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(A) The given differential equation is $(x^2+1) \frac{dy}{dx} - 2xy = (x^2+1)^2 \cos x$. Dividing by $(x^2+1)$,we get $\frac{dy}{dx} - \frac{2x}{x^2+1

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$24$

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(A) The given differential equation is $(x^2+1) \frac{dy}{dx} - 2xy = (x^2+1)^2 \cos x$. Dividing by $(x^2+1)$,we get $\frac{dy}{dx} - \frac{2x}{x^2+1} y = (x^2+1) \cos x$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2x}{x^2+1}$ and $Q(x) = (x^2+1) \cos x$. The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int -\frac{2x}{x^2+1} dx} = e^{-\ln(x^2+1)} = \frac{1}{x^2+1}$. The solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$. $y \cdot \frac{1}{x^2+1} = \int (x^2+1) \cos x \cdot \frac{1}{x^2+1} dx = \int \cos x dx = \sin x + C$. Given $y(0) = 1$,we have $\frac{1}{0^2+1} = \sin(0) + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$. Thus,$y = (x^2+1)(\sin x + 1)$. Now,$\int_{-3}^3 y(x) dx = \int_{-3}^3 (x^2+1)(\sin x + 1) dx = \int_{-3}^3 (x^2 \sin x + x^2 + \sin x + 1) dx$. Since $x^2 \sin x$ and $\sin x$ are odd functions,their integral over $[-3, 3]$ is $0$. So,$\int_{-3}^3 y(x) dx = \int_{-3}^3 x^2 dx + \int_{-3}^3 1 dx = 2 \int_{0}^3 x^2 dx + 2 \int_{0}^3 1 dx = 2 [\frac{x^3}{3}]_0^3 + 2[x]_0^3 = 2(9) + 2(3) = 18 + 6 = 24$.

Let $y=y(x)$ be the solution of the differential equation $(x^2+1) y^{\prime}-2 x y=(x^4+2 x^2+1) \cos x$,with $y(0)=1$. Then $\int_{-3}^3 y(x) d x$ is :

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