If the solution of the differential equation | vedclass

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(C) The given equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = e^x(x^2 - 2)$ and $Q(x) = (x^2 - 2x)

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(C) The given equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = e^x(x^2 - 2)$ and $Q(x) = (x^2 - 2x)(x^2 - 2)e^{2x}$.First,we find the Integrating Factor ($I$.$F$.):$	ext{I.F.} = e^{\int P(x) dx} = e^{\int e^x(x^2 - 2) dx}$.Note that $\frac{d}{dx}(e^x(x^2 - 2x)) = e^x(x^2 - 2x) + e^x(2x - 2) = e^x(x^2 - 2)$.Thus,$	ext{I.F.} = e^{e^x(x^2 - 2x)}$.The general solution is $y \cdot 	ext{I.F.} = \int Q(x) \cdot 	ext{I.F.} dx$.$y \cdot e^{e^x(x^2 - 2x)} = \int (x^2 - 2x)(x^2 - 2)e^{2x} \cdot e^{e^x(x^2 - 2x)} dx$.Let $t = e^x(x^2 - 2x)$. Then $dt = e^x(x^2 - 2x + 2x - 2) dx = e^x(x^2 - 2) dx$.Substituting this into the integral:$y \cdot e^t = \int t e^t dt = t e^t - e^t + C$.Given $y(0) = 0$,at $x = 0$,$t = e^0(0^2 - 0) = 0$.$0 \cdot e^0 = 0 \cdot e^0 - e^0 + C \implies 0 = -1 + C \implies C = 1$.So,$y \cdot e^t = t e^t - e^t + 1$.At $x = 2$,$t = e^2(2^2 - 2(2)) = 0$.$y(2) \cdot e^0 = 0 \cdot e^0 - e^0 + 1 \implies y(2) = -1 + 1 = 0$.

If the solution of the differential equation $\frac{dy}{dx} + e^x(x^2 - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x}$ satisfies $y(0) = 0$,then the value of $y(2)$ is

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