Find the general solution to the differential equation dydt=t3+2t2−8t Also, part 8B. asks: Show that...

Parts of problems that say ''show that [explicitly given function] is a solution'' can be solved by simply ''plugging in''. E.g., if you're supposed to show that ${\displaystyle y\left(t\right)=\frac{1}{t-t}}$ is a solution to the differential equation ${\displaystyle \frac{dy}{dt}=y^2}$, then you should compute ${\displaystyle \frac{dy}{dt}=\frac{1}{{(1-t)}^2}}$, compute ${\displaystyle y^2=\frac{1}{{(1-t)}^2}}$, and ''plug them in'' to check that the differential equation is satisfied for this choice of y.
You're correct that the difference in the second problem is that you no longer have the derivative with respect to t expressed as a function of t. When the equation involves both a function y and its derivative (or derivatives), it can generally be more difficult to solve. In this case, a hint I'll give you is that ${\displaystyle \frac{dy}{dt}=\frac{1}{\frac{dt}{dy}}}$
You can first turn things around and think of t as a function of y. You can find t(y) using similar methods to the first problem (but with a more complicated function), and then find the inverse function to get y(t).

The difference is that the main variable is y and you separate the variables to integrate and obtain the solution
Hint 1:
The first equation is very easy, so:
${\displaystyle \frac{dy}{dt}=t^3+2t^2-8t\Rightarrow dy=(t^3+2t^2-8t)dt\Rightarrow \int dy=\int (t^3+2t^2-8t)dt}$
Hint 2:
For the other.
${\displaystyle \frac{dy}{dt}=y^3+2y^2-8y\Rightarrow \frac{1}{y^3+2y^2-8y}dy=dt}$
Then proceed integrating.

In the first question, you are given the derivative in terms of the variable. But in the second question, you are given an expression for the derivative that involves the function. For instance, it would be one thing if you were told $\frac{dy}{dx}=x$ (which would mean that $y=\frac{1}{2}{x}^2+C$), and a completely different thing if you are told $\frac{dy}{dx}=y$ (this tells you that the function y is equal to its derivative; which means that $y=A{e}^x$ for some constant A). Actually, we can solve the second differential equation; but we dont