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Differential Equations question

2018 · Shift 0 · Q38
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  5. /2018 · Shift 0 · Q38

Differential Equations question

2018 · Shift 0 · Q38

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation sin⁡xdydx+ycos⁡x=4x\sin x{{dy} \over {dx}} + y\cos x = 4xsinxdxdy​+ycosx=4x, x∈(0,π)x \in \left( {0,\pi } \right)x∈(0,π). If y(π2)=0y\left( {{\pi \over 2}} \right) = 0y(2π​)=0, then y(π6)y\left( {{\pi \over 6}} \right)y(6π​) is equal to :
  1. A
    −49π2- {4 \over 9}{\pi ^2}−94​π2
  2. B
    493π2{4 \over {9\sqrt 3 }}{\pi ^2}93​4​π2
  3. C
    −893π2- {8 \over {9\sqrt 3 }}{\pi ^2}−93​8​π2
  4. D
    −89π2- {8 \over 9}{\pi ^2}−98​π2
View written solutionFree

Correct answer: D

  1. Write the differential equation in standard linear form

Given

sin⁡x dydx+ycos⁡x=4x,x∈(0,π).\sin x\,\frac{dy}{dx}+y\cos x=4x, \qquad x\in(0,\pi).sinxdxdy​+ycosx=4x,x∈(0,π).

Since sin⁡x>0\sin x>0sinx>0 on (0,π)(0,\pi)(0,π), divide throughout by sin⁡x\sin xsinx:

dydx+ycot⁡x=4xcsc⁡x.\frac{dy}{dx}+y\cot x=4x\csc x.dxdy​+ycotx=4xcscx.

This is a linear differential equation:

dydx+P(x)y=Q(x),\frac{dy}{dx}+P(x)y=Q(x),dxdy​+P(x)y=Q(x),

with

P(x)=cot⁡x.P(x)=\cot x.P(x)=cotx.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫cot⁡x dx=eln⁡(sin⁡x)=sin⁡x.\text{I.F.}=e^{\int \cot x\,dx}=e^{\ln(\sin x)}=\sin x.I.F.=e∫cotxdx=eln(sinx)=sinx.
  1. Multiply the equation by the integrating factor

Multiplying by sin⁡x\sin xsinx:

sin⁡xdydx+ysin⁡xcot⁡x=4x.\sin x\frac{dy}{dx}+y\sin x\cot x=4x.sinxdxdy​+ysinxcotx=4x.

Since sin⁡xcot⁡x=cos⁡x\sin x\cot x=\cos xsinxcotx=cosx, this becomes

sin⁡xdydx+ycos⁡x=4x.\sin x\frac{dy}{dx}+y\cos x=4x.sinxdxdy​+ycosx=4x.

But the left-hand side is exactly

ddx(ysin⁡x).\frac{d}{dx}(y\sin x).dxd​(ysinx).

Hence,

ddx(ysin⁡x)=4x.\frac{d}{dx}(y\sin x)=4x.dxd​(ysinx)=4x.
  1. Integrate

Integrating with respect to xxx:

ysin⁡x=∫4x dx=2x2+C.y\sin x=\int 4x\,dx=2x^2+C.ysinx=∫4xdx=2x2+C.

So,

y=2x2+Csin⁡x.y=\frac{2x^2+C}{\sin x}.y=sinx2x2+C​.
  1. Use the given condition

Given

y(π2)=0.y\left(\frac{\pi}{2}\right)=0.y(2π​)=0.

Substitute x=π2x=\frac{\pi}{2}x=2π​:

0=2(π2)2+Csin⁡(π2).0=\frac{2\left(\frac{\pi}{2}\right)^2+C}{\sin\left(\frac{\pi}{2}\right)}.0=sin(2π​)2(2π​)2+C​.

Since sin⁡(π/2)=1\sin(\pi/2)=1sin(π/2)=1,

2⋅π24+C=02\cdot\frac{\pi^2}{4}+C=02⋅4π2​+C=0 π22+C=0\frac{\pi^2}{2}+C=02π2​+C=0 C=−π22.C=-\frac{\pi^2}{2}.C=−2π2​.

Thus,

y=2x2−π22sin⁡x.y=\frac{2x^2-\frac{\pi^2}{2}}{\sin x}.y=sinx2x2−2π2​​.
  1. Find y(π6)y\left(\frac{\pi}{6}\right)y(6π​)

Substitute x=π6x=\frac{\pi}{6}x=6π​:

y(π6)=2(π6)2−π22sin⁡(π6).y\left(\frac{\pi}{6}\right)=\frac{2\left(\frac{\pi}{6}\right)^2-\frac{\pi^2}{2}}{\sin\left(\frac{\pi}{6}\right)}.y(6π​)=sin(6π​)2(6π​)2−2π2​​.

Now,

2\left(\frac{\pi}{6}\right)^2=2\cdot\frac{\pi^2}{36}=\frac\pi^2{18}.

So the numerator is

π218−π22=π2(118−918)=−8π218=−4π29.\frac{\pi^2}{18}-\frac{\pi^2}{2}=\pi^2\left(\frac{1}{18}-\frac{9}{18}\right)=-\frac{8\pi^2}{18}=-\frac{4\pi^2}{9}.18π2​−2π2​=π2(181​−189​)=−188π2​=−94π2​.

Also,

sin⁡(π6)=12.\sin\left(\frac{\pi}{6}\right)=\frac12.sin(6π​)=21​.

Therefore,

y(π6)=−4π2912=−8π29.y\left(\frac{\pi}{6}\right)=\frac{-\frac{4\pi^2}{9}}{\frac12}=-\frac{8\pi^2}{9}.y(6π​)=21​−94π2​​=−98π2​.
  1. Compare with the options

Thus,

y(π6)=−89π2.y\left(\frac{\pi}{6}\right)=-\frac{8}{9}\pi^2.y(6π​)=−98​π2.

This matches Option D.

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