For x0, y=y(x) solves (x+1),dy=((x+1)^2+y-3),dx, y(2)=0.

Question

For $x\ge0$, $y=y(x)$ solves $(x+1)\,dy=((x+1)^2+y-3)\,dx$, $y(2)=0$. Then $y(3)$ is \_\_\_\_\_\_.

Answer: 3

Accepted Answer

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

I.F.$=\dfrac{1}{x+1}$.
$$\frac{y}{x+1} = x+\frac{3}{x+1}+C.$$

Using $y(2)=0$: $0=3(2+1+C) \Rightarrow C=-3$. At $x=3$: $y=4\left(3+\dfrac34-3\right)=3$.

Solutions · 1