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Differential Equations question

2025 · 3 Apr · Shift 1 · Q39
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  5. /2025 · 3 Apr · Shift 1 · Q39

Differential Equations question

2025 · 3 Apr · Shift 1 · Q39

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let ggg be a differentiable function such that ∫0xg(t)dt=x−∫0xtg⁡(t)dt,x≥0\int_0^x g(t) d t=x-\int_0^x \operatorname{tg}(t) d t, x \geq 0∫0x​g(t)dt=x−∫0x​tg(t)dt,x≥0 and let y=y(x)y=y(x)y=y(x) satisfy the differential equation dydx−ytan⁡x=2(x+1)sec⁡xg(x),x∈[0,π2)\frac{d y}{d x}-y \tan x=2(x+1) \sec x g(x), x \in\left[0, \frac{\pi}{2}\right)dxdy​−ytanx=2(x+1)secxg(x),x∈[0,2π​). If y(0)=0y(0)=0y(0)=0, then y(π3)y\left(\frac{\pi}{3}\right)y(3π​) is equal to
  1. A
    4π3\frac{4 \pi}{3}34π​
  2. B
    2π3\frac{2 \pi}{3}32π​
  3. C
    2π33\frac{2 \pi}{3 \sqrt{3}}33​2π​
  4. D
    4π33\frac{4 \pi}{3 \sqrt{3}}33​4π​
View written solutionFree

Correct answer: A

  1. Interpret the given integral equation for ggg.

We are given

∫0xg(t) dt=x−∫0xtg(t) dt,x≥0.\int_0^x g(t)\,dt = x - \int_0^x t g(t)\,dt, \qquad x\ge 0.∫0x​g(t)dt=x−∫0x​tg(t)dt,x≥0.

Differentiate both sides with respect to xxx using the Fundamental Theorem of Calculus:

g(x)=1−xg(x).g(x)=1-xg(x).g(x)=1−xg(x).

So,

g(x)(1+x)=1  ⟹  g(x)=1x+1.g(x)(1+x)=1 \implies g(x)=\frac{1}{x+1}.g(x)(1+x)=1⟹g(x)=x+11​.
  1. Substitute g(x)g(x)g(x) into the differential equation.

Given

dydx−ytan⁡x=2(x+1)sec⁡x g(x).\frac{dy}{dx}-y\tan x=2(x+1)\sec x\, g(x).dxdy​−ytanx=2(x+1)secxg(x).

Using g(x)=1x+1g(x)=\frac{1}{x+1}g(x)=x+11​,

dydx−ytan⁡x=2sec⁡x.\frac{dy}{dx}-y\tan x=2\sec x.dxdy​−ytanx=2secx.

So the differential equation becomes

dydx−ytan⁡x=2sec⁡x.\frac{dy}{dx}-y\tan x=2\sec x.dxdy​−ytanx=2secx.
  1. Solve the linear differential equation.

Write it in standard 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)=2sec⁡x.P(x)=-\tan x, \qquad Q(x)=2\sec x.P(x)=−tanx,Q(x)=2secx.

The integrating factor is

I.F.=e∫−tan⁡x dx=eln⁡(cos⁡x)=cos⁡x.\text{I.F.}=e^{\int -\tan x\,dx}=e^{\ln(\cos x)}=\cos x.I.F.=e∫−tanxdx=eln(cosx)=cosx.

Multiply the equation by cos⁡x\cos xcosx:

cos⁡x dydx−ysin⁡x=2.\cos x\,\frac{dy}{dx}-y\sin x=2.cosxdxdy​−ysinx=2.

But

ddx(ycos⁡x)=cos⁡x dydx−ysin⁡x.\frac{d}{dx}(y\cos x)=\cos x\,\frac{dy}{dx}-y\sin x.dxd​(ycosx)=cosxdxdy​−ysinx.

Hence,

ddx(ycos⁡x)=2.\frac{d}{dx}(y\cos x)=2.dxd​(ycosx)=2.

Integrate:

ycos⁡x=2x+C.y\cos x=2x+C.ycosx=2x+C.

Using y(0)=0y(0)=0y(0)=0,

0⋅cos⁡0=2(0)+C  ⟹  C=0.0\cdot \cos 0=2(0)+C \implies C=0.0⋅cos0=2(0)+C⟹C=0.

Thus,

y=2xcos⁡x=2xsec⁡x.y=\frac{2x}{\cos x}=2x\sec x.y=cosx2x​=2xsecx.
  1. Evaluate y(π3)y\left(\frac{\pi}{3}\right)y(3π​).
y(π3)=2⋅π3⋅sec⁡π3.y\left(\frac{\pi}{3}\right)=2\cdot \frac{\pi}{3}\cdot \sec\frac{\pi}{3}.y(3π​)=2⋅3π​⋅sec3π​.

Since

sec⁡π3=2,\sec\frac{\pi}{3}=2,sec3π​=2,

we get

y(π3)=2⋅π3⋅2=4π3.y\left(\frac{\pi}{3}\right)=2\cdot \frac{\pi}{3}\cdot 2=\frac{4\pi}{3}.y(3π​)=2⋅3π​⋅2=34π​.
  1. Compare with the options.
y(π3)=4π3\boxed{y\left(\frac{\pi}{3}\right)=\frac{4\pi}{3}}y(3π​)=34π​​

So the correct option is A.

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