Question

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

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

Answer 1

Answer: (C)

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