I came across this question and just want to make sure my understanding is correct.

I need to find the general solution of:

$$ \frac{dx}{dt} = a(1 - x) $$

In this case, I'm finding the how $x$ changes with respect to $t$ so I'm integrating with respect to $t$. Does that mean the answer is $at - xat + C$?

Thanks :)

  • 2
    $\begingroup$ You cannot integrate over $t$ if you have $x$ in the integral, which depends on $t$ in a yet unknown way. Separate the variables, it's the first elementary procedure you must learn to solve differential equations. Once you have $x$ on the left and $t$ on the right, you can proceed. $\endgroup$ – orion Aug 21 '18 at 9:27
  • $\begingroup$ If $x(t) = t^2$, for example, then $\int x\, dt$ is clearly not $tx$. $\endgroup$ – anomaly Aug 21 '18 at 19:57

You can't say that, simply because the function $x$ has a dependence in $t$ !


If you are given, for $x=x(t)$ $$\frac{dx}{dt} = a(1-x)$$

Then, one can re-arrange (by abuse of notation), giving \begin{align} \frac{dx}{1-x} &= a\, dt \\ \implies \int \left(\frac{1}{1-x}\right)dx &= a \int dt \end{align} The right hand side is straightforward for you, the left requires some knowledge of standard integrals.

Does this help?

  • $\begingroup$ Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"? $\endgroup$ – Hews Aug 21 '18 at 9:40
  • $\begingroup$ Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^{-C}$ to something else, as in this form has clearer meaning in the final function compared to just $C$. $\endgroup$ – orion Aug 21 '18 at 9:45
  • $\begingroup$ @orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^{-C}$ can’t reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation. $\endgroup$ – Roman Odaisky Aug 21 '18 at 15:33
  • $\begingroup$ @Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= \ldots$ $\endgroup$ – Kevin Aug 22 '18 at 13:41
  • $\begingroup$ So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone! $\endgroup$ – Hews Aug 22 '18 at 23:25

You must write

$$\frac{dx}{1-x}=a dt$$


If you write it as $dx = a(1-x)dt$ then it would be correct to integrate $x$ with respect to $t$. The trouble is, to perform the integration correctly, you would need to know what the dependence of $x$ on $t$ is, which is what you're trying to find out in the first place. To see what the problem with $\int xdt =xt+C$ is, consider any function, e.g. $x=t$. If we substitute $t$ in for $x$ before integrating, we get $\int tdt =\frac {t^2}2+C$. But if we use $\int xdt =xt+C$ and substitute $t$ in for $x$ afterwards, we get $\int xdt =t^2+C$, which is off by a factor of 2. Or if $x = \sin(t)$, then we would have $\int \sin(t)dt=t\sin(t)+C$ instead of $\int \sin(t)dt = \cos(t)+C$. If we had that $\int f(t)dt =tf(t)+C$, that would make the whole concept of an integral rather trivial; the integral of any function would just be that function times the independent variable. The identity $\int xdt=xt+C$ works only if $x$ doesn't depend on $t$. Remember, an integral can be interpreted as the area under a curve. If $x$ is a constant, then we just have a rectangle with width $t$ and height $x$, so the area is $xt$. But if $x$ is varying with $t$, then we can't just take the value of $x$ at the end of the interval; clearly the area is going to depend on what $x$ is doing in between.

Note that if you get familiar with basic differential forms, you should get to a point where you recognize that when the derivative is proportional to the function value, you have an exponential function. In this case, the differential equation is modified by a constant term. So if you take the test solution $x = c_1e^{c_2t}+c_3$, solve for the derivative in terms of $c_1$, $c_2$, and $c_3$, and then plug that into the differential equation, then you can solve for $c_1$, $c_2$, and $c_3$. Note that one degree of freedom will remain, since this is a first-order equation and no initial condition is given.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.