
$\frac{dy}{dx}+\frac{1}{x^2}.y=\frac{1}{x^3}\cdots \left(i\right)$
$IF=e^{\int \frac{1}{x^2}dx}=e^{\frac{-1}{x}}$
Therefore, the solution is
$y.e^{\frac{-1}{x}}=\int e^{\frac{-1}{x}}\times \frac{1}{x^3}dx$
Let $\frac{-1}{x}=t,\frac{1}{x^2}dx=dt$
$y.e^{\frac{-1}{x}}=-\int e^{\left(t\right)}tdt=-e^t(t-1)+c$
$y{e}^{\frac{-1}{x}}=-e^{\frac{-1}{x}}(\frac{-1}{x}-1)+c$
$y=\frac{1}{x}+1+c{e}^{\frac{1}{x}}$
$y\left(1\right)=1\Rightarrow 1=2+ce\Rightarrow c=\frac{-1}{e}$
$y\left(\frac{1}{2}\right)=2+1+\left(\frac{-1}{e}\right).e^2$
$=3-e$