Math 555: Differential Equations I

Lecture Companion

§2.1: Linear First Order DE

A linear first order differential equation for an unknown function $y = y(t)$ can be written in the form

$$ y' + p(t)y = q(t). \tag{1} $$
In general, linear equations are not separable. However, they can be made separable by a simple procedure. Suppose $\mu = \mu(t)$ is a positive differentiable function of $t$. Since $\mu(t) > 0$ for all $t$, then multiplying equation $(1)$ by $\mu$ will not change the solution of the DE, and yields

$$ \mu(t) y' + \mu(t) p(t)y = \mu(t) q(t). \tag{2} $$
Now, consider the function $\mu(t)y(t)$. This is the product of two differentiable functions of $t$, hence it is differentiable itself, and its derivative is

$$ \frac{d}{dt}\Big[ \mu(t)y(t) \Big] = \mu(t)y'(t) + \mu'(t) y(t). \tag{3} $$
Comparing the lefthand side of equation $(2)$ with the righthand side of equation $(3)$ we see that these will be equal if $\mu$ is a solution of the differential equation

$$ \mu' = p\mu. \tag{4} $$
One solution of this equation is

$$ \mu(t) = e^{\int p(t)\, dt}. \tag{5} $$
This function is indeed positive as it is an exponential, and it is differentiable. The function $\mu$ in equation $(5)$ is called the integrating factor of the differential equation $(1)$.

Now that we have found a function $\mu$ that satisfies the criteria for both equations $(2)$ and $(3)$, we may combine them to become

$$ \frac{d}{dt}\Big[\mu(t)y(t)\Big] = \mu(t)q(t). \tag{6} $$
This equation can be separated and integrated to give the desired solution,

$$ y(t) = \frac{1}{\mu(t)} \int\mu(t)q(t)\, dt + \frac{C}{\mu(t)}. \tag{7} $$
Once you understand this procedure and why it works, you don't need to reproduce the argument for every linear FODE. The first step is to identify that your equation is linear and that it's in the form of equation $(1)$, but then you can just plug into equations $(5)$ and $(7)$.

We look at some examples below.

Example 1. $\ \ y' = t^2 - 2y$

This equation is a first order linear DE, but it is not in the form of equation $(1)$. Rearranging,

$$ y' + 2y = t^2. $$
Now that it's in the correct form, we see that $p(t) = 2$ and $q(t) = t^2$. It's easy to compute $\mu(t) = e^{2t}$, so that the solution is given by

$$ y(t) = e^{-2t}\int t^2e^{2t}\, dt + Ce^{-2t}. $$ Integrating and simplifying yields

$$ y(t) = \frac{1}{2}t^2 - \frac{1}{2}t + \frac{1}{4} + Ce^{-2t}. $$
We plot the slope field and some solution curves.

Example 2. $\ \ y' + \frac{2y}{t} = e^{t^3}$

This equation is linear and it is in the correct form. The integrating factor is

$$ \mu = t^2, $$
and the general solution is

$$ y = \frac{e^{t^3}}{t^2} + \frac{C}{t^2}. $$
Again we plot the slope field together with some solution curves.

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