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The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a first-order linear differential equation.
The integrating factor is given by $IF = e^{\int P(x) dx}$. In this case, $P(x) = -2x$. So, $IF = e^{\int -2x dx} = e^{-x^2}$.
Multiplying the given differential equation by the integrating factor $e^{-x^2}$, we get: $e^{-x^2}\frac{dy}{dx} - 2xe^{-x^2}y = 3x^2e^{x^2}e^{-x^2}$ $e^{-x^2}\frac{dy}{dx} - 2xe^{-x^2}y = 3x^2$
The left-hand side is the derivative of $y \cdot IF$ with respect to $x$. So, we have: $\int \frac{d}{dx}(ye^{-x^2}) dx = \int 3x^2 dx$ $ye^{-x^2} = \int 3x^2 dx$ $ye^{-x^2} = x^3 + C$, where $C$ is the constant of integration.
Multiply both sides by $e^{x^2}$ to get the general solution: $y = e^{x^2}(x^3 + C)$
Given $y(0) = 5$, substitute $x = 0$ and $y = 5$ into the general solution: $5 = e^{0^2}(0^3 + C)$ $5 = 1(0 + C)$ $C = 5$
Substitute $C = 5$ into the general solution: $y = e^{x^2}(x^3 + 5)$
Final Answer: $y = e^{x^2}(x^3 + 5)$
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