Question

Let $y=f(x)$ and $xy(1+y)dx=dy$ . If $f(0)=1$ and $kf(2)=(1+f(2))e^2,$ $k\in N,$ then $k$ is equal to
[Note : e denotes Napier's constant]

Answer 1

Answer: 2

We have, $xdx=\frac{dy}{y}-\frac{dy}{1+y}$
$\therefore$ On integration, we get
$C+\frac{x^2}{2}=Iny-In(1+y)$
$f(0)=1\Rightarrow C=-In2$
$\Rightarrow \frac{2y}{1+y}=e^{\frac{x^2}{2}}$
$\therefore 2f(x)=(1+(f(x))e^{\frac{x^2}{2}}$
Put $x=2$ . we get $2f(2)=(1+f(2))e^2$
$\therefore k=2$