Maths Q19 – JEE-MAIN 2025

Home Maths Differential EquationsQuestion #187

Let

y = y (x)

be the solution of the differential equation

(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x

y (0) = 1

. Then

\int_{- 3}^{3} y (x) d x

is :

Clue 1 of 3: Differential Equation Type

Recognize the given differential equation as a first-order linear differential equation.

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Step 1: Identify and Solve the Linear Differential Equation✦ Active

The given differential equation is $(x^{2} + 1) y^{'} - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x$ . Dividing by $(x^{2} + 1)$ , we get $y^{'} - \frac{2 x}{x^{2} + 1} y = (x^{2} + 1) \cos x$ , which is a first-order linear differential equation of the form $y^{'} + P (x) y = Q (x)$ .

P (x) = - \frac{2 x}{x^{2} + 1}, Q (x) = (x^{2} + 1) \cos x

Integrating Factor (IF) = e^{\int P (x) d x} = e^{\int - \frac{2 x}{x^{2} + 1} d x} = e^{- \ln (x^{2} + 1)} = e^{\ln ((x^{2} + 1)^{- 1})} = \frac{1}{x^{2} + 1}

Multiplying by IF: \frac{d}{d x} (y \cdot \frac{1}{x^{2} + 1}) = \cos x

Integrating both sides: \frac{y}{x^{2} + 1} = \int \cos x d x = \sin x + C

y (x) = (x^{2} + 1) (\sin x + C)

💡 Teacher's Secret Hint

Remember to correctly identify $P (x)$ and $Q (x)$ and calculate the integrating factor carefully.

Step 2: Determine the Constant of Integration○ Expand

Use the initial condition $y (0) = 1$ to find the value of $C$ .

y (0) = (0^{2} + 1) (\sin 0 + C) = 1 (0 + C) = C

Given y (0) = 1 ⟹ C = 1

Thus, the solution is y (x) = (x^{2} + 1) (\sin x + 1)

💡 Teacher's Secret Hint

Do not forget to use the initial condition to find the particular solution.

Step 3: Evaluate the Definite Integral○ Expand

We need to evaluate $\int_{- 3}^{3} y (x) d x = \int_{- 3}^{3} (x^{2} + 1) (\sin x + 1) d x$ . This can be split into two integrals: $\int_{- 3}^{3} (x^{2} + 1) \sin x d x + \int_{- 3}^{3} (x^{2} + 1) d x$ .

For the first integral, let f (x) = (x^{2} + 1) \sin x .

f (- x) = ((- x)^{2} + 1) \sin (- x) = (x^{2} + 1) (- \sin x) = - f (x) .

Since f (x) is an odd function, \int_{- 3}^{3} (x^{2} + 1) \sin x d x = 0.

For the second integral, let g (x) = x^{2} + 1.

g (- x) = (- x)^{2} + 1 = x^{2} + 1 = g (x) .

Since g (x) is an even function, \int_{- 3}^{3} (x^{2} + 1) d x = 2 \int_{0}^{3} (x^{2} + 1) d x .

2 \int_{0}^{3} (x^{2} + 1) d x = 2 {[\frac{x^{3}}{3} + x]}_{0}^{3} = 2 ((\frac{3^{3}}{3} + 3) - (0))

= 2 (\frac{27}{3} + 3) = 2 (9 + 3) = 2 (12) = 24.

Therefore, \int_{- 3}^{3} y (x) d x = 0 + 24 = 24.

💡 Teacher's Secret Hint

Utilize the properties of even and odd functions for definite integrals over symmetric intervals to simplify calculations.

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Maths Question 19 – JEE-MAIN 2025

Clue 1 of 3: Differential Equation Type

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