First Order Differential Equations

First Order DFQ with constants

We can write a general first order differential equation (DFQ) as:

$$\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y)$$

We will consider solutions to only linear equations such as

$$\frac{\mathrm{d}y}{\mathrm{d}x}+g(x)\cdot y=f(x)$$

When $g(x)$ and $f(x)$ are constants the solution is straightforward, but let’s just do it. Off topic, but I actually used to hate when textbooks would say something like that, but now writing this I see why they do it. Anyway,

$$\begin{array}{c}\frac{\mathrm{d}y}{\mathrm{d}x}+a\cdot y=b\\ \frac{\mathrm{d}y}{\mathrm{d}x}=b-ay\\ \frac{\mathrm{d}y}{b-ay}=dx\text{Yeah, we can do that!}\\ \int \frac{\mathrm{d}y}{b-ay}=\int dx\\ \frac{\ln (b-ay)}{-a}=x+C\\ \ln (b-ay)=-ax-aC\\ b-ay=C_1{e}^{-ax}\\ y=C_2{e}^{-ax}+b/a\end{array}$$

and check that it indeed satisfies the initial DFQ.

f(x) != constant

When the terms are not constant, the solution is a little trickier. The trick is to arrange the left side of the equation such that it is a derivative of $y(x)\cdot \mu (x)$. So consider this example:

$$\begin{array}{c}\frac{\mathrm{d}y}{\mathrm{d}x}+a\cdot y=f(x)\\ \mu (x)\cdot \frac{\mathrm{d}y}{\mathrm{d}x}+\mu (x)\cdot a\cdot y=\mu (x)\cdot f(x)\end{array}$$

Now, we have to cook up $\mu (x)$ such that its derivative is $\mu (x)\cdot a$, an obvious candidate is $\mu (x)=e^{ax}$ !

$$\begin{array}{c}{e}^{ax}\cdot \frac{\mathrm{d}y}{\mathrm{d}x}+a{e}^{ax}\cdot y=e^{ax}\cdot f(x)\\ \frac{\mathrm{d}(y\cdot e^{ax})}{\mathrm{d}x}=e^{ax}\cdot f(x)\\ y\cdot e^{ax}=\int e^{ax}\cdot f(x)dx\\ y=C{e}^{-ax}+e^{-ax}\int e^{ax}\cdot f(x)dx\end{array}$$

Where we only have to evaluate the integral. In most cases, it will require to do Integration by parts.

g(x), f(x) != constants

It gets a little harder when $g(x)$ is not constant: one additional step (but the general idea is the same).

$$\begin{array}{c}\frac{\mathrm{d}y}{\mathrm{d}x}+g(x)\cdot y=f(x)\\ \mu (x)\cdot \frac{\mathrm{d}y}{\mathrm{d}x}+\mu (x)\cdot g(x)\cdot y=\mu (x)\cdot f(x)\end{array}$$

Similarly, we have to cook up $\mu (x)$ such that its derivative is $\mu (x)\cdot g(x)$. That’s a differential equation itself:

$$\begin{array}{c}\frac{\mathrm{d}\mu (x)}{\mathrm{d}x}=\mu (x)\cdot g(x)\\ \frac{\mathrm{d}\mu (x)}{\mu (x)}=g(x)\mathrm{d}x\\ \int \frac{\mathrm{d}\mu (x)}{\mu (x)}=\int g(x)\mathrm{d}x\\ \ln (\mu (x))=\int g(x)\mathrm{d}x\\ \mu (x)=\exp \left(\int g(x)\mathrm{d}x\right)\end{array}$$

(I dropped an integration constant above because it will be “eaten up” by another constant further down the line). Using this result, we will obtain

$$\begin{array}{c}y\cdot \mu (x)=\int f(x)\cdot \mu (x)\mathrm{d}x+C\\ y=\frac{1}{\mu (x)}\int f(x)\cdot \mu (x)\mathrm{d}x+C\mu (x{)}^{-1}\end{array}$$

The next interesting topic is the actual application of these tools to real life problems - Modeling with First Order DFQ