The general solution of the differential equa | vedclass

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(C) Given differential equation: $(x^2+2) dy + 2xy dx = e^x(x^2+2) dx$ Divide by $(x^2+2) dx$: $\frac{dy}{dx} + \frac{2x}{x^2+2} y = e^x$ This is a li

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$(x^2+2) y=e^x(x^2-2x+4)+c$

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(C) Given differential equation: $(x^2+2) dy + 2xy dx = e^x(x^2+2) dx$ Divide by $(x^2+2) dx$: $\frac{dy}{dx} + \frac{2x}{x^2+2} y = e^x$ This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{2x}{x^2+2}$ and $Q(x) = e^x$. Integrating Factor $(IF)$ = $e^{\int P(x) dx} = e^{\int \frac{2x}{x^2+2} dx} = e^{\ln(x^2+2)} = x^2+2$. The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$. $y(x^2+2) = \int e^x(x^2+2) dx + C$. Using integration by parts for $\int e^x(x^2+2) dx$: $y(x^2+2) = (x^2+2)e^x - \int 2x e^x dx + C$. $y(x^2+2) = (x^2+2)e^x - 2[x e^x - \int e^x dx] + C$. $y(x^2+2) = (x^2+2)e^x - 2x e^x + 2e^x + C$. $y(x^2+2) = e^x(x^2+2 - 2x + 2) + C$. $y(x^2+2) = e^x(x^2-2x+4) + C$.

The general solution of the differential equation $(x^2+2) dy + 2xy dx = e^x(x^2+2) dx$ is

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