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2.2Constant coeﬃcient second order linear ODEs

Note: more than 1 lecture, second part of §3.1 in [EP], §3.1 in [BD]

Suppose we have the problem

This is a second order linear homogeneous equation with constant coeﬃcients. Constant coeﬃcients means that the functions in front of , , and are constants, not depending on .

To guess a solution, think of a function that you know stays essentially the same when we diﬀerentiate it, so that we can take the function and its derivatives, add some multiples of these together, and end up with zero.

Let us try a solution of the form . Then and . Plug in to get

Hence, if or , then is a solution. So let and .

Exercise 2.2.1: Check that and are solutions.

The functions and are linearly independent. If they were not linearly independent we could write for some constant , implying that for all , which is clearly not possible. Hence, we can write the general solution as

We need to solve for and . To apply the initial conditions we ﬁrst ﬁnd . We plug in and solve.

Either apply some matrix algebra, or just solve these by high school math. For example, divide the second equation by 2 to obtain , and subtract the two equations to get . Then as . Hence, the solution we are looking for is

Let us generalize this example into a method. Suppose that we have an equation

 (2.3)

where are constants. Try the solution to obtain

The equation is called the characteristic equation of the ODE. Solve for the by using the quadratic formula.

Therefore, we have and as solutions. There is still a diﬃculty if , but it is not hard to overcome.

Theorem 2.2.1. Suppose that and are the roots of the characteristic equation.

(i)
If and are distinct and real (when ), then (2.3) has the general solution

(ii)
If (happens when ), then (2.3) has the general solution

For another example of the ﬁrst case, take the equation . Here the characteristic equation is or . Consequently, and are the two linearly independent solutions.

Example 2.2.1: Find the general solution of

The characteristic equation is . The equation has a double root . The general solution is, therefore,

Exercise 2.2.2: Check that and are linearly independent.

That solves the equation is clear. If solves the equation, then we know we are done. Let us compute and . Plug in

We should note that in practice, doubled root rarely happens. If coeﬃcients are picked truly randomly we are very unlikely to get a doubled root.

Let us give a short proof for why the solution works when the root is doubled. This case is really a limiting case of when the two roots are distinct and very close. Note that is a solution when the roots are distinct. When we take the limit as goes to , we are really taking the derivative of using as the variable. Therefore, the limit is , and hence this is a solution in the doubled root case.

2.2.1Complex numbers and Euler’s formula

It may happen that a polynomial has some complex roots. For example, the equation has no real roots, but it does have two complex roots. Here we review some properties of complex numbers.

Complex numbers may seem a strange concept, especially because of the terminology. There is nothing imaginary or really complicated about complex numbers. A complex number is simply a pair of real numbers, . We can think of a complex number as a point in the plane. We add complex numbers in the straightforward way: . We deﬁne multiplication by

It turns out that with this multiplication rule, all the standard properties of arithmetic hold. Further, and most importantly .

Generally we just write as , and we treat as if it were an unknown. When is zero, then is just the number . We do arithmetic with complex numbers just as we would with polynomials. The property we just mentioned becomes . So whenever we see , we replace it by . The numbers and are the two roots of .

Note that engineers often use the letter instead of for the square root of . We will use the mathematicians’ convention and use .

Exercise 2.2.3: Make sure you understand (that you can justify) the following identities:

• , , ,
• ,
• ,
• ,
• .

We also deﬁne the exponential of a complex number. We do this by writing down the Taylor series and plugging in the complex number. Because most properties of the exponential can be proved by looking at the Taylor series, these properties still hold for the complex exponential. For example the very important property: . This means that . Hence if we can compute , we can compute . For we use the so-called Euler’s formula.

Theorem 2.2.2 (Euler’s formula).

In other words, .

Exercise 2.2.4: Using Euler’s formula, check the identities:

Exercise 2.2.5: Double angle identities: Start with . Use Euler on each side and deduce:

For a complex number we call the real part and the imaginary part of the number. Often the following notation is used,

2.2.2Complex roots

Suppose the equation has the characteristic equation that has complex roots. By the quadratic formula, the roots are . These roots are complex if . In this case the roots are

As you can see, we always get a pair of roots of the form . In this case we can still write the solution as

However, the exponential is now complex valued. We would need to allow and to be complex numbers to obtain a real-valued solution (which is what we are after). While there is nothing particularly wrong with this approach, it can make calculations harder and it is generally preferred to ﬁnd two real-valued solutions.

Here we can use Euler’s formula. Let

Then

Linear combinations of solutions are also solutions. Hence,

are also solutions. Furthermore, they are real-valued. It is not hard to see that they are linearly independent (not multiples of each other). Therefore, we have the following theorem.

Theorem 2.2.3. Take the equation

If the characteristic equation has the roots (when ), then the general solution is

Example 2.2.2: Find the general solution of , for a constant .

The characteristic equation is . Therefore, the roots are , and by the theorem, we have the general solution

Example 2.2.3: Find the solution of , , .

The characteristic equation is . By completing the square we get and hence the roots are . By the theorem we have the general solution

To ﬁnd the solution satisfying the initial conditions, we ﬁrst plug in zero to get

Hence and . We diﬀerentiate

We again plug in the initial condition and obtain , or . Hence the solution we are seeking is

2.2.3Exercises

Exercise 2.2.6: Find the general solution of .

Exercise 2.2.7: Find the general solution of .

Exercise 2.2.8: Solve for , .

Exercise 2.2.9: Solve for , .

Exercise 2.2.10: Find the general solution of .

Exercise 2.2.11: Find the general solution of .

Exercise 2.2.12: Find the general solution of using the methods of this section.

Exercise 2.2.13: The method of this section applies to equations of other orders than two. We will see higher orders later. Try to solve the ﬁrst order equation using the methods of this section.

Exercise 2.2.14: Let us revisit the Cauchy-Euler equations of Exercise 2.1.6. Suppose now that . Find a formula for the general solution of . Hint: Note that .

Exercise 2.2.15: Find the solution to , , , where , , and are real numbers.

Exercise 2.2.16: Construct an equation such that is the general solution.

Exercise 2.2.101: Find the general solution to .

Exercise 2.2.102: Find the general solution to .

Exercise 2.2.103: Find the solution to , , .

Exercise 2.2.104: Find the solution to , , .

Exercise 2.2.105: Find the solution to , , .

Exercise 2.2.106: Find the solution to , , , where , , , and are real numbers, and .

Exercise 2.2.107: Construct an equation such that is the general solution.