To solve the differential equation $\frac{dy}{dx} = 2xy + 3x^2 e^{x^2}$ with the initial condition $y(0) = 5$, we can use an integrating factor method or attempt to solve it as a first-order linear differential equation.

Step 1: Rearrange the equation

The equation can be written in the form:

$\frac{dy}{dx} - 2xy = 3x^2 e^{x^2}$

This is a linear differential equation of the form:

$\frac{dy}{dx} + P(x)y = Q(x)$

where $P(x) = -2x$ and $Q(x) = 3x^2 e^{x^2}$.

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int P(x) dx} = e^{\int -2x dx} = e^{-x^2}$

Step 3: Multiply the differential equation by the integrating factor

Multiply the entire differential equation by $\mu(x) = e^{-x^2}$:

$e^{-x^2} \frac{dy}{dx} - 2xe^{-x^2}y = 3x^2 e^{x^2} e^{-x^2}$

which simplifies to:

$e^{-x^2} \frac{dy}{dx} - 2xe^{-x^2}y = 3x^2$

The left-hand side can be recognized as the derivative of $y \cdot e^{-x^2}$:

$\frac{d}{dx}\left( y \cdot e^{-x^2} \right) = 3x^2$

Step 4: Integrate both sides

Integrate both sides with respect to $x$:

$y \cdot e^{-x^2} = \int 3x^2 dx$

The integral on the right-hand side is:

$\int 3x^2 dx = x^3 + C$

So, the equation becomes:

$y \cdot e^{-x^2} = x^3 + C$

Step 5: Solve for $y$

Multiply both sides by $e^{x^2}$:

$y = e^{x^2} \left( x^3 + C \right)$

Step 6: Apply the initial condition

Use the initial condition $y(0) = 5$ to find the constant $C$:

$5 = e^{0} \cdot (0^3 + C)$

$C = 5$

Final Solution

The solution to the differential equation is:

$y = e^{x^2} \left( x^3 + 5 \right)$

Would you like more details or have any questions about this solution?

Here are 5 related questions for further exploration:

1. How would the solution change if the initial condition were $y(1) = 5$ instead of $y(0) = 5$?
2. What is the general solution for $\frac{dy}{dx} = 2xy + 3x^2 e^{x^2}$ without any initial condition?
3. How would you find the particular solution if $P(x)$ were not linear in $x$?
4. Can this method be applied to non-linear differential equations?
5. How would the solution be affected if the function $Q(x)$ had a different form?

Tip: Always check your integrating factor calculation, as it is crucial for simplifying the equation correctly.