Ordinary differential equations: Solving ODEs by an integrating factor
Known examples of ODEs solved by an integrating factor
We have previously determined the general solution of an exponential growth model by separation of variables. However, there are other methods for solving ODEs, for example by a so-called integrating factor. Actually, we have already met this method of solving differential equations in several proofs of growth models, including for the ODE of exponential growth. In this theory page, we summarise earlier examples.
Exponential growth The first-order ODE \[\frac{\dd y }{\dd t}=r\,y\] where \(r\) is a nonzero constant can be rewritten in differential form as \[\dd y -r\,y\,\dd t=0\] Now we multiply both sides of the equation by \(e^{-r\,t}\): \[e^{-r\,t}\,\dd y -re^{-r\,t}\,y\,\dd t=0\] On the one hand, the left-hand side has become more complicated, but on the other hand, it is not: on the left-hand side you may recognise, via the product rule of differentials, a differential of a function: \[\dd\left(e^{-r\,t}\,y\right)=0.\] If a differential of a function is equal to 0, then it is a constant function. So \[e^{-r\,t}\,y=c,\] for some constant \(c\). This result can be rewritten as \[y=c\,e^{r\,t}.\]
The most crucial step in the above approach was multiplying both sides of the equation by \(e^{-r\,t}\), so that the left-hand side of the ODE can be simplified to a differential of a function. The multiplier is called an integrating factor. It is a sport to detect such an integrating factor.
Limited exponential growth The first-order ODE \[\frac{\dd y }{\dd t}=r\,y+s\] where \(r\) and \(s\) are nonzero constants can be rewritten in differential form as \[\dd y -r\,y\,\dd t=s\,\dd t\] Multiplying both sides of the equation by \(e^{-r\,t}\) gives \[e^{-r\,t}\,\dd y -re^{-r\,t}\,y\,\dd t=s\,e^{-r\,t}\,\dd t\] On the left-hand side you may recognize, via the product rule of differentials, a differential of a function and conclude that the following equation holds: \[\dd\left(e^{-r\,t}\,y\right)=s\,e^{-r\,t}\,\dd t\] So: \[e^{-r\,t}\,y=\int s\,e^{-r\,t}\,\dd t=-\frac{s}{r}e^{-r\,t}+c\] for some constant \(c\). This result can be rewritten as \[y=c\,e^{r\,t}-\frac{s}{r}\] Note that this solution is equal to the sum of the equilibrium \(y(t)=-\frac{s}{r}\), and the general solution \(y=c\,e^{r\,t}\) of the homogeneous equation \(\frac{\dd y }{\dd t}=r\,y\).