The Method of Direct Integration

The simplest differential equations to solve are those in the form of $\frac{dy}{dt} = f(t)$. These differential equations can be solved by directly integrating both sides to get that for any antiderivative $F$ and for $C$ as a constant we have that:

(1)

\begin{align} \quad \int \frac{dy}{dt} \: dt = \int f(t) \: dt \\ \quad y = F(t) + C \end{align}

For example, suppose that we wanted to solve the differential equation $\frac{dy}{dt} = t^2 + \cos 2t$. Then we have that for $C$ as a constant:

(2)

\begin{align} \quad \int \frac{dy}{dt} \:dt = \int t^2 + \cos 2t \: dt \\ \quad y = \frac{t^3}{3} + \sin 2t + C \end{align}

Another type of differential equation that can be solved by direct substitution are differential equation for which $a$ and $b$ are constants, in the following form:

(3)

\begin{align} \frac{dy}{dt} = ay + b \end{align}

For $a \neq 0$ and $y \neq -\frac{b}{a}$ we can rewrite the differential equation above as:

(4)

\begin{align} \frac{dy}{dt} = a\left (y + \frac{b}{a} \right) \\ \frac{\left ( \frac{dy}{dt} \right )}{y + \frac{b}{a}} = a \end{align}

Notice that if $p(t) = \ln \biggr \rvert y + \frac{b}{a} \biggr \rvert$ then $\frac{dp}{dt} = \frac{d}{dt} \ln \biggr \rvert y + \frac{b}{a} \biggr \rvert = \frac{\left ( \frac{dy}{dt} \right )}{y+ \frac{b}{a}}$ and so we have:

(5)

\begin{align} \quad \frac{d}{dt} \ln \biggr \rvert y + \frac{b}{a} \biggr \rvert = a \\ \quad \int \frac{d}{dt} \ln \biggr \rvert y + \frac{b}{a} \biggr \rvert \: dt = \int a \: dt \\ \quad \ln \biggr \rvert y + \frac{b}{a} \biggr \rvert = at + C \\ \quad \biggr \rvert y + \frac{b}{a} \biggr \rvert = e^{at + C} \\ \quad y + \frac{b}{a} = \pm e^{at + C} \\ \quad y = \pm e^{at + C} - \frac{b}{a} \\ \quad y = \pm e^{at}e^C - \frac{b}{a} \end{align}

Generally, since $C$ is an arbitrary constant, we have that $\pm e^C$ is also an arbitrary constant which we will denote as $D$, and hence, the solutions to $\frac{dy}{dt} = ay + b$ are given by $y = De^{at} - \frac{b}{a}$. Notice that we must proceed with some caution, since we must check if $D = 0$ yields a solution. In this case, since if $D = 0$ then $y = -\frac{b}{a}$ which was omitted earlier. In most cases, $y = -\frac{b}{a}$ will be a solution to the differential equation.

Let's look at an example of using the method if direction integration described above.

Example 1

Find all solutions to the differential equation $\frac{dy}{dt} = 3y - 210$.

We will start by factoring the righthand side of this equation to get $\frac{dy}{dt} = 3(y - 70)$. We now divide both sides of this equation by $y - 70$, and so for $y \neq 70$ we have rewritten the above differential equation as:

(6)

\begin{align} \quad \frac{\left ( \frac{dy}{dt} \right )}{y - 70} = 3 \end{align}

Now we can rewrite our differential equation above as follows:

(7)

\begin{align} \quad \frac{d}{dt} \ln \mid y - 70 \mid = \frac{\frac{dy}{dt}}{\mid y - 70 \mid} \\ \quad \frac{d}{dt} \ln \mid y - 70 \mid = 3 \end{align}

We have now reduced our original differential equation to the first type of differential equation we mentioned at the beginning of this page. We will now integrate both sides with respect to $t$ and apply the Fundamental Theorem of Calculus to get:

(8)

\begin{align} \quad \int \frac{d}{dt} \ln \mid y - 70 \mid \: dt = \int 3 \: dt \\ \quad \ln \mid y - 70 \mid = 3t + C \\ \mid y - 70 \mid = e^{3t + C} \\ y - 70 = \pm e^{3t + C} \\ y = 70 \pm e^{3t + C} \\ y = 70 \pm e^{3t}e^C \end{align}

Cleaning up our solution, we get that $y = 70 + De^{3t}$, for $D$ as a constant (noting that if $D = 0$ then $y = 70$ is a solution to $\frac{dy}{dt} = 3y - 210$.