Question

Let the solution curve $y=f(x)$ of the differential equation $\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}}, x \in(-1,1)$ pass through the origin. Then $\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$ is equal to

• A. $\frac{\pi}{3}-\frac{1}{4}$
• B. $\frac{\pi}{3}-\frac{\sqrt{3}}{4}$
• C. $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$
• D. $\frac{\pi}{6}-\frac{\sqrt{3}}{2}$

Answer 1

Answer: (B)

$ \frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}} $
I.F $=e^{\int \frac{x}{x^2-1} d x} $
I.F $=\sqrt{1-x^2}$
Solution of D.E.
$y \cdot \sqrt{1-x^2}=\int \frac{x^4+2 x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} d x $
$y \cdot \sqrt{1-x^2}=\int\left(x^4+2 x\right) d x $
$ y \cdot \sqrt{1-x^2}=\frac{x^5}{5}+x^2+C $
At $ x=0, y=0, $ get $ C=0 $
$y=\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}$
Now,
$ \int\limits_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x=\int\limits_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^5}{5 \sqrt{1-x^2}} d x+\int\limits_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} d x $
$ \int\limits_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x=0+2 \int\limits_0^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} d x $
$ \int\limits_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x=\frac{\pi}{3}-\frac{\sqrt{3}}{4}$