If for x≥ 0,y=y(x) is the solution of the differential equation, (x+1)dy=((x+1)^2+y-3)dx,y(2)=0 then y(3) is equal to

$$\frac{dy}{dx}=(1+x)+(\frac{y-3}{1+x})$$dxdy​=(1+x)+(1+xy−3​)

$$\frac{dy}{dx}-\frac{1}{(1+x)}y=(1+x)-\frac{3}{(1+x)}$$dxdy​−(1+x)1​y=(1+x)−(1+x)3​

I.F$\displaystyle ={e}^{-\int \frac{1}{(1+x)}dx}=\frac{1}{(1+x)}$=e−∫(1+x)1​dx=(1+x)1​

$$\therefore \frac{d}{dx}(\frac{y}{1+x})=x+3{(1+x)}^{-1}+c$$∴dxd​(1+xy​)=x+3(1+x)−1+c

$$\Rightarrow \frac{y}{1+x}=x+3{(1+x)}^{-1}+c$$⇒1+xy​=x+3(1+x)−1+c

$\Rightarrow y=(1+x)[x+\frac{3}{(1+x)}+c]$⇒y=(1+x)[x+(1+x)3​+c] , $\therefore$∴ at $x=2,y=0$x=2,y=0

$\Rightarrow 0=3(2+1+c)$⇒0=3(2+1+c) $\Rightarrow c=-3$⇒c=−3

At $x=3,y=3$x=3,y=3