Question

Let $y=y(x)$ be the solution of the differential equation $x^{3}dy+(xy-1)dx=0,x>0$, $y\left(\frac{1}{2}\right)=3-e$. Then $y(1)$ is equal to

• A. $2-e$
• B. e
• C. 1
• D. 3

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{1-xy}{x^3}=\frac{1}{x^3}-\frac{y}{x^2}$
$\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}$
$\text{ If }=e^{\int \frac{1}{x^2}dx}=e^{-\frac{1}{x}}$
$y\cdot e^{-\frac{1}{x}}=\int e^{-\frac{1}{x}}\cdot \frac{1}{x^3}dx\left(\text{ put }-\frac{1}{x}=t\right)$
$y\cdot e^{-\frac{1}{x}}=-\int e^t\cdot tdt$
$y=\frac{1}{x}+1+C{e}^{\frac{1}{x}}$
Where $C$ is constant
Put $x=\frac{1}{2}$
$3-e=2+1+C{e}^2$
$C=-\frac{1}{e}$
$y(1)=1$