Review 6: Solutions to Differential Equations

As with any equation, one is interested in solving, or finding the solutions to, a differential equation. There are many methods of doing so, and in class we will talk about two (that only work for some particular types of differential equations).

In this module, we will discuss something more basic: given an equation and a function, how can we check if that function is a solution to the given equation?

Example 1 Consider the differential equation

$LaTeX: \begin{equation}\label{eq:ed1} \frac{dy}{dt} = 5y-3. \end{equation}$ $\begin{equation}\label{eq:ed1} \frac{dy}{dt} = 5y-3. \end{equation}$

Check that the function $LaTeX: y(t) = e^{5t} + \frac{3}{5}$ $y(t) = e^{5t} + \frac{3}{5}$ is a solution to the differential equation above.

Solution

We need to plug in LaTeX: y(t) $y(t)$ into the left hand side and the right hand side of the equation and check to see that they are equal.

On the left hand side, we get
$LaTeX: \begin{equation}\label{eq:left} \frac{dy}{dt} = 5e^{5t}. \end{equation}$ $\begin{equation}\label{eq:left} \frac{dy}{dt} = 5e^{5t}. \end{equation}$

On the right hand side, we get

$LaTeX: \begin{equation}\label{eq:right} 5y-3 = 5(e^{5t} + \frac{3}{5}) - 3 = 5e^{5t}. \end{equation}$ $\begin{equation}\label{eq:right} 5y-3 = 5(e^{5t} + \frac{3}{5}) - 3 = 5e^{5t}. \end{equation}$

Since the left hand side and the right hand side are equal, we conclude that LaTeX: y(t) $y(t)$ is a solution to the differential equation.

Example 2

Consider the differential equation

$LaTeX: \begin{equation}\label{eq:ed2} \frac{dy}{dx} = \frac{1}{x}(2-y). \end{equation}$ $\begin{equation}\label{eq:ed2} \frac{dy}{dx} = \frac{1}{x}(2-y). \end{equation}$

Show that $LaTeX: y = \frac{C}{x}+2$ $y = \frac{C}{x}+2$ is a solution to the differential equation, where LaTeX: C $C$ is a constant.

Solution

We need to plug in LaTeX: y(x) $y(x)$ into the left hand side and the right hand side of the equation and check to see that they are equal.

On the left hand side, we get

$LaTeX: \begin{equation}\label{eq:left2} \frac{dy}{dx} = \frac{-C}{x^2} . \end{equation}$ $\begin{equation}\label{eq:left2} \frac{dy}{dx} = \frac{-C}{x^2} . \end{equation}$

On the right hand side, we get

$LaTeX: \begin{equation}\label{eq:right2} \frac{1}{x}(2-y) = \frac{1}{x} \left(2- \left(\frac{C}{x}+2\right)\right) = \frac{-C}{x^2}. \end{equation}$ $\begin{equation}\label{eq:right2} \frac{1}{x}(2-y) = \frac{1}{x} \left(2- \left(\frac{C}{x}+2\right)\right) = \frac{-C}{x^2}. \end{equation}$

Since the left hand side and the right hand side are equal, we conclude that LaTeX: y(x) $y(x)$ is a solution to the differential equation.

Note: It turns out that finding solutions to differential equations is often quite hard. Even when a solution can not be found, people study the question of whether there exists a solution, and if a solution exists, whether it is unique (the only solution). People also study the properties the solutions ought to have, even when they don't have an explicit formula for the solutions.