Let the solution curve y=f(x) of dydx+xyx^2-1=x^4+2x1-x^2, x(-1,1), pass through the origi…

Question

Let the solution curve $y=f(x)$ of $\dfrac{dy}{dx}+\dfrac{xy}{x^2-1}=\dfrac{x^4+2x}{1-x^2}$, $x\in(-1,1)$, pass through the origin. Then $\int_{-\sqrt3/2}^{\sqrt3/2}f(x)\,dx$ is:

Accepted Answer

I.F. $=\sqrt{1-x^2}$.
$$y\sqrt{1-x^2} = \int(x^4+2x)\,dx = \frac{x^5}{5}+x^2+c.$$

Using $y(0)=0 \Rightarrow c=0$. So $y=\dfrac{x^5+5x^2}{5\sqrt{1-x^2}}$.

$$\int_{-\sqrt3/2}^{\sqrt3/2}y\,dx = 2\int_0^{\sqrt3/2}\frac{x^2}{\sqrt{1-x^2}}\,dx.$$

With $x=\sin\theta$: $=2\int_0^{\pi/3}\sin^2\theta\,d\theta = \left[\theta-\frac{\sin2\theta}{2}\right]_0^{\pi/3} = \frac{\pi}{3}-\frac{\sqrt3}{4}$.

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