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Differential Equations question

2020 · 5 Sep · Shift 2 · Q20
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  5. /2020 · 5 Sep · Shift 2 · Q20

Differential Equations question

2020 · 5 Sep · Shift 2 · Q20

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation cosx dydx{{dy} \over {dx}}dxdy​+ 2ysinx = sin2x, x ∈\in∈ (0,π2)\left( {0,{\pi \over 2}} \right)(0,2π​). If y (π3)\left( {{\pi \over 3}} \right)(3π​) = 0, then y (π4)\left( {{\pi \over 4}} \right)(4π​) is equal to :
  1. A
    12−1{1 \over {\sqrt 2 }} - 12​1​−1
  2. B
    2−2{\sqrt 2 - 2}2​−2
  3. C
    2−2{2 - \sqrt 2 }2−2​
  4. D
    2+2{2 + \sqrt 2 }2+2​
View written solutionFree

Correct answer: B

  1. Write the differential equation in standard linear form

Given

cos⁡x dydx+2ysin⁡x=sin⁡2x,x∈(0,π2).\cos x\,\frac{dy}{dx}+2y\sin x=\sin 2x, \qquad x\in\left(0,\frac{\pi}{2}\right).cosxdxdy​+2ysinx=sin2x,x∈(0,2π​).

Using sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos xsin2x=2sinxcosx, divide throughout by cos⁡x\cos xcosx (valid since x∈(0,π/2)x\in(0,\pi/2)x∈(0,π/2), so cos⁡x>0\cos x>0cosx>0):

dydx+2ytan⁡x=2sin⁡x.\frac{dy}{dx}+2y\tan x=2\sin x.dxdy​+2ytanx=2sinx.

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)=2tan⁡x,Q(x)=2sin⁡x.P(x)=2\tan x, \qquad Q(x)=2\sin x.P(x)=2tanx,Q(x)=2sinx.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫2tan⁡x dx.\text{I.F.}=e^{\int 2\tan x\,dx}.I.F.=e∫2tanxdx.

Now,

∫tan⁡x dx=−ln⁡(cos⁡x),\int \tan x\,dx = -\ln(\cos x),∫tanxdx=−ln(cosx),

so

∫2tan⁡x dx=−2ln⁡(cos⁡x)=ln⁡(sec⁡2x).\int 2\tan x\,dx = -2\ln(\cos x)=\ln(\sec^2 x).∫2tanxdx=−2ln(cosx)=ln(sec2x).

Hence,

I.F.=eln⁡(sec⁡2x)=sec⁡2x.\text{I.F.}=e^{\ln(\sec^2 x)}=\sec^2 x.I.F.=eln(sec2x)=sec2x.
  1. Multiply the equation by the integrating factor

Multiplying by sec⁡2x\sec^2 xsec2x:

sec⁡2xdydx+2ytan⁡xsec⁡2x=2sin⁡xsec⁡2x.\sec^2 x\frac{dy}{dx}+2y\tan x\sec^2 x=2\sin x\sec^2 x.sec2xdxdy​+2ytanxsec2x=2sinxsec2x.

The left side becomes

ddx(ysec⁡2x).\frac{d}{dx}(y\sec^2 x).dxd​(ysec2x).

Therefore,

ddx(ysec⁡2x)=2sin⁡xsec⁡2x.\frac{d}{dx}(y\sec^2 x)=2\sin x\sec^2 x.dxd​(ysec2x)=2sinxsec2x.

Simplify the right side:

2sin⁡xsec⁡2x=2sin⁡xcos⁡2x=2tan⁡xsec⁡x.2\sin x\sec^2 x = 2\frac{\sin x}{\cos^2 x}=2\tan x\sec x.2sinxsec2x=2cos2xsinx​=2tanxsecx.

So,

ddx(ysec⁡2x)=2tan⁡xsec⁡x.\frac{d}{dx}(y\sec^2 x)=2\tan x\sec x.dxd​(ysec2x)=2tanxsecx.
  1. Integrate both sides

Integrating,

ysec⁡2x=∫2tan⁡xsec⁡x dx.y\sec^2 x=\int 2\tan x\sec x\,dx.ysec2x=∫2tanxsecxdx.

Since

ddx(sec⁡x)=sec⁡xtan⁡x,\frac{d}{dx}(\sec x)=\sec x\tan x,dxd​(secx)=secxtanx,

we get

∫2tan⁡xsec⁡x dx=2sec⁡x+C.\int 2\tan x\sec x\,dx = 2\sec x + C.∫2tanxsecxdx=2secx+C.

Thus,

ysec⁡2x=2sec⁡x+C.y\sec^2 x=2\sec x + C.ysec2x=2secx+C.

Multiply by cos⁡2x\cos^2 xcos2x:

y=2cos⁡x+Ccos⁡2x.y = 2\cos x + C\cos^2 x.y=2cosx+Ccos2x.
  1. Use the given condition y(π3)=0y\left(\frac{\pi}{3}\right)=0y(3π​)=0

Substitute x=π3x=\frac{\pi}{3}x=3π​:

0=2cos⁡π3+Ccos⁡2π3.0 = 2\cos\frac{\pi}{3} + C\cos^2\frac{\pi}{3}.0=2cos3π​+Ccos23π​.

Now,

cos⁡π3=12,cos⁡2π3=14.\cos\frac{\pi}{3}=\frac12, \qquad \cos^2\frac{\pi}{3}=\frac14.cos3π​=21​,cos23π​=41​.

So,

0=2⋅12+C⋅14=1+C4.0 = 2\cdot\frac12 + C\cdot\frac14 = 1+\frac{C}{4}.0=2⋅21​+C⋅41​=1+4C​.

Hence,

C=−4.C=-4.C=−4.

Therefore,

y=2cos⁡x−4cos⁡2x.y = 2\cos x - 4\cos^2 x.y=2cosx−4cos2x.
  1. Find y(π4)y\left(\frac{\pi}{4}\right)y(4π​)

Substitute x=π4x=\frac{\pi}{4}x=4π​:

y(π4)=2cos⁡π4−4cos⁡2π4.y\left(\frac{\pi}{4}\right)=2\cos\frac{\pi}{4}-4\cos^2\frac{\pi}{4}.y(4π​)=2cos4π​−4cos24π​.

Using

cos⁡π4=12,cos⁡2π4=12,\cos\frac{\pi}{4}=\frac{1}{\sqrt2}, \qquad \cos^2\frac{\pi}{4}=\frac12,cos4π​=2​1​,cos24π​=21​,

we get

y(π4)=2⋅12−4⋅12=2−2.y\left(\frac{\pi}{4}\right)=2\cdot\frac{1}{\sqrt2}-4\cdot\frac12 =\sqrt2-2.y(4π​)=2⋅2​1​−4⋅21​=2​−2.
  1. Match with the options
2−2\sqrt2-22​−2

corresponds to Option B.

  1. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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