Question

If for $x\ge 0$ , $y=y\left(x\right)$ is the solution of the differential equation $(1+x)dy=\left[\right(1+x{)}^2+y-3]dx,y(2)=0,$ then y(3) is equal to

Answer 1

$(1+x)\frac{dy}{dx}=\left[\right(1+x{)}^2+(y-3)]$
$\Rightarrow (1+x)\frac{dy}{dx}-y=(1+x{)}^2-3$
$\Rightarrow \frac{dy}{dx}-\frac{1}{(1+x)}y=1+x-\frac{3}{1+x}$

$I.F=e^{-\int \frac{1}{1+x}dx}=\frac{1}{1+x}$

$\Rightarrow y\times \frac{1}{1+x}=\int 1-\frac{3}{(1+x{)}^2}dx$
$\Rightarrow \frac{y}{1+x}=x+\frac{3}{1+x}+c$
$\Rightarrow y=x(1+x)+3+c(1+x)$

At $x=2,y=0,$ we get
$0=6+3+3c$
$\Rightarrow c=-3$
At $x=3,$
$y=x^2-2x=9-6=3\Rightarrow y\left(3\right)=3$