Math 555: Differential Equations

2.1 First Order Linear Differential Equations;
Method of Integrating Factors

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2.1.1 First-Order Differential Equations

The primary focus of this chapter will be studying different varieties of first-order differential equations, those of the form

$$ \dfrac{dy}{dt} = f(t,y) $$
where $t$ is the independent variable, $y$ is the dependent variable, and $f$ is a function of these two variables.

We seek a differentiable function $y = \phi (t)$ that satisfies this equation for all $t$ on some interval. Any such $\phi$ will be called a solution to the differential equation.

In general, there is no explicit formula for $\phi$ that works for an arbitrary function $f(t,y)$, so it is necessary to classify the differential equations we study and examine them case by case. Some such classifications we will discuss this chapter are linear (this section), separable, and exact.

2.1.2 First-Order Linear Differential Equations

This section focuses on equations of the form

$$ P(t)\dfrac{dy}{dt} + Q(t)y = G(t), $$
called first-order linear differential equations .

The designation of first-order comes from the fact that the highest-order derivative present in the equation is a first derivative. It is linear because it is possible to write the equation as depicted above in a "polynomial" like form, where the derivatives are the "powers" of $y$. (See Section 1.3)

$$ P(t)y' + Q(t)y - G(t) = 0 $$
Often, it is useful to write a first-order linear equation in standard form

$$\dfrac{dy}{dt} + p(t)y = g(t) \quad\text{or}\quad y' + p(t)y = g(t),$$
which is possible whenever $P(t)\neq 0$ (because you have to divide by $P$).

To solve equations of this form, we will be using something called an integrating factor . An integrating factor is a tool that changes the form of the differential equation in a particular, convenient way. Let's begin by studying an example that demonstrates what we're looking for out of our integrating factor.

2.1.3 A Convenient Example

Solve the differential equation

$$ \left(5 + t^4\right)\dfrac{dy}{dt} + 4t^3 y = 2t. $$

Video Example

2.1.4 Integrating Factor

Clearly, the key to our previous example is that the left hand side can be rewritten as a product rule. A natural question to ask is if it would be possible to ensure that every first-order linear equation can be manipulated to also be a product rule. This is the purpose of an integrating factor. It takes any differential equation of the form

$$\dfrac{dy}{dt} + p(t)y = g(t)$$
and makes it possible to write it as a product rule on the left hand side.

Derivation of the Integrating Factor

Integrating Factor Formula

$$ \mu (t) = \displaystyle\exp\left(\int\! p(t)\, dt\right) $$

2.1.5 Practice Using Integrating Factors

Work out the following examples for equations of the form

$$ P(t)\dfrac{dy}{dt} + Q(t)y = G(t) $$
by first converting them to standard form and then applying an integrating factor.

Example 1

Find the general solution to the differential equation

$$ y' - 2y = 4e^t $$

Follow Along

The first thing to check for a first-order linear differential equation is that it is in standard form:
$$ y' + p(t)y = g(t). $$
We see that this is already the case, with $p(t) = -2$ and $g(t) = 4e^t$. Next, we find the integrating factor
$$ \begin{aligned} \mu (t) &= \exp\left(\int\! p(t)\, dt\right) \\ \\ &= \exp\left(\int\! -2\, dt\right) \\ \\ &= \exp(-2t) \\ \\ &= e^{-2t} \end{aligned} $$
and multiply our differential equation by it.
$$ e^{-2t}y' + e^{-2t}(-2t)y = e^{-2t}(4e^t) $$
On the left hand side, we now have a product rule. This allows us to rewrite the differential equation as
$$ \dfrac{d}{dt}\left(e^{-2t}y\right) = 4e^{-t}. $$
Everything from here on is a calculus exercise. We integrate both sides with respect to $t$
$$ \begin{aligned} e^{-2t}y &= \int 4e^{-t}\, dt \\ \\ &= -4e^{-t} + C \end{aligned} $$
and then isolate our function $y(t)$
$$ \begin{aligned} y(t) &= \dfrac{-4e^{-t} + C}{e^{-2t}} \\ \\ &= -4e^{t} + Ce^{2t}. \end{aligned} $$

Example 2

Solve the initial value problem for

$$ ty' + (t+1)y = 2te^{-t},\quad y(1) = 3,\quad t>0 $$

Check Your Work

$$ y(t) = te^{-t} + \dfrac{3e-1}{t}e^{-t} $$

Video Solution

Example 3

Find the general solution to the differential equation

$$ y' + y = 3\sin\left(2t\right) $$

Check Your Work

$$ y(t) = \dfrac{3}{5}\Big(\sin(2t) - 2\cos(2t)\Big) + Ce^{-t} $$

Video Solution

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