Let y = y(x) be the solution of the differential equation , y(0) = 1. Then is : [2025]

Question

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 : [2025]

Smart Achievers · Accepted Answer

(4)

We have, $(x^{2} + 1) \frac{d y}{d x} - 2 x y = (x^{4} + 2 x^{2} + 1) \cos x$

$\int_{- 3}^{3} y (x) d x$ (Linear D.E.)

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

$I . F . = e^{\int P d x} = e^{\int \frac{- 2 x}{(x^{2} + 1)} d x} = \frac{1}{(x^{2} + 1)}$

Thus, the solution of linear D.E. is given by

$\Rightarrow y \cdot \frac{1}{(x^{2} + 1)} = \int (x^{2} + 1) \cos x \cdot \frac{1}{(x^{2} + 1)} d x + c$

$\Rightarrow \frac{y}{x^{2} + 1} = \sin x + c \Rightarrow \frac{1}{1} = 0 + c \Rightarrow c = 1$ [ $∵$ y(0) = 1]

$∴ y = (x^{2} + 1) (\sin x + 1)$

$\int_{- 3}^{3} y (x) d x = \int_{- 3}^{3} (x^{2} + 1) (\sin x + 1) d x$

$= \int_{- 3}^{3} (x^{2} \sin x + x^{2} + \sin x + 1) d x$

$\int_{- 3}^{3} x^{2} \sin x d x + \int_{- 3}^{3} x^{2} d x + \int_{- 3}^{3} \sin x d x + \int_{- 3}^{3} 1 d x$

= 0 + 18 + 0 + 6 = 24

Solution

Q.
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 : [2025]

1 18

2 36

3 30

4 24

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