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Differential Equations question

2021 · 16 Mar · Shift 2 · Q24
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  5. /2021 · 16 Mar · Shift 2 · Q24

Differential Equations question

2021 · 16 Mar · Shift 2 · Q24

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If y = y(x) is the solution of the differential equation dydx{{dy} \over {dx}}dxdy​+ (tan x) y = sin x, 0≤x≤π30 \le x \le {\pi \over 3}0≤x≤3π​, with y(0) = 0, then y(π4)y\left( {{\pi \over 4}} \right)y(4π​) equal to :
  1. A
    12{1 \over 2}21​ loge 2
  2. B
    (122)\left( {{1 \over {2\sqrt 2 }}} \right)(22​1​) loge 2
  3. C
    loge 2
  4. D
    14{1 \over 4}41​ loge 2
View written solutionFree

Correct answer: B

  1. Given differential equation

We need to solve

dydx+(tan⁡x)y=sin⁡x,\frac{dy}{dx} + (\tan x) y = \sin x,dxdy​+(tanx)y=sinx,

with initial condition

y(0)=0.y(0)=0.y(0)=0.

Then find

y(π4).y\left(\frac{\pi}{4}\right).y(4π​).
  1. Identify it as a linear differential equation

This is of the form

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

where

P(x)=tan⁡x,Q(x)=sin⁡x.P(x)=\tan x, \qquad Q(x)=\sin x.P(x)=tanx,Q(x)=sinx.

The integrating factor is

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

Now,

∫tan⁡x dx=∫sin⁡xcos⁡x dx=−ln⁡(cos⁡x).\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = -\ln(\cos x).∫tanxdx=∫cosxsinx​dx=−ln(cosx).

So,

I.F.=e−ln⁡(cos⁡x)=sec⁡x.\text{I.F.} = e^{-\ln(\cos x)} = \sec x.I.F.=e−ln(cosx)=secx.
  1. Multiply the equation by the integrating factor

Multiplying throughout by sec⁡x\sec xsecx,

sec⁡xdydx+sec⁡xtan⁡x y=sec⁡xsin⁡x.\sec x\frac{dy}{dx} + \sec x\tan x\, y = \sec x\sin x.secxdxdy​+secxtanxy=secxsinx.

The left-hand side becomes

ddx(ysec⁡x),\frac{d}{dx}(y\sec x),dxd​(ysecx),

so

ddx(ysec⁡x)=sec⁡xsin⁡x=tan⁡x.\frac{d}{dx}(y\sec x) = \sec x\sin x = \tan x.dxd​(ysecx)=secxsinx=tanx.

Thus,

ysec⁡x=∫tan⁡x dx+C=−ln⁡(cos⁡x)+C.y\sec x = \int \tan x\,dx + C = -\ln(\cos x) + C.ysecx=∫tanxdx+C=−ln(cosx)+C.

Hence,

y=cos⁡x [−ln⁡(cos⁡x)+C].y = \cos x\,[-\ln(\cos x)+C].y=cosx[−ln(cosx)+C].
  1. Use the initial condition y(0)=0y(0)=0y(0)=0

Substitute x=0x=0x=0:

y(0)=cos⁡0 [−ln⁡(cos⁡0)+C]=1⋅[−ln⁡1+C]=C.y(0)=\cos 0\,[-\ln(\cos 0)+C] = 1\cdot[-\ln 1 + C] = C.y(0)=cos0[−ln(cos0)+C]=1⋅[−ln1+C]=C.

Since y(0)=0y(0)=0y(0)=0, we get

C=0.C=0.C=0.

So the solution is

y=−cos⁡xln⁡(cos⁡x).y = -\cos x\ln(\cos x).y=−cosxln(cosx).

This can also be written as

y=cos⁡xln⁡(sec⁡x).y = \cos x\ln(\sec x).y=cosxln(secx).
  1. Evaluate at x=π4x=\frac{\pi}{4}x=4π​

Since

cos⁡π4=12,\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}},cos4π​=2​1​,

we get

y(π4)=−12ln⁡(12).y\left(\frac{\pi}{4}\right)= -\frac{1}{\sqrt{2}}\ln\left(\frac{1}{\sqrt{2}}\right).y(4π​)=−2​1​ln(2​1​).

Now,

ln⁡(12)=−ln⁡(2)=−12ln⁡2.\ln\left(\frac{1}{\sqrt{2}}\right)= -\ln(\sqrt{2})= -\frac{1}{2}\ln 2.ln(2​1​)=−ln(2​)=−21​ln2.

Therefore,

y(π4)=−12(−12ln⁡2)=122ln⁡2.y\left(\frac{\pi}{4}\right)= -\frac{1}{\sqrt{2}}\left(-\frac{1}{2}\ln 2\right) = \frac{1}{2\sqrt{2}}\ln 2.y(4π​)=−2​1​(−21​ln2)=22​1​ln2.
  1. Compare with options

Thus,

y(π4)=122ln⁡2.y\left(\frac{\pi}{4}\right)= \frac{1}{2\sqrt{2}}\ln 2.y(4π​)=22​1​ln2.

This matches Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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