[Notes on Diffy Qs home] [PDF version] [Buy cheap paperback version]

[next] [prev] [prev-tail] [tail] [up]

Note: less than 1 lecture, first part of §3.1 in [EP], parts of §3.1 and §3.2 in [BD]

Let us consider the general second order linear differential equation

We usually divide through by to get

(2.1) |

where , , and . The word linear means that the equation contains no powers nor functions of , , and .

In the special case when , we have a so-called homogeneous equation

(2.2) |

We have already seen some second order linear homogeneous equations.

If we know two solutions of a linear homogeneous equation, we know a lot more of them.

Theorem 2.1.1 (Superposition). Suppose and are two solutions of the homogeneous equation (2.2). Then

also solves (2.2) for arbitrary constants and .

That is, we can add solutions together and multiply them by constants to obtain new and different solutions. We call the expression a linear combination of and . Let us prove this theorem; the proof is very enlightening and illustrates how linear equations work.

Proof: Let . Then

The proof becomes even simpler to state if we use the operator notation. An operator is an object that eats functions and spits out functions (kind of like what a function is, but a function eats numbers and spits out numbers). Define the operator by

The differential equation now becomes . The operator (and the equation) being linear means that . The proof above becomes

Two different solutions to the second equation are and . Let us remind ourselves of the definition, and . Therefore, these are solutions by superposition as they are linear combinations of the two exponential solutions.

The functions and are sometimes more convenient to use than the exponential. Let us review some of their properties.

Linear equations have nice and simple answers to the existence and uniqueness question.

Theorem 2.1.2 (Existence and uniqueness). Suppose are continuous functions on some interval , is a number in , and are constants. The equation

has exactly one solution defined on the same interval satisfying the initial conditions

For example, the equation with and has the solution

The equation with and has the solution

Using and in this solution allows us to solve for the initial conditions in a cleaner way than if we have used the exponentials.

The initial conditions for a second order ODE consist of two equations. Common sense tells us that if we have two arbitrary constants and two equations, then we should be able to solve for the constants and find a solution to the differential equation satisfying the initial conditions.

Question: Suppose we find two different solutions and to the homogeneous equation (2.2). Can every solution be written (using superposition) in the form ?

Answer is affirmative! Provided that and are different enough in the following sense. We will say and are linearly independent if one is not a constant multiple of the other.

Theorem 2.1.3. Let be continuous functions. Let and be two linearly independent solutions to the homogeneous equation (2.2). Then every other solution is of the form

That is, is the general solution.

For example, we found the solutions and for the equation . It is not hard to see that sine and cosine are not constant multiples of each other. If for some constant , we let and this would imply . But then for all , which is preposterous. So and are linearly independent. Hence

is the general solution to .

We will study the solution of nonhomogeneous equations in § 2.5. We will first focus on finding general solutions to homogeneous equations.

Exercise 2.1.4: Prove the superposition principle for nonhomogeneous equations. Suppose that is a solution to and is a solution to (same linear operator ). Show that solves .

Exercise 2.1.5: For the equation , find two solutions, show that they are linearly independent and find the general solution. Hint: Try .

Note that equations of the form are called Euler’s equations or Cauchy-Euler equations. They are solved by trying and solving for (we can assume that for simplicity).

Exercise 2.1.6: Suppose that . a) Find a formula for the general solution of . Hint: Try and find a formula for . b) What happens when or ?

We will revisit the case when later.

Exercise 2.1.7: Same equation as in Exercise 2.1.6. Suppose . Find a formula for the general solution of . Hint: Try for the second solution.

If you have one solution to a second order linear homogeneous equation you can find another one. This is the reduction of order method.

Note: If you wish to come up with the formula for reduction of order yourself, start by trying . Then plug into the equation, use the fact that is a solution, substitute , and you have a first order linear equation in . Solve for and then for . When solving for , make sure to include a constant of integration. Let us solve some famous equations using the method.

Exercise 2.1.9 (Chebyshev’s equation of order 1): Take . a) Show that is a solution. b) Use reduction of order to find a second linearly independent solution. c) Write down the general solution.