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Differential Equations question

2021 · 27 Aug · Shift 1 · Q27
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  5. /2021 · 27 Aug · Shift 1 · Q27

Differential Equations question

2021 · 27 Aug · Shift 1 · Q27

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation dydx=2(y+2sin⁡x−5)x−2cos⁡x{{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos xdxdy​=2(y+2sinx−5)x−2cosx such that y(0) = 7. Then y(π\piπ) is equal to :
  1. A
    2eπ2+52{e^{{\pi ^2}}} + 52eπ2+5
  2. B
    eπ2+5{e^{{\pi ^2}}} + 5eπ2+5
  3. C
    3eπ2+53{e^{{\pi ^2}}} + 53eπ2+5
  4. D
    7eπ2+57{e^{{\pi ^2}}} + 57eπ2+5
View written solutionFree

Correct answer: A

  1. Given differential equation

We have

dydx=2x(y+2sin⁡x−5)−2cos⁡x,\frac{dy}{dx}=2x(y+2\sin x-5)-2\cos x,dxdy​=2x(y+2sinx−5)−2cosx,

with initial condition

y(0)=7.y(0)=7.y(0)=7.

We need to find y(π)y(\pi)y(π).

  1. Rewrite the equation in a suggestive form

Expand the right-hand side:

dydx=2xy+4xsin⁡x−10x−2cos⁡x.\frac{dy}{dx}=2xy+4x\sin x-10x-2\cos x.dxdy​=2xy+4xsinx−10x−2cosx.

Bring the 2xy2xy2xy term to the left:

dydx−2xy=4xsin⁡x−10x−2cos⁡x.\frac{dy}{dx}-2xy=4x\sin x-10x-2\cos x.dxdy​−2xy=4xsinx−10x−2cosx.

This is a linear differential equation.

  1. Look for a useful substitution

Notice the combination 2sin⁡x−52\sin x-52sinx−5 appears in the original equation. Let

z=y+2sin⁡x−5.z=y+2\sin x-5.z=y+2sinx−5.

Then

dzdx=dydx+2cos⁡x.\frac{dz}{dx}=\frac{dy}{dx}+2\cos x.dxdz​=dxdy​+2cosx.

Using the given differential equation,

dydx=2x(y+2sin⁡x−5)−2cos⁡x=2xz−2cos⁡x.\frac{dy}{dx}=2x(y+2\sin x-5)-2\cos x=2xz-2\cos x.dxdy​=2x(y+2sinx−5)−2cosx=2xz−2cosx.

So,

dzdx=(2xz−2cos⁡x)+2cos⁡x=2xz.\frac{dz}{dx}=(2xz-2\cos x)+2\cos x=2xz.dxdz​=(2xz−2cosx)+2cosx=2xz.

Thus the equation becomes

dzdx=2xz.\frac{dz}{dx}=2xz.dxdz​=2xz.
  1. Solve the separable equation

We have

dzdx=2xz.\frac{dz}{dx}=2xz.dxdz​=2xz.

Separate variables:

dzz=2x dx.\frac{dz}{z}=2x\,dx.zdz​=2xdx.

Integrate:

∫1z dz=∫2x dx\int \frac{1}{z}\,dz=\int 2x\,dx∫z1​dz=∫2xdx ln⁡∣z∣=x2+C.\ln|z|=x^2+C.ln∣z∣=x2+C.

Hence,

z=Cex2.z=Ce^{x^2}.z=Cex2.

Since z=y+2sin⁡x−5z=y+2\sin x-5z=y+2sinx−5, we get

y+2sin⁡x−5=Cex2.y+2\sin x-5=Ce^{x^2}.y+2sinx−5=Cex2.

Therefore,

y=Cex2−2sin⁡x+5.y=Ce^{x^2}-2\sin x+5.y=Cex2−2sinx+5.
  1. Use the initial condition

Given y(0)=7y(0)=7y(0)=7:

7=Ce0−2sin⁡0+5=C+5.7=Ce^{0}-2\sin 0+5=C+5.7=Ce0−2sin0+5=C+5.

So,

C=2.C=2.C=2.

Thus the solution is

y=2ex2−2sin⁡x+5.y=2e^{x^2}-2\sin x+5.y=2ex2−2sinx+5.
  1. Evaluate at x=πx=\pix=π

Since sin⁡π=0\sin \pi=0sinπ=0,

y(π)=2eπ2−2sin⁡π+5=2eπ2+5.y(\pi)=2e^{\pi^2}-2\sin \pi+5=2e^{\pi^2}+5.y(π)=2eπ2−2sinπ+5=2eπ2+5.
  1. Compare with options

The value is

2eπ2+5.\boxed{2e^{\pi^2}+5}.2eπ2+5​.

This matches Option A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So the stored answer is correct.

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