Let y = y(x) be the solution curve of the dif | vedclass

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(D) The given differential equation is $x(x^2 + e^x) dy + (e^x(x-2)y - x^3) dx = 0$.Rearranging it into the linear form $\frac{dy}{dx} + P(x)y = Q(x)$

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$\frac{4}{4+e^2}$

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(D) The given differential equation is $x(x^2 + e^x) dy + (e^x(x-2)y - x^3) dx = 0$.Rearranging it into the linear form $\frac{dy}{dx} + P(x)y = Q(x)$:$x(x^2 + e^x) \frac{dy}{dx} + e^x(x-2)y = x^3$$\frac{dy}{dx} + \frac{e^x(x-2)}{x(x^2 + e^x)} y = \frac{x^2}{x^2 + e^x}$.Integrating Factor ($I$.$F$.) $= e^{\int \frac{e^x(x-2)}{x(x^2 + e^x)} dx}$.Let $u = 1 + \frac{e^x}{x^2}$. Then $du = \frac{x^2 e^x - e^x(2x)}{x^4} dx = \frac{e^x(x-2)}{x^3} dx$.Thus,$I$.$F$. $= e^{\int \frac{1}{u} du} = u = 1 + \frac{e^x}{x^2}$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y(1 + \frac{e^x}{x^2}) = \int \frac{x^2}{x^2 + e^x} \cdot (\frac{x^2 + e^x}{x^2}) dx + C$.$y(1 + \frac{e^x}{x^2}) = \int 1 dx + C = x + C$.Since the curve passes through $(1, 0)$,we have $0(1 + e) = 1 + C$,so $C = -1$.Thus,$y = \frac{x-1}{1 + \frac{e^x}{x^2}}$.For $x = 2$,$y(2) = \frac{2-1}{1 + \frac{e^2}{4}} = \frac{1}{\frac{4+e^2}{4}} = \frac{4}{4+e^2}$.

Let $y = y(x)$ be the solution curve of the differential equation $x(x^2 + e^x) dy + (e^x(x-2)y - x^3) dx = 0, x > 0$,passing through the point $(1, 0)$. Then $y(2)$ is equal to:

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