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Differential Equations question

2025 · 3 Apr · Shift 2 · Q33
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  5. /2025 · 3 Apr · Shift 2 · Q33

Differential Equations question

2025 · 3 Apr · Shift 2 · Q33

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx+3(tan⁡2x)y+3y=sec⁡2x,y(0)=13+e3\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\frac{1}{3}+e^3dxdy​+3(tan2x)y+3y=sec2x,y(0)=31​+e3. Then y(π4)y\left(\frac{\pi}{4}\right)y(4π​) is equal to :
  1. A
    43\frac{4}{3}34​
  2. B
    23+e3\frac{2}{3}+e^332​+e3
  3. C
    43+e3\frac{4}{3}+e^334​+e3
  4. D
    23\frac{2}{3}32​
View written solutionFree

Correct answer: A

  1. Write the differential equation in standard linear form

Given

dydx+3(tan⁡2x)y+3y=sec⁡2x.\frac{dy}{dx}+3(\tan^2 x)y+3y=\sec^2 x.dxdy​+3(tan2x)y+3y=sec2x.

Combine the yyy-terms:

3(tan⁡2x)y+3y=3(tan⁡2x+1)y=3sec⁡2x y.3(\tan^2 x)y+3y=3(\tan^2 x+1)y=3\sec^2 x\, y.3(tan2x)y+3y=3(tan2x+1)y=3sec2xy.

So the equation becomes

dydx+3sec⁡2x y=sec⁡2x.\frac{dy}{dx}+3\sec^2 x\, y=\sec^2 x.dxdy​+3sec2xy=sec2x.

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)=3sec⁡2x,Q(x)=sec⁡2x.P(x)=3\sec^2 x, \qquad Q(x)=\sec^2 x.P(x)=3sec2x,Q(x)=sec2x.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(x) dx=e∫3sec⁡2x dx=e3tan⁡x.\text{I.F.}=e^{\int P(x)\,dx}=e^{\int 3\sec^2 x\,dx}=e^{3\tan x}.I.F.=e∫P(x)dx=e∫3sec2xdx=e3tanx.
  1. Multiply the equation by the integrating factor

Multiplying throughout by e3tan⁡xe^{3\tan x}e3tanx:

e3tan⁡xdydx+3sec⁡2x e3tan⁡xy=sec⁡2x e3tan⁡x.e^{3\tan x}\frac{dy}{dx}+3\sec^2 x\, e^{3\tan x} y=\sec^2 x\, e^{3\tan x}.e3tanxdxdy​+3sec2xe3tanxy=sec2xe3tanx.

The left side is

ddx(ye3tan⁡x).\frac{d}{dx}\left(y e^{3\tan x}\right).dxd​(ye3tanx).

Hence,

ddx(ye3tan⁡x)=sec⁡2x e3tan⁡x.\frac{d}{dx}\left(y e^{3\tan x}\right)=\sec^2 x\, e^{3\tan x}.dxd​(ye3tanx)=sec2xe3tanx.
  1. Integrate both sides
ye3tan⁡x=∫sec⁡2x e3tan⁡x dx+C.y e^{3\tan x}=\int \sec^2 x\, e^{3\tan x}\,dx + C.ye3tanx=∫sec2xe3tanxdx+C.

Let

t=3tan⁡x⇒dt=3sec⁡2x dx,t=3\tan x \quad \Rightarrow \quad dt=3\sec^2 x\,dx,t=3tanx⇒dt=3sec2xdx,

so

sec⁡2x dx=dt3.\sec^2 x\,dx=\frac{dt}{3}.sec2xdx=3dt​.

Therefore,

∫sec⁡2x e3tan⁡x dx=13∫et dt=13et=13e3tan⁡x.\int \sec^2 x\, e^{3\tan x}\,dx =\frac{1}{3}\int e^t\,dt =\frac{1}{3}e^t =\frac{1}{3}e^{3\tan x}.∫sec2xe3tanxdx=31​∫etdt=31​et=31​e3tanx.

Thus,

ye3tan⁡x=13e3tan⁡x+C.y e^{3\tan x}=\frac{1}{3}e^{3\tan x}+C.ye3tanx=31​e3tanx+C.

So,

y=13+Ce−3tan⁡x.y=\frac{1}{3}+Ce^{-3\tan x}.y=31​+Ce−3tanx.
  1. Use the initial condition

Given

y(0)=13+e3.y(0)=\frac{1}{3}+e^3.y(0)=31​+e3.

Since tan⁡0=0\tan 0=0tan0=0,

y(0)=13+Ce0=13+C.y(0)=\frac{1}{3}+Ce^0=\frac{1}{3}+C.y(0)=31​+Ce0=31​+C.

Hence,

13+C=13+e3⇒C=e3.\frac{1}{3}+C=\frac{1}{3}+e^3 \Rightarrow C=e^3.31​+C=31​+e3⇒C=e3.

Therefore the solution is

y=13+e3e−3tan⁡x.y=\frac{1}{3}+e^3 e^{-3\tan x}.y=31​+e3e−3tanx.
  1. Evaluate at x=π4x=\frac{\pi}{4}x=4π​

Since

tan⁡π4=1,\tan\frac{\pi}{4}=1,tan4π​=1,

we get

y\left(\frac{\pi}{4}\right)=\frac{1}{3}+e^3 e^{-3}= rac{1}{3}+1= rac{4}{3}.
  1. Check options

The value is

43.\boxed{\frac{4}{3}}.34​​.

So the correct option is A.

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