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Differential Equations question

2025 · 2 Apr · Shift 2 · Q49
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  5. /2025 · 2 Apr · Shift 2 · Q49

Differential Equations question

2025 · 2 Apr · Shift 2 · Q49

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dy dx+2ysec⁡2x=2sec⁡2x+3tan⁡x⋅sec⁡2x\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \sec ^2 x=2 \sec ^2 x+3 \tan x \cdot \sec ^2 x dxdy​+2ysec2x=2sec2x+3tanx⋅sec2x such that y(0)=54y(0)=\frac{5}{4}y(0)=45​. Then 12(y(π4)−e−2)12\left(y\left(\frac{\pi}{4}\right)-\mathrm{e}^{-2}\right)12(y(4π​)−e−2) is equal to ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 21

  1. We are given the linear differential equation
dydx+2ysec⁡2x=2sec⁡2x+3tan⁡x sec⁡2x\frac{dy}{dx}+2y\sec^2 x=2\sec^2 x+3\tan x\,\sec^2 xdxdy​+2ysec2x=2sec2x+3tanxsec2x

with initial condition

y(0)=54.y(0)=\frac54.y(0)=45​.
  1. Write it in standard linear form:
dydx+P(x)y=Q(x),\frac{dy}{dx}+P(x)y=Q(x),dxdy​+P(x)y=Q(x),

where

P(x)=2sec⁡2x,Q(x)=2sec⁡2x+3tan⁡xsec⁡2x.P(x)=2\sec^2 x, \qquad Q(x)=2\sec^2 x+3\tan x\sec^2 x.P(x)=2sec2x,Q(x)=2sec2x+3tanxsec2x.
  1. Find the integrating factor:
I.F.=e∫2sec⁡2x dx=e2tan⁡x.\text{I.F.}=e^{\int 2\sec^2 x\,dx}=e^{2\tan x}.I.F.=e∫2sec2xdx=e2tanx.
  1. Multiply the differential equation by the integrating factor:
e2tan⁡xdydx+2sec⁡2x e2tan⁡xy=(2sec⁡2x+3tan⁡xsec⁡2x)e2tan⁡x.e^{2\tan x}\frac{dy}{dx}+2\sec^2 x\,e^{2\tan x}y =\left(2\sec^2 x+3\tan x\sec^2 x\right)e^{2\tan x}.e2tanxdxdy​+2sec2xe2tanxy=(2sec2x+3tanxsec2x)e2tanx.

The left-hand side becomes

ddx(ye2tan⁡x).\frac{d}{dx}\left(ye^{2\tan x}\right).dxd​(ye2tanx).

So,

ddx(ye2tan⁡x)=(2sec⁡2x+3tan⁡xsec⁡2x)e2tan⁡x.\frac{d}{dx}\left(ye^{2\tan x}\right)=\left(2\sec^2 x+3\tan x\sec^2 x\right)e^{2\tan x}.dxd​(ye2tanx)=(2sec2x+3tanxsec2x)e2tanx.
  1. Integrate both sides. Let
t=2tan⁡x⇒dt=2sec⁡2x dx.t=2\tan x \Rightarrow dt=2\sec^2 x\,dx.t=2tanx⇒dt=2sec2xdx.

Then

∫(2sec⁡2x+3tan⁡xsec⁡2x)e2tan⁡xdx\int \left(2\sec^2 x+3\tan x\sec^2 x\right)e^{2\tan x}dx∫(2sec2x+3tanxsec2x)e2tanxdx

can be simplified using u=tan⁡xu=\tan xu=tanx, so du=sec⁡2x dxdu=\sec^2 x\,dxdu=sec2xdx:

∫(2+3u)e2u du.\int (2+3u)e^{2u}\,du.∫(2+3u)e2udu.

Now evaluate:

∫(2+3u)e2u du.\int (2+3u)e^{2u}\,du.∫(2+3u)e2udu.

Notice that

ddu(32ue2u+14e2u)=32e2u+3ue2u+12e2u=(2+3u)e2u.\frac{d}{du}\left(\frac{3}{2}ue^{2u}+\frac14 e^{2u}\right) =\frac32 e^{2u}+3ue^{2u}+\frac12 e^{2u} =(2+3u)e^{2u}.dud​(23​ue2u+41​e2u)=23​e2u+3ue2u+21​e2u=(2+3u)e2u.

Hence,

∫(2+3u)e2u du=32ue2u+14e2u+C.\int (2+3u)e^{2u}\,du=\frac{3}{2}ue^{2u}+\frac14 e^{2u}+C.∫(2+3u)e2udu=23​ue2u+41​e2u+C.

Substituting back u=tan⁡xu=\tan xu=tanx,

ye2tan⁡x=32tan⁡x e2tan⁡x+14e2tan⁡x+C.ye^{2\tan x}=\frac32 \tan x\,e^{2\tan x}+\frac14 e^{2\tan x}+C.ye2tanx=23​tanxe2tanx+41​e2tanx+C.

Therefore,

y=32tan⁡x+14+Ce−2tan⁡x.y=\frac32 \tan x+\frac14+Ce^{-2\tan x}.y=23​tanx+41​+Ce−2tanx.
  1. Use the initial condition y(0)=54y(0)=\frac54y(0)=45​.

Since tan⁡0=0\tan 0=0tan0=0,

54=32(0)+14+C\frac54=\frac32(0)+\frac14+C45​=23​(0)+41​+C

so

C=1.C=1.C=1.

Thus the solution is

y(x)=32tan⁡x+14+e−2tan⁡x.y(x)=\frac32\tan x+\frac14+e^{-2\tan x}.y(x)=23​tanx+41​+e−2tanx.
  1. Now compute y(π4)y\left(\frac\pi4\right)y(4π​).

Since tan⁡π4=1\tan \frac\pi4=1tan4π​=1,

y(π4)=32(1)+14+e−2=74+e−2.y\left(\frac\pi4\right)=\frac32(1)+\frac14+e^{-2}=\frac74+e^{-2}.y(4π​)=23​(1)+41​+e−2=47​+e−2.
  1. Therefore,
y(π4)−e−2=74.y\left(\frac\pi4\right)-e^{-2}=\frac74.y(4π​)−e−2=47​.

So,

12(y(π4)−e−2)=12⋅74=21.12\left(y\left(\frac\pi4\right)-e^{-2}\right)=12\cdot \frac74=21.12(y(4π​)−e−2)=12⋅47​=21.

Hence the required integer is

21.\boxed{21}.21​.
  1. Comparison with stored answer:

Stored correct answer = 212121, which matches our result.

Next

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