Question

The function $y=f(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-4}=\frac{7x^6+2x}{\sqrt{4-x^2}}$ in $(-2,2)$ satisfying $f(0)=1$. If ${\int }_{-}^{\sqrt{3}}f(x)dx=a\pi +b\sqrt{3},a,b$ are rational numbers, then value of $3a+2b$ is

• A. 4
• B. 5
• C. 6
• D. none of these

Answer 1

Answer: (C)

$\frac{dy}{dx}\sqrt{4-x^2}-\frac{x}{\sqrt{4-x^2}}y=7x^6+2x$
$d\left(\sqrt{4-x^2}\cdot y\right)=\left(7x^6+2x\right)dx$
$\Rightarrow \sqrt{4-x^2}y=x^7+x^2+C$
$\because f(0)=1\Rightarrow c=2$
$\therefore y=\frac{x^7+x^2+2}{\sqrt{4-x^2}}$
$\underset{-\sqrt{3}}{\overset{\sqrt{3}}{\int }}\left(\frac{x^7}{\sqrt{4-x^2}}+\frac{x^2+2}{\sqrt{4-x^2}}\right)dx$
$=2\underset{0}{\overset{\sqrt{3}}{\int }}\frac{x^2+2}{\sqrt{4-x^2}}dx$
put $x=2\sin \theta$
$\Rightarrow dx=2\cos \theta d\theta$
$=2\underset{0}{\overset{\pi /3}{\int }}\frac{4{\sin }^2\theta +2}{2\cos \theta }\cdot 2\cos \theta d\theta$
$=2\underset{0}{\overset{\pi /3}{\int }}[4-2\cos 2\theta ]d\theta$
$=2\left[\frac{4\pi }{3}-\frac{\sqrt{3}}{2}\right]=\frac{8\pi }{3}-\sqrt{3}$
$\therefore a=\frac{8}{3},b=-1$