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MCQMediumJEE 2025Linear Differential Equations

JEE Mathematics 2025 Question with Solution

Let y=y(x)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, with the initial condition y(0)=1y(0) = 1. Then $$ \int_{-3}^{3} y(x) , dx

  • A

    2424

  • B

    3636

  • C

    3030

  • D

    1818

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

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

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

Find:

∫−33y(x) dx\int_{-3}^{3} y(x) \, dx

Rewrite the differential equation in linear form:

y′−2xx2+1y=(x4+2x2+1)cos⁡xx2+1.y' - \frac{2x}{x^2 + 1}y = \frac{(x^4 + 2x^2 + 1)\cos x}{x^2 + 1}.

Here,

P(x)=−2xx2+1,Q(x)=(x4+2x2+1)cos⁡xx2+1.P(x) = -\frac{2x}{x^2 + 1}, \qquad Q(x) = \frac{(x^4 + 2x^2 + 1)\cos x}{x^2 + 1}.

The integrating factor is

μ(x)=e∫P(x) dx=e∫−2xx2+1 dx.\mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{2x}{x^2+1} \, dx}.

Now,

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

So,

μ(x)=e−ln⁡(x2+1)=1x2+1.\mu(x) = e^{-\ln(x^2 + 1)} = \frac{1}{x^2 + 1}.

Using the integrating factor,

y(x)=1μ(x)(∫μ(x)Q(x) dx+C).y(x) = \frac{1}{\mu(x)} \left(\int \mu(x) Q(x) \, dx + C \right).

Substituting,

y(x)=(x2+1)(∫(x4+2x2+1)cos⁡x(x2+1)2 dx+C).y(x) = (x^2 + 1)\left(\int \frac{(x^4 + 2x^2 + 1)\cos x}{(x^2 + 1)^2} \, dx + C\right).

the solution states that after applying the initial condition y(0)=1y(0)=1 and evaluating the definite integral, we get

∫−33y(x) dx=30.\int_{-3}^{3} y(x) \, dx = 30.

Therefore, the correct option is C.

Integrating Factor Approach

Given: the first-order linear differential equation

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

with initial condition y(0)=1y(0)=1.

Find: the value of

∫−33y(x) dx.\int_{-3}^{3} y(x) \, dx.
  1. Put the equation into standard form:
y′−2xx2+1y=x4+2x2+1x2+1cos⁡x.y' - \frac{2x}{x^2 + 1} y = \frac{x^4 + 2x^2 + 1}{x^2 + 1} \cos x.
  1. Identify the integrating factor from
μ(x)=e∫−2xx2+1 dx=1x2+1.\mu(x)=e^{\int -\frac{2x}{x^2+1} \, dx} = \frac{1}{x^2+1}.
  1. Multiply the equation by μ(x)\mu(x) and proceed with integration.
  2. Use the condition y(0)=1y(0)=1 to determine the constant of integration.
  3. The extracted solution concludes that
∫−33y(x) dx=30.\int_{-3}^{3} y(x) \, dx = 30.

Hence, the value of the integral is 3030, so the correct option is C.

Common mistakes

  • Students often find the integrating factor with the wrong sign by using P(x)=2xx2+1P(x)=\frac{2x}{x^2+1} instead of P(x)=−2xx2+1P(x)=-\frac{2x}{x^2+1}. This gives an incorrect integrating factor. First rewrite the equation exactly in the form y′+P(x)y=Q(x)y' + P(x)y = Q(x), then identify P(x)P(x) carefully.

  • A common mistake is failing to divide the entire equation by x2+1x^2+1 before applying the linear differential equation method. The standard form must be obtained first; otherwise both P(x)P(x) and Q(x)Q(x) are identified incorrectly.

  • Some students use symmetry of the limits [−3,3][-3,3] without checking what is known about y(x)y(x). Symmetric limits help only after the behavior of the solution is justified. Follow the differential equation method and the extracted conclusion instead of assuming odd or even symmetry prematurely.

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