Question

Let the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}},x\in (-1,1)$ pass through the origin. Then $\underset{-\frac{\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}f(x)dx$ 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)

$\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$
I.F $=e^{\int \frac{x}{x^2-1}dx}$
I.F $=\sqrt{1-x^2}$
Solution of D.E.
$y\cdot \sqrt{1-x^2}=\int \frac{x^4+2x}{\sqrt{1-x^2}}\cdot \sqrt{1-x^2}dx$
$y\cdot \sqrt{1-x^2}=\int \left(x^4+2x\right)dx$
$y\cdot \sqrt{1-x^2}=\frac{x^5}{5}+x^2+C$
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,
$\underset{\frac{-\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}f(x)dx=\underset{\frac{-\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}\frac{x^5}{5\sqrt{1-x^2}}dx+\underset{\frac{-\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}\frac{x^2}{\sqrt{1-x^2}}dx$
$\underset{\frac{-\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}f(x)dx=0+2\underset{0}{\overset{\frac{\sqrt{3}}{2}}{\int }}\frac{x^2}{\sqrt{1-x^2}}dx$
$\underset{\frac{-\sqrt{3}}{2}}{\overset{\frac{\sqrt{3}}{2}}{\int }}f(x)dx=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$