What Are Initial Conditions for Differential Equations?

(Video) Basic Differential Equations with an Initial Condition

by PatrickJMT

An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.

The video above uses the example $$\begin{cases} \frac{dy}{dx} = \text{cos}(x) \\ y(0) = -1 \end{cases}$$ to illustrate a simple initial value problem. Solving the differential equation without the initial condition gives you $y = \text{sin}(x) + C$.

Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem. In this case, plugging in $0$ for $x$ and $-1$ for $y$ gives us $-1 = C$, meaning that the particular solution must be $y = \text{sin}(x) - 1$.

So the general way to solve initial value problems is:

• First, find the general solution while ignoring the initial condition.
• Then, use the initial condition to plug in values and find a particular solution.

Two additional things to keep in mind: First, the initial value doesn't necessarily have to just be $y$-values. Higher-order equations might have an initial value for both $y$ and $y'$, for example.

Second, an initial value problem doesn't always have a unique solution. It's possible for an initial value problem to have multiple solutions, or even no solution at all.

An initial condition is a "starting point" for a differential equation. That is, it gives a coordinate $(x, y)$ that satisfies the general solution to the differential equation, thus allowing one to find the value of the generic constant $C$. Here is a sample differential equation problem:

Solve the differential equation $x\frac{dy}{dx} = e^{-y}$.

There is no initial condition present. Rather, we find the general solution, which is written in the form $y = f(x) + C$ for an unknown constant $C$. We cannot find $C$ without an initial condition.

We now integrate both sides of the equation with respect to the given differentials:

$e^y + C_1 = \ln(x) + C_2$

Move the constants to the same side and rename $C_2 - C_1 = C$:

$e^y = \ln(x) + C$

Take the natural log of both sides of the equation:

$y = \ln(\ln(x) + C)$

This is the general solution to the differential equation. To satisfy the initial condition, we plug in the given $(x,y)$ coordinate, $(1,1)$:

$1 = \ln(\ln(1) + C)$

Simplify and solve for $C$:

$1 = \ln(0 + C) \Rightarrow$ $1 = \ln(C) \Rightarrow$ $C = e$

We plug the value of $C$ back into the general solution of the differential equation to get the desired specific solution:

$y = \ln(\ln(x) + e)$.

A differential equation will often have a family of general solutions, so to specify a unique solution we'll usually need initial conditions or other data in addition to the differential equation. The problem of finding the unique solution, given initial conditions, is sometimes called an initial value problem.