Question

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

Answer 1

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