We are given the first-order linear differential equation:

$\frac{dy}{dt} - 2ty = 9t^2e^{t^2}$ with the initial condition $y(0) = -2$.

Step-by-step solution:

This is a linear first-order differential equation in the form: $\frac{dy}{dt} + P(t)y = Q(t)$ where $P(t) = -2t$ and $Q(t) = 9t^2e^{t^2}$.

1. Find the integrating factor (IF):

The integrating factor is found using: $IF = e^{\int P(t) dt} = e^{\int -2t \, dt} = e^{-t^2}$

2. Multiply both sides of the equation by the integrating factor:

Multiply the entire equation by $e^{-t^2}$: $e^{-t^2} \frac{dy}{dt} - 2t e^{-t^2} y = 9t^2e^{t^2} e^{-t^2}$ This simplifies to: $e^{-t^2} \frac{dy}{dt} - 2t e^{-t^2} y = 9t^2$ The left-hand side becomes: $\frac{d}{dt} \left( y e^{-t^2} \right) = 9t^2$

3. Integrate both sides:

Now, integrate both sides with respect to $t$: $\int \frac{d}{dt} \left( y e^{-t^2} \right) dt = \int 9t^2 dt$ This gives: $y e^{-t^2} = 3t^3 + C$ where $C$ is the constant of integration.

4. Solve for $y(t)$:

Now solve for $y(t)$ by multiplying both sides by $e^{t^2}$: $y(t) = e^{t^2} (3t^3 + C)$

5. Use the initial condition to find $C$:

We are given that $y(0) = -2$. Substitute $t = 0$ and $y(0) = -2$ into the equation: $-2 = e^{0^2}(3(0)^3 + C) = C$ So, $C = -2$.

6. Final solution:

Substitute $C = -2$ into the equation for $y(t)$: $y(t) = e^{t^2}(3t^3 - 2)$

Thus, the function $y(t)$ that satisfies the differential equation is: $y(t) = e^{t^2}(3t^3 - 2)$

Let me know if you'd like further clarification on any step.

Here are 5 related questions for deeper understanding:

1. What is the general form of a first-order linear differential equation?
2. How do you determine the integrating factor for linear differential equations?
3. Can you explain the importance of the initial condition in solving differential equations?
4. What happens to the solution if the initial condition $y(0) = -2$ is changed?
5. How would the solution change if the differential equation were slightly modified?

Tip: In first-order linear differential equations, always check if multiplying by the integrating factor simplifies the equation into an easily integrable form.