How to Solve Second Order Differential Equations with Initial Conditions

Just like with the first order differential equations, you often want to find a function which is the solution to a given differential equation with certain initial conditions. However, for second order differential equations, you need two initial conditions instead of one.

You’ll often be given a point—that is, a functional value at a given x, and a value for the derivative at another point. With these two conditions, you can find the constants of the general solution and get a particular solution. You determine the particular solution by inserting the values for the initial conditions and solve the resulting system of equations with two unknowns.

Example 1

Solve the differential equation y″ ⁡ + y′ − y = 0 with the initial conditions y(0) = 2 and y′(0) = 1

This differential equation has the solution

y(x) = C1ex + C 2e−2x

The derivative of () is

y′(x) = C 1ex − 2C 2e−2x

To find C1 and C2, you make two equations with two unknowns from the initial conditions and solve the system of equations:

2 = y(0) = C1e0 + C 2e0 = C1 + C2 C1 = 2 − C2 C1 = 2 −1 3 = 5 3 1 = y′(0) = C1e0 − 2C 2e0 = C1 − 2C2 1 = (2 − C2) − 2C2 = 2 − 3C2 C2 = 1 3

2 = y(0) 1 = y′(0) = C1e0 + C 2e0 = C 1e0 − 2C 2e0 = C1 + C2 = C1 − 2C2 C1 = 2 − C2 1 = (2 − C2) − 2C2 = 2 − 3C2 C2 = 1 3 C1 = 2 −1 3 = 5 3

Insert the values for C1 and C2 into the general solution () to find the particular solution:

y(x) = 5 3ex + 1 3e−2x