$\text{ If for }x\ge 0,y=y\left(x\right)\text{ is the solution of the differential equation, }$

$(x+1)dy=\left((x+1{)}^2+y-3\right)dx,y\left(2\right)=0,\text{ then }y\left(3\right)\text{ is equal to }$

$\begin{array}{l}\frac{dy}{dx}=(1+x)+\left(\frac{y-3}{1+x}\right)\\ \frac{dy}{dx}-\frac{1}{(1+x)}y=(1+x)-\frac{3}{(1+x)}\\ \text{ I.F. }=e^{-\int \frac{1}{(1+x)}dx}=\frac{1}{(1+x)}\end{array}$

$\begin{array}{l}\text{ So, the solution will be }\frac{y}{1+x}=\int \left[1-\frac{3}{(1+x{)}^2}\right]dx\\ \frac{y}{1+x}=x+3(1+x{)}^{-1}+c\\ y=(1+x)\left[x+\frac{3}{(1+x)}+c\right]\\ \text{ At }x=2,y=0\\ 0=3(2+1+c)\Rightarrow c=-3\end{array}$

$\text{ Hence the solution is }y=(1+x)\left(x+\frac{3}{1+x}-3\right)$

$$\begin{gathered} Substitute x=3 \\ It gives \end{gathered}$$

$\begin{array}{c}y=\left(1+3\right)\left(3+\frac{3}{1+3}-3\right)\\ =4\left(\frac{3}{4}\right)\\ =3\end{array}$