If for x geq 0,y=y(x) is the solution of the | vedclass

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(D) The given differential equation is $(x+1) dy = ((x+1)^{2} + y - 3) dx$.Rearranging the terms,we get $(x+1) dy - y dx = ((x+1)^{2} - 3) dx$.Dividin

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$3$

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(D) The given differential equation is $(x+1) dy = ((x+1)^{2} + y - 3) dx$.Rearranging the terms,we get $(x+1) dy - y dx = ((x+1)^{2} - 3) dx$.Dividing both sides by $(x+1)^{2}$,we obtain $\frac{(x+1) dy - y dx}{(x+1)^{2}} = \left(1 - \frac{3}{(x+1)^{2}}\right) dx$.This is equivalent to $d\left(\frac{y}{x+1}\right) = \left(1 - \frac{3}{(x+1)^{2}}\right) dx$.Integrating both sides,we get $\frac{y}{x+1} = x + \frac{3}{x+1} + C$.Given $y(2) = 0$,we substitute $x=2$ and $y=0$: $0 = 2 + \frac{3}{3} + C \Rightarrow 0 = 3 + C \Rightarrow C = -3$.Thus,the solution is $\frac{y}{x+1} = x + \frac{3}{x+1} - 3$.Multiplying by $(x+1)$,we get $y = x(x+1) + 3 - 3(x+1) = x^{2} + x + 3 - 3x - 3 = x^{2} - 2x$.For $y(3)$,we substitute $x=3$: $y(3) = 3^{2} - 2(3) = 9 - 6 = 3$.

If for $x \geq 0$,$y=y(x)$ is the solution of the differential equation $(x+1) dy = ((x+1)^{2} + y - 3) dx$ with $y(2) = 0$,then $y(3)$ is equal to:

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