Solve the dy/dt = t2(1+y) equation.

The given differential equation is:

$$\frac{dy}{dt} = t^2(1+y)$$

To separate the variables, divide both sides by $(1+y)$ and multiply by $dt$:

$$\frac{dy}{1+y} = t^2 dt$$

Integrate the left side with respect to $y$ and the right side with respect to $t$:

$$\int \frac{1}{1+y} dy = \int t^2 dt$$

The integral of $\frac{1}{1+y}$ is $\ln|1+y|$.
The integral of $t^2$ is $\frac{t^3}{3}$.
So, we get:

$$\ln|1+y| = \frac{t^3}{3} + C$$

where $C$ is the constant of integration.

To isolate $y$, exponentiate both sides using $e$:

$$e^{\ln|1+y|} = e^{\frac{t^3}{3} + C}$$ $$|1+y| = e^{\frac{t^3}{3}} e^C$$

Let $A = e^C$. Since $C$ is an arbitrary constant, $A$ is an arbitrary positive constant. We can remove the absolute value by allowing $A$ to be any non-zero constant (positive or negative).

$$1+y = A e^{\frac{t^3}{3}}$$

Finally, solve for $y$:

$$y = A e^{\frac{t^3}{3}} - 1$$