Let y=y(x) be the solution of (x^2+1)y'-2xy=(x^4+2x^2+1) x, y(0)=1.

Question

Let $y=y(x)$ be the solution of $(x^2+1)y'-2xy=(x^4+2x^2+1)\cos x$, $y(0)=1$. Then $\int_{-3}^3 y(x)\,dx$ is:

Accepted Answer

$$\frac{dy}{dx}-\frac{2x}{x^2+1}y = (x^2+1)\cos x.$$

I.F. $=\dfrac{1}{x^2+1}$.
$$\frac{y}{x^2+1} = \sin x+c.$$

Using $y(0)=1 \Rightarrow c=1$. So $y=(x^2+1)(\sin x+1)$.

$$\int_{-3}^3 y\,dx = \int_{-3}^3(x^2+1)\,dx = 0+18+0+6 = 24$$
(odd terms vanish).

Solutions · 1