If for x >0, y=y(x) is the solution


If for $x \geq 0, y=y(x)$ is the solution of the differential equation,

$(x+1) d y=\left((x+1)^{2}+y-3\right) d x, y(2)=0$

then $y(3)$ is equal to__________.



$(x+1) d y \quad((x+1)+(y \quad 3)) d x \quad 0$

$\Rightarrow \frac{d y}{d x}=(1+x)+\left(\frac{y-3}{1+x}\right)$

$\frac{d y}{d x}-\frac{1}{(1+x)} y=(1+x)-\frac{3}{(1+x)}$

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

$\therefore \quad \frac{d}{d x}\left(\frac{y}{1+x}\right)=1-\frac{3}{(1+x)^{2}}$



$\because \quad$ At $x=2, y=0$

$\therefore \quad 0=3(2+1+C) \Rightarrow C=-3$

Then, $y=(1+x)\left[x+\frac{3}{1+x}-3\right]$

Now, at $x=3, y=(1+3)\left[3+\frac{3}{1+3}-3\right]=3$


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