Discussion Forum

1)

Let y= y(x) be the solution  of the different equation 

$\sin x\frac{dy}{dx}+ycosx=4x,x\epsilon (0,\pi)$. If  $y(\frac{\pi}{2})=0$, then $y(\frac{\pi}{6})$ is equal to

Answer:

Option C

Explanation:

We have

    $\sin x\frac{dy}{dx}+y\cos x=4x$

$\Rightarrow$          $\frac{dy}{dx}+y\cot x=4x\times cosec x$

This is a linear differential equation of form

         $\frac{dy}{dx}+Py=Q$

where P= cot x, Q= 4x cosec x

Now,    IF= $e^{\int_{}^{} Pdx}=e^{\cot x dx}=e^{\log_{}{} \sin x}=\sin x$

Solution of the differential equation is

                      $y. \sin x= \int_{}^{} 4x cosecx\sin xdx +C$

$\Rightarrow$      $y \sin x=\int_{}^{} 4xdx+C=2x^{2}+C$

 Put   $x=\frac{\pi}{2},y=0,$ we get

        $C= \frac{-\pi^{2}}{2}$

$\Rightarrow$     $y \sin x=2x^{2}-\frac{\pi^{2}}{2}$

Put       $x=\frac{\pi}{6}$

  $y(\frac{1}{2})=2(\frac{\pi^{2}}{36})-\frac{\pi^{2}}{2}$

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

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

Alternative method

We have ,  $\sin x\frac{dy}{dx}+y\cos x=4x$ ,

which can be written as $\frac{d}{dx}(\sin x.y)=4x$

On integrating both sides, we get

      $\int_{}^{}\frac{d}{dx}(\sin x.y)dx= \int_{}^{} 4x.dx$

$\Rightarrow$     $y.\sin x=\frac{4x^{2}}{2}+C$

$\Rightarrow$      $y.\sin x=2x^{2}+C$

Now, as y=0 when $x=\frac{\pi}{2}$

$\therefore$  C= $-\frac{\pi^{2}}{2}$

$\Rightarrow$   $y.\sin x=2x^{2}-\frac{\pi^{2}}{2}$

Now , putting  $x=\frac{\pi}{6}$, we get

$y(\frac{1}{2})=2(\frac{\pi^{2}}{36})-\frac{\pi^{2}}{2}$

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