Question

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

• A. $2-e$
• B. e
• C. 1
• D. 3

Answer 1

Answer: (C)

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