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Differential Equations question

2024 · 6 Apr · Shift 1 · Q46
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  5. /2024 · 6 Apr · Shift 1 · Q46

Differential Equations question

2024 · 6 Apr · Shift 1 · Q46

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1+x2)dydx+y=etan⁡−1x\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}(1+x2)dxdy​+y=etan−1x, y(1)=0y(1)=0y(1)=0. Then y(0)y(0)y(0) is
  1. A
    14(eπ/2−1)\frac{1}{4}\left(e^{\pi / 2}-1\right)41​(eπ/2−1)
  2. B
    12(1−eπ/2)\frac{1}{2}\left(1-e^{\pi / 2}\right)21​(1−eπ/2)
  3. C
    14(1−eπ/2)\frac{1}{4}\left(1-e^{\pi / 2}\right)41​(1−eπ/2)
  4. D
    12(eπ/2−1)\frac{1}{2}\left(e^{\pi / 2}-1\right)21​(eπ/2−1)
View written solutionFree

Correct answer: B

  1. Rewrite the differential equation in linear form

Given

(1+x2)dydx+y=etan⁡−1x.(1+x^2)\frac{dy}{dx}+y=e^{\tan^{-1}x}.(1+x2)dxdy​+y=etan−1x.

Divide throughout by 1+x21+x^21+x2:

dydx+11+x2y=etan⁡−1x1+x2.\frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{e^{\tan^{-1}x}}{1+x^2}.dxdy​+1+x21​y=1+x2etan−1x​.

This is a linear differential equation of the form

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

with

P(x)=11+x2,Q(x)=etan⁡−1x1+x2.P(x)=\frac{1}{1+x^2}, \qquad Q(x)=\frac{e^{\tan^{-1}x}}{1+x^2}.P(x)=1+x21​,Q(x)=1+x2etan−1x​.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(x) dx=e∫11+x2 dx=etan⁡−1x.\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \frac{1}{1+x^2}\,dx}=e^{\tan^{-1}x}.I.F.=e∫P(x)dx=e∫1+x21​dx=etan−1x.
  1. Multiply the equation by the integrating factor

Multiplying both sides by etan⁡−1xe^{\tan^{-1}x}etan−1x:

etan⁡−1xdydx+etan⁡−1x1+x2y=e2tan⁡−1x1+x2.e^{\tan^{-1}x}\frac{dy}{dx}+\frac{e^{\tan^{-1}x}}{1+x^2}y =\frac{e^{2\tan^{-1}x}}{1+x^2}.etan−1xdxdy​+1+x2etan−1x​y=1+x2e2tan−1x​.

The left side becomes

ddx(yetan⁡−1x)=e2tan⁡−1x1+x2.\frac{d}{dx}\left(y e^{\tan^{-1}x}\right)=\frac{e^{2\tan^{-1}x}}{1+x^2}.dxd​(yetan−1x)=1+x2e2tan−1x​.

So,

ddx(yetan⁡−1x)=e2tan⁡−1x1+x2.\frac{d}{dx}\left(y e^{\tan^{-1}x}\right)=\frac{e^{2\tan^{-1}x}}{1+x^2}.dxd​(yetan−1x)=1+x2e2tan−1x​.
  1. Integrate

Let

t=tan⁡−1x  ⟹  dt=dx1+x2.t=\tan^{-1}x \implies dt=\frac{dx}{1+x^2}.t=tan−1x⟹dt=1+x2dx​.

Then

∫e2tan⁡−1x1+x2 dx=∫e2t dt=12e2t+C.\int \frac{e^{2\tan^{-1}x}}{1+x^2}\,dx=\int e^{2t}\,dt=\frac{1}{2}e^{2t}+C.∫1+x2e2tan−1x​dx=∫e2tdt=21​e2t+C.

Substituting back t=tan⁡−1xt=\tan^{-1}xt=tan−1x,

yetan⁡−1x=12e2tan⁡−1x+C.y e^{\tan^{-1}x}=\frac{1}{2}e^{2\tan^{-1}x}+C.yetan−1x=21​e2tan−1x+C.

Hence,

y=12etan⁡−1x+Ce−tan⁡−1x.y=\frac{1}{2}e^{\tan^{-1}x}+Ce^{-\tan^{-1}x}.y=21​etan−1x+Ce−tan−1x.
  1. Use the initial condition y(1)=0y(1)=0y(1)=0

Since

tan⁡−1(1)=π4,\tan^{-1}(1)=\frac{\pi}{4},tan−1(1)=4π​,

we get

0=12eπ/4+Ce−π/4.0=\frac{1}{2}e^{\pi/4}+Ce^{-\pi/4}.0=21​eπ/4+Ce−π/4.

Thus,

Ce−π/4=−12eπ/4Ce^{-\pi/4}=-\frac{1}{2}e^{\pi/4}Ce−π/4=−21​eπ/4

which gives

C=−12eπ/2.C=-\frac{1}{2}e^{\pi/2}.C=−21​eπ/2.

So the solution is

y=12etan⁡−1x−12eπ/2e−tan⁡−1x.y=\frac{1}{2}e^{\tan^{-1}x}-\frac{1}{2}e^{\pi/2}e^{-\tan^{-1}x}.y=21​etan−1x−21​eπ/2e−tan−1x.
  1. Find y(0)y(0)y(0)

Now,

tan⁡−1(0)=0.\tan^{-1}(0)=0.tan−1(0)=0.

Therefore,

y(0)=12e0−12eπ/2e0=12−12eπ/2=12(1−eπ/2).y(0)=\frac{1}{2}e^0-\frac{1}{2}e^{\pi/2}e^0 =\frac{1}{2}-\frac{1}{2}e^{\pi/2} =\frac{1}{2}(1-e^{\pi/2}).y(0)=21​e0−21​eπ/2e0=21​−21​eπ/2=21​(1−eπ/2).
  1. Match with the options
y(0)=12(1−eπ/2)y(0)=\frac{1}{2}(1-e^{\pi/2})y(0)=21​(1−eπ/2)

which is Option B.

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