Ordinary Differential Equations/Initial Value Problems (IVP)

In [the previous article], we saw how to verify whether or not a function solved an ODE, and while doing so, found that many differential equations have infinitely many solutions. This means that as it stands, it is not meaningful to speak of "the" solution to an ODE, which is a big restriction on their utility. This suggests that additional information needs to be specified in order to distinguish between solutions of an ODE and select a unique one for a particular situation.

The most common additional data that are specified are the initial conditions. An initial condition is an equation that specifies the value of the solution, and possibly its derivatives, at some point. In the context of problems where the dependent variable is time, an initial condition specifies the value of the solution and its derivatives at 0, or in other words, specifies the initial value of the solution, which motivates the following: An Initial Value Problem (IVP) is a differential equation combined with initial conditions.

Here is an example of an IVP:

Find a function $y(x)$ such that

$${\begin{cases}{\frac {dy}{dx}}=y\\y(0)=1.\end{cases}}$$

The first line is the ODE, and the second line is the initial condition. To begin solving this IVP, we will begin by solving the ODE. Later in the course, it will be justified that our proposed solutions are the only solutions, but we can verify that functions of the form $y(x)=Ce^{x}$ for some real constant $C$ solve this differential equation. We compute $${\frac {dy}{dx}}={\frac {d}{dx}}(Ce^{x})=C{\frac {d}{dx}}(e^{x})=Ce^{x}=y,$$ so this solves the ODE. Now, we need to find which of these solutions also satisfy the initial condition. To do this, we assume $y$ satisfies the initial conditions and find out for which value of $C$ this is the case.

If $y$ satisfies the initial condition, then $y(0)=1$, so

$$y(0)=Ce^{0}=C=1.$$

Then, the solution to the IVP is $$y(x)=e^{x}.$$

The number of initial conditions needed to specify a unique solution depends on the order of the ODE. In general, $n$ initial conditions need to be specified to create an IVP for an $n^{\text{th}}$ order ODE with a unique solution.