Differential Equations Workbook For Dummies®

4.2. Finding the Solution When Constant Coefficients Come into Play

After you know that the second order differential equation you're working with is both linear and homogeneous, the next step is to work it out. This section offers you some practice doing just that.

Following is an example of a second order differential equation that's both linear and homogeneous:

y″ − y = 0

where

y(0) = 9

and

y′(0) = −1

To solve this differential equation, you need a solution y = f(x) whose second derivative is the same as f(x) itself because subtracting the f(x) from f″(x) gives you 0. I bet you can think of one such solution: y = e^x. Substituting y = e^x gives you

e^x − e^x = 0

so y = e^x is a solution.

In fact, y = c₁e^x is also a solution, because y″ still equals c₁e^x, which means that substituting y = c₁e^x gives you

c₁e^x − c₁e^x = 0

Guess what? That means y = c₁e^x is also a solution. In fact, it's more general than just y = e^x, because y = c₁e^x represents an infinite number of solutions, depending on the value of c₁.

You can keep on going if you note that y = e^−x is also a solution because

y″ − y = e^−x − e^−x = 0

Of course, that realization alerts you to the fact that y = c₂e^−x is yet another solution because

y″ − y = c₂e^−x − c₂e^−x = 0

Hmmm. If y = c₁e^x is a solution and y = c₂e^−x is a solution, then the sum of these two solutions must also be a solution:

y = c₁e^x + c₂e^−x

To match the initial conditions, you can use the form of the solution y = c₁e^x + c₂e^−x, which means that y′ = c₁e^x − c₂e^−x