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MCQMediumJEE 2024Separation of Variables

JEE Mathematics 2024 Question with Solution

If the solution y=y(x)y = y(x) of the differential equation (x4+2x3+3x2+2x+2) dy=(2x2+2x+3) dx(x^4 + 2x^3 + 3x^2 + 2x + 2) \, dy = (2x^2 + 2x + 3) \, dx satisfies y(−1)=−π/4y(-1) = -\pi/4, then y(0)y(0) is equal to:

  • A

    −π/12-\pi/12

  • B

    00

  • C

    π/4\pi/4

  • D

    π/2\pi/2

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

  • The differential equation is
(x4+2x3+3x2+2x+2) dy=(2x2+2x+3) dx(x^4 + 2x^3 + 3x^2 + 2x + 2) \, dy = (2x^2 + 2x + 3) \, dx
  • The condition is y(−1)=−π/4y(-1) = -\pi/4.

Find: y(0)y(0).

The solution is unrelated to this question, so the working for the given differential equation could not be extracted from it. Using the differential equation itself,

dydx=2x2+2x+3x4+2x3+3x2+2x+2\frac{dy}{dx} = \frac{2x^2 + 2x + 3}{x^4 + 2x^3 + 3x^2 + 2x + 2}

Factor the denominator:

(x2+1)(x2+2x+2)=(x2+1)((x+1)2+1)(x^2 + 1)(x^2 + 2x + 2) = (x^2 + 1)\big((x+1)^2 + 1\big)

Now observe that

ddx[tan⁡−1(x)+tan⁡−1(x+1)]=1x2+1+1(x+1)2+1\frac{d}{dx}\left[\tan^{-1}(x) + \tan^{-1}(x+1)\right] = \frac{1}{x^2+1} + \frac{1}{(x+1)^2+1}

Combining these,

1x2+1+1x2+2x+2=2x2+2x+3(x2+1)(x2+2x+2)\frac{1}{x^2+1} + \frac{1}{x^2+2x+2} = \frac{2x^2 + 2x + 3}{(x^2+1)(x^2+2x+2)}

Hence,

dydx=ddx[tan⁡−1(x)+tan⁡−1(x+1)]\frac{dy}{dx} = \frac{d}{dx}\left[\tan^{-1}(x) + \tan^{-1}(x+1)\right]

So,

y=tan⁡−1(x)+tan⁡−1(x+1)+Cy = \tan^{-1}(x) + \tan^{-1}(x+1) + C

Use the condition y(−1)=−π/4y(-1) = -\pi/4:

−π/4=tan⁡−1(−1)+tan⁡−1(0)+C-\pi/4 = \tan^{-1}(-1) + \tan^{-1}(0) + C −π/4=−π/4+0+C-\pi/4 = -\pi/4 + 0 + C

Therefore,

C=0C = 0

Now evaluate at x=0x=0:

y(0)=tan⁡−1(0)+tan⁡−1(1)=0+π/4=π/4y(0) = \tan^{-1}(0) + \tan^{-1}(1) = 0 + \pi/4 = \pi/4

Therefore, the correct option is C.

Common mistakes

  • Splitting the fraction incorrectly. The denominator factors as (x2+1)(x2+2x+2)(x^2+1)(x^2+2x+2), and missing this prevents recognition of inverse tangent derivatives. Factor the quartic first.

  • Forgetting that x2+2x+2=(x+1)2+1x^2+2x+2 = (x+1)^2+1. Without completing the square, the standard form for tan⁡−1\tan^{-1} is not visible. Rewrite the quadratic before integrating.

  • Using the initial condition at the wrong point. The condition is y(−1)=−π/4y(-1) = -\pi/4, not y(0)y(0). First determine the constant CC from x=−1x=-1, then compute y(0)y(0).

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