Question

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

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

Answer 1

Answer: (D)

$x^{2}dy+\left(y-\frac{1}{x}\right)dx=0:x>0,y(1)=1$
$x^{2}dy+\frac{(xy-1)}{x}dx=0$
$x^{2}dy=\frac{(xy-1)}{x}dx$
$\frac{dy}{dx}=\frac{1-xy}{x^3}$
$\frac{dy}{dx}=\frac{1}{x^3}-\frac{y}{x^2}$
$\frac{dy}{dx}=\frac{1}{x^2}\cdot y=\frac{1}{x^3}$
If $e^{\int \frac{1}{x^2}dx}=e^{\frac{1}{x}}$
$y{e}^{-\frac{1}{x}}=\int \frac{1}{x^3}\cdot e^{-\frac{1}{x}}$
$y{e}^{-\frac{1}{x}}=e^{-x}\left(1+\frac{1}{x}\right)+C$
$1.e^{-1}=e^{-1}(2)+C$
$C=-e^{-1}=-\frac{1}{e}$
$y{e}^{-\frac{1}{x}}=e^{-\frac{1}{x}}\left(1+\frac{1}{x}\right)-\frac{1}{e}$
$y\left(\frac{1}{2}\right)=3-\frac{1}{e}\times e^2$
$y\left(\frac{1}{2}\right)=3-e$