I am taking my first class in differential equations right now, and I've been given the following equation to solve for the general solution:
$$\frac{du}{dt} = (u^2 - u)(u-0.2)$$
I rearranged the equation into separable form, integrated by partial fractions, and got the following result:
$$5\,ln(u) + 1.25\,ln(u-1) - 6.25\,ln(u-0.2) = t + C$$
In class, my professor said something about leaving a general solution as is (in certain situations), so I am wondering what specific types of situations you would not try to simplify by exponentiating both sides (not sure if that's the right terminology but I mean to write raise $e$ to the power of LHS and RHS).
My feeling is that it would probably make the solution more complicated to try to simplify this solution, but I'm hoping to gain a more clear definition of when you should vs. should not do that (if there are clear-cut rules).
In this problem, we can combine the logarithms. $$ \ln \frac{u^5(u-1)^{5/4}}{(u-0.2)^{25/4}} = t+C \text{.} $$ Many people are more comfortable with the exponential than with the logarithm, so would replace this with $$ \frac{u^5(u-1)^{5/4}}{(u-0.2)^{25/4}} = C \mathrm{e}^t \text{,} $$ (where this "$C$" is not the same as the prior "$C$", but this is a typical manipulation, typically justified by handwaving while saying something like "the exponential of a constant is some constant, which we'll just call '$C$' again." As long as you don't confuse yourself into thinking these different constants are actually numerically equal, there's no problem. Alternatively, give different constants different names.) Raising both sides to the $4/5$ power, this can be rearranged into a quintic for $u$, but that polynomial has no useful factors and its coefficients are sprinkled with "$\mathrm{e}^{4t/5}$"s. That is, continuing to try to isolate $u$ quickly worsens the representation of the result.
A somewhat unrelated observation: The equation forces three "obvious" constant solutions: $u = 0$, $u = 1$, and $u = 0.2$. None of these are produced by your general solution. (Each of these causes one (or more) of the (real) logarithms to be undefined.) There is nothing wrong with your method; nonlinear equations are irritating. You'll probably eventually be taught that this sort of autonomous equation has a general solution for "general initial conditions". But this equation also has solutions that cannot be made to agree with arbitrary initial data; agreement only occurs for three specific choices of initial data.