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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).

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  • $\begingroup$ You can apply $\exp $ in both sides. You won't be able to get the $u$ explicitly but it's at least better $\endgroup$ – Gonzalo Benavides Jul 5 '18 at 4:23
  • $\begingroup$ If you can isolate $u$ don't stop, if you see you can't, get to the simplest looking equation you got when you tried and stop there. In this case I would rase $e$ to the power of everything(although I still won't be able to isolate $u$). I guess the professor did this beforehand and to save some time s/he decided to stop there instead of doing algebra you can do yourself without getting explicit function of $u$ $\endgroup$ – ℋolo Jul 5 '18 at 4:32
  • $\begingroup$ So I have raised e to both sides and have this after some simplification: $$\frac{(u^5)(u-1)^{5/4}}{(u-0.2)^{25/4}} = Ke^t $$ where $K = e^c$. Is this the most I should simplify? Or is there anything better I could do? $\endgroup$ – ECF Jul 5 '18 at 4:37
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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.

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  • $\begingroup$ No need for handwaving when it can be done rigorously: just put in the absolute values that should have been in the logarithms to begin with and say $\ln|\cdots|=t+C \iff \cdots = \pm e^C e^t = D e^t$, where $D:=\pm e^C$ is an arbitrary nonzero real constant (since $C$ was an arbitrary real constant). $\endgroup$ – Hans Lundmark Jul 5 '18 at 5:09
  • $\begingroup$ Note that the constant in your answer becomes strictly positive now. $\endgroup$ – user370967 Jul 5 '18 at 7:12
  • $\begingroup$ @Math_QED : Only if we stop ignoring the absolute values in the logarithm. Either we keep the bars and have a positive constant or ignore the bars and allow negative constants. $\endgroup$ – Eric Towers Jul 5 '18 at 7:24
  • $\begingroup$ Yes, you are right. Apologies! $\endgroup$ – user370967 Jul 5 '18 at 7:24

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