Solutions to Simple Differential Equations

# Solutions to Simple Differential Equations

We will now look at two different types of elementary differential equations that we can solve without must hassle.

## Differential Equations with One Isolated Derivative

The simplest type of differential equations to solve are in the form:

(1)
\begin{align} \quad \frac{dy}{dx} = f(x) \end{align}

In fact, you have more than likely solved many of this type of differential equations already. If we integrate both sides of the equation above with respect to $x$, then by the Fundamental Theorem of Calculus Part 1, we have that:

(2)
$$y = F(x)$$

We note that $F(x)$ is any antiderivative of $f$. Thus, the process of antidifferentiation solves differential equations in the form $\frac{dy}{dx} = f(x)$. We summarize all of this in the following theorem.

 Theorem 1: If $\frac{dy}{dx} = f(x)$, then the set of solutions for this differential equation is the set of antiderivatives $F(x)$ of $f$ which can be obtained by integrating both sides of this differential equation with respect to $x$.

Let's look at an some examples.

### Example 1

Solve the differential equation $\frac{dy}{dx} = x^2 + \cos (2x)$.

If we integrate both sides with respect to $x$ and apply the Fundamental Theorem of Calculus, we get that:

(3)
\begin{align} \quad \int \frac{dy}{dx} \: dx = \int x^2 + \cos (2x) \: dx \\ \quad y = \frac{x^3}{3} + \frac{1}{2} \sin (2x) + C \end{align}

Therefore all solutions to $\frac{dy}{dx} = x^2 + \cos (2x)$ are given by $y = \frac{x^3}{3} + \frac{1}{2} \sin (2x) + C$ for $C \in \mathbb{R}$.

### Example 2

Solve the differential equation $\frac{d^2y}{dx^2} = 3e^x - \sin x$.

If we integrate both sides with respect to $x$ twice and apply the Fundamental Theorem of Calculus, we get that:

(4)
\begin{align} \quad \int \frac{d^2y}{dx^2} \: dx = \int 3e^x - \sin x \\ \quad \frac{dy}{dx} = 3e^x + \cos x + C \\ \quad \int \frac{dy}{dx} \: dx = \int 3e^x + \cos x + C \: dx \\ \quad y = 3e^x + \sin x + Cx + D \end{align}

Therefore all solutions to $\frac{d^2y}{dx^2} = 3e^x - \sin x$ are given by $y = 3e^x + \sin x + Cx + D$ for $C, D \in \mathbb{R}$.

## Basic Linear Differential Equations

We will now look at another type of differential equations that are relatively easy to solve. Consider a differential equation in the following form (where both $a$ and $b$ are constants):

(5)
\begin{align} \frac{dy}{dx} = ay + b \end{align}

The process for solving a differential equation in the form above is best explained with an example. Consider the differential equation $\frac{dy}{dx} = 3y - 210$. We will now begin to find the solutions to this differential equation. We will start by factoring the righthand side of this equation to get $\frac{dy}{dx} = 3(y - 70)$. 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}{dx} \right )}{y - 70} = 3 \end{align}

We ultimately want to get rid of $\frac{dy}{dx}$ in the above equation. To do so, we notice that if $p(x) = \ln \mid y - 70 \mid$ then by applying the differentiation rule for the natural logarithm function, we get that $p'(x) = \frac{\frac{dy}{dx}}{\mid y - 70 \mid}$. Therefore we see that for our particular differential equation example, we have that:

(7)
\begin{align} \quad \frac{d}{dx} \ln \mid y - 70 \mid = \frac{\frac{dy}{dx}}{\mid y - 70 \mid} \\ \quad \frac{d}{dx} \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 $x$ and apply the Fundamental Theorem of Calculus to get:

(8)
\begin{align} \quad \int \frac{d}{dx} \ln \mid y - 70 \mid \: dx = \int 3 \: dx \\ \quad \ln \mid y - 70 \mid = 3x + C \\ \mid y - 70 \mid = e^{3x + C} \\ y - 70 = \pm e^{3x + C} \\ y = 70 \pm e^{3x + C} \\ y = 70 \pm e^{3x}e^C \end{align}

We will now clean up and condense our solution. Note that since $C \in \mathbb{R}$ is a constant that was introduced after integration, then $e^C$ will be some positive constant $e^C = D > 0$, and so we can write our solution as $y = 70 \pm De^{3x}$ where $D > 0$. Now notice that we have a "$\pm$" in our solution. To get rid of this, if we instead let $G \in \mathbb{R} \setminus \{ 0 \}$ then $y = 70 \pm De^{3x}$ is equivalent to $y = 70 + Ge^{3x}$. Lastly, we should note that if $G = 0$, then we get that $y = 70$ which is in fact a solution to our differential equation (verify this), and so the best way to write all of the solutions to $\frac{dy}{dx} = 3y - 210$ is $y = 70 + Ke^{3x}$, $K \in \mathbb{R}$.