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Let $y=y(x)$ be the solution of the differential equation

$\sin x\frac{\mathrm{d} y}{\mathrm{d} x}+y\cos x= 4x$ ,

$x\epsilon \left ( 0,\pi \right )$ . If $y\left ( \frac{\pi }{2} \right )=0$ , then $y\left ( \frac{\pi }{6} \right )$ is equal to:

Option 1)
$\frac{-4}{9}\pi ^{2}$

Option 2)
$\frac{4}{9\sqrt3}\pi ^{2}$

Option 3)
$\frac{-8}{9\sqrt3}\pi ^{2}$

Option 4)
$\frac{-8}{9}\pi ^{2}$

Answers (2)

best_answer

$\frac{dy}{dx}+y\cot x= 4x \csc x$

IF = $e^{\int P(dx)}$

$=e^{\int \cot x dx}$ = $\sin x$

Thus $y \sin x = \int 4x \csc x \sin x dx$

$y \sin x = \int 4x dx = 2 x^{2}+c$

also $x= \pi /2 , y= 0$

$0= 2\left ( \frac{\pi }{2} \right )+c\Rightarrow c= -\pi ^{2}/2$

Thus $y \sin x = 2x^{2}- \pi ^{2}/2$

put $x= \pi /6$

and $y= -\frac{8}{9}\pi ^{2}$

Linear Differential Equation -

$\frac{dy}{dx}+Py= Q$

- wherein

P, Q are functions of x alone.

Option 1)

$\frac{-4}{9}\pi ^{2}$

Option 2)

$\frac{4}{9\sqrt3}\pi ^{2}$

Option 3)

$\frac{-8}{9\sqrt3}\pi ^{2}$

Option 4)

$\frac{-8}{9}\pi ^{2}$

Posted by

gaurav

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