Let v be the solution of the differential equation (1-x^ 2 ) d y=… - Vidyadip

MCQ

Let $v$ be the solution of the differential equation $\left(1-x^{2}\right) d y=\left(xy+\left(x^{3}+2\right) \sqrt{1-x^{2}}\right) d x,-1 < x < 1$ and $y(0)=0$ if $\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^{2}} y(x) d x=k$ then $k^{-1}$ is equal to:

A
$320$
B
$321$
✓
$322$
D
$323$

✓

Answer

Correct option: C.

$322$

c
$\left(1-x^{2}\right) \frac{d y}{d x}=x y+\left(x^{3}+2\right) \sqrt{1-x^{2}}$

$\Rightarrow \frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{x^{3}+2}{\sqrt{1-x^{2}}}$

$I F=e^{\int \frac{-x}{1-x^{2}} d x}=\sqrt{1-x^{2}}$

$y(x) \cdot \sqrt{1-x^{2}}=\frac{x^{4}}{4}+2 x+c$

$y (0)=0 \Rightarrow c =0$

$\sqrt{1-x^{2}} y(x)=\frac{x^{4}}{4}+2 x$

required value $=\int\limits_{-1 / 2}^{1 / 2}\left(\frac{x^{4}}{4}+2 x\right) d x-\frac{1}{4} \cdot 2 \int\limits_{0}^{1 / 2} x^{4} d x$

$=\frac{1}{10}\left(x^{5}\right)_{0}^{1 / 2}=\frac{1}{320}$

$k ^{-1}=320$

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