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Differential Equations question

2025 · 7 Apr · Shift 2 · Q27
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  5. /2025 · 7 Apr · Shift 2 · Q27

Differential Equations question

2025 · 7 Apr · Shift 2 · Q27

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation (x2+1)y′−2xy=(x4+2x2+1)cos⁡x(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x(x2+1)y′−2xy=(x4+2x2+1)cosx, y(0)=1y(0) = 1y(0)=1. Then ∫−33y(x) dx\int\limits_{-3}^{3} y(x) \, dx−3∫3​y(x)dx is :
  1. A
    36
  2. B
    24
  3. C
    18
  4. D
    30
View written solutionFree

Correct answer: B

  1. Rewrite the differential equation

Given

(x2+1)y′−2xy=(x4+2x2+1) cos⁡x,(x^2+1)y' - 2xy = (x^4+2x^2+1)\,\cos x,(x2+1)y′−2xy=(x4+2x2+1)cosx,

with

y(0)=1.y(0)=1.y(0)=1.

Notice that

x4+2x2+1=(x2+1)2.x^4+2x^2+1=(x^2+1)^2.x4+2x2+1=(x2+1)2.

So the equation becomes

(x2+1)y′−2xy=(x2+1)2cos⁡x.(x^2+1)y' - 2xy = (x^2+1)^2\cos x.(x2+1)y′−2xy=(x2+1)2cosx.
  1. Put it in linear form

Divide throughout by x2+1x^2+1x2+1:

y′−2xx2+1y=(x2+1)cos⁡x.y' - \frac{2x}{x^2+1}y = (x^2+1)\cos x.y′−x2+12x​y=(x2+1)cosx.

This is a linear differential equation:

y′+P(x)y=Q(x),y' + P(x)y = Q(x),y′+P(x)y=Q(x),

where

P(x)=−2xx2+1.P(x)=-\frac{2x}{x^2+1}.P(x)=−x2+12x​.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫−2xx2+1 dx.\text{I.F.} = e^{\int -\frac{2x}{x^2+1}\,dx}.I.F.=e∫−x2+12x​dx.

Since

∫−2xx2+1 dx=−ln⁡(x2+1),\int -\frac{2x}{x^2+1}\,dx = -\ln(x^2+1),∫−x2+12x​dx=−ln(x2+1),

we get

I.F.=e−ln⁡(x2+1)=1x2+1.\text{I.F.} = e^{-\ln(x^2+1)}=\frac{1}{x^2+1}.I.F.=e−ln(x2+1)=x2+11​.
  1. Multiply the equation by the integrating factor

Multiplying by 1x2+1\frac{1}{x^2+1}x2+11​,

1x2+1y′−2x(x2+1)2y=cos⁡x.\frac{1}{x^2+1}y' - \frac{2x}{(x^2+1)^2}y = \cos x.x2+11​y′−(x2+1)22x​y=cosx.

The left-hand side is

ddx(yx2+1).\frac{d}{dx}\left(\frac{y}{x^2+1}\right).dxd​(x2+1y​).

Therefore,

ddx(yx2+1)=cos⁡x.\frac{d}{dx}\left(\frac{y}{x^2+1}\right)=\cos x.dxd​(x2+1y​)=cosx.
  1. Integrate

Integrating,

yx2+1=sin⁡x+C.\frac{y}{x^2+1}=\sin x + C.x2+1y​=sinx+C.

Hence,

y=(x2+1)(sin⁡x+C).y=(x^2+1)(\sin x + C).y=(x2+1)(sinx+C).
  1. Use the initial condition

Given y(0)=1y(0)=1y(0)=1:

1=(02+1)(sin⁡0+C)=1⋅(0+C)=C.1=(0^2+1)(\sin 0 + C)=1\cdot(0+C)=C.1=(02+1)(sin0+C)=1⋅(0+C)=C.

So

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

Thus,

y=(x2+1)(sin⁡x+1).y=(x^2+1)(\sin x+1).y=(x2+1)(sinx+1).
  1. Evaluate the integral

We need

∫−33y(x) dx=∫−33(x2+1)(sin⁡x+1) dx.\int_{-3}^{3} y(x)\,dx = \int_{-3}^{3} (x^2+1)(\sin x+1)\,dx.∫−33​y(x)dx=∫−33​(x2+1)(sinx+1)dx.

Expand:

∫−33(x2+1)sin⁡x dx+∫−33(x2+1) dx.\int_{-3}^{3}(x^2+1)\sin x\,dx + \int_{-3}^{3}(x^2+1)\,dx.∫−33​(x2+1)sinxdx+∫−33​(x2+1)dx.

Now,

  • (x2+1)(x^2+1)(x2+1) is an even function,
  • sin⁡x\sin xsinx is an odd function,
  • so (x2+1)sin⁡x(x^2+1)\sin x(x2+1)sinx is odd.

Hence,

∫−33(x2+1)sin⁡x dx=0.\int_{-3}^{3}(x^2+1)\sin x\,dx = 0.∫−33​(x2+1)sinxdx=0.

So only the second integral remains:

∫−33(x2+1) dx.\int_{-3}^{3}(x^2+1)\,dx.∫−33​(x2+1)dx.

Since x2+1x^2+1x2+1 is even,

∫−33(x2+1) dx=2∫03(x2+1) dx.\int_{-3}^{3}(x^2+1)\,dx = 2\int_{0}^{3}(x^2+1)\,dx.∫−33​(x2+1)dx=2∫03​(x2+1)dx.

Compute:

2[x33+x]03=2(273+3)=2(9+3)=24.2\left[\frac{x^3}{3}+x\right]_0^3 =2\left(\frac{27}{3}+3\right) =2(9+3)=24.2[3x3​+x]03​=2(327​+3)=2(9+3)=24.
  1. Final answer

Therefore,

∫−33y(x) dx=24.\int_{-3}^{3} y(x)\,dx = 24.∫−33​y(x)dx=24.

So the correct option is B.

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