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Home , Equation solving

Higher-Order Linear Differential

Sean Z. Roberson March 29, 2014

In Chapter 2, we examined a linear differential of the form y0 + P y = Q, where P and Q were functions of x (or whatever independent is used in the context of the problem). The solution was simple to construct by using an integrating factor. In Chapter 4, our focus shifted to higher-order linear differential equations (which I will abbreviate as DE from now on) which took a bit more work to solve. This paper will summarize the methods discussed in this chapter and will show how to apply each method.

1 Reduction of Order

To begin, let’s say that one (fundamental) solution of a differential equation is given. How can you find the rest? Doing so requires a substitution. In order to find a second solution to a second-order linear DE, assume that this second solution y2 is the product of some unknown u and the known solution y1. With this assumption, we take derivatives (quite unfortunately, using the product rule multiple times) and substitute into the DE. Afterwards, the equation is now in terms of the unknown function u, but solely in its derivatives. The equation can then be solved by making the substitution w = u0, which reduces into a first-order DE. We illustrate this with an example.

Example 1.

x 00 Given that y1 = e is a solution to y − y = 0, find another solution. Solution 1.

x 0 x 0 x Let us begin by assuming that this second solution has the form y2 = ue . Then we have y2 = ue + u e 00 x 0 x 00 x 00 0 x 00 x x 00 0 and y2 = ue +2u e +u e (verify). Substituting into the DE, we have y −y = 2u e +u e = e (u +2u ) = 0. The exponential is never zero, so we must the second factor equal to zero and solve. That is, we must solve u00 + 2u0 = 0. Note that the equation is expressed only in terms of derivatives. So, let us make the substitution w = u0, from which it follows that w0 = u00. Thus, our equation can be rewritten as w0 +2w = 0, −2x 0 which is a first-order . Solving for w gives w = c1e . But, this is actually u , so another 1 c e−x integration gives u = − c e−2x + c . It follows that our second solution y is − 1 + c ex. But notice that 2 1 2 2 2 −x our first solution appears in this solution, so by an appropriate choice of constants, we have that y2 = e .

The method, in essence, revolves around substitutions and reducing the equation into a first-order equa- tion. Once must be careful when taking derivatives, since products are involved and terms may be difficult to keep track of.

2 The Auxiliary Equation

The solution of a linear ODE is related to its auxiliary equation (other texts may refer to this as the characteristic equation; I prefer this name for the associated equation in in finding eigenvalues). Suppose we have the differential equation

y00 + 6y0 + 8y = 0.

1 To build the auxiliary equation, we make the substitution y(n) = mn, where n denotes the order of the derivative. In this case, the equation becomes

m2 + 6m + 8 = 0.

This equation can easily be solved by factoring. Doing so gives m = −4 and m = −2. Now, these roots will give us our solutions to the associated DE. An explanation follows.

Let a, b, c be constants. Consider the differential equation ay00 + by0 + cy = 0. Now, let y = erx. Then we have

ar2erx + brerx + cerx = 0

which factors as erx(ar2 + br + c) = 0. Now, since the exponential is never zero, it must be the case that the quadratic factor must be zero somewhere. Hence, the solutions of the quadratic give the multiplier of x in the exponential, and we may write a solution y = erx, where r is one root of the auxiliary equation.

−4x −2x So, in our example, two solutions to the differential equation are y1(x) = e and y2(x) = e . Now, the key idea in this chapter is that any solution to a linear differential equation is a linear combination of the fundamental solutions. (Students of linear algebra will understand this concept.) A fundamental solution is one of the solutions obtained by solving the auxiliary equation. So, for constants c1, c2, we may write our general solution to the DE as

−4x −2x y(x) = c1e + c2e .

Any choice of constants c1, c2 will create a new solution to the DE (this is left to the reader to verify).

What happens if roots repeat? What about complex roots? We treat each case.

2.1 Repeated Roots Suppose we have the DE y00 − 4y0 + 4y = 0, whose auxiliary equation is m2 − 4m + 4 = 0. Note that this factors as (m − 2)2 = 0, and so the solution to the auxiliary equation is m = 2. But notice that this root has multiplicity 2. If we were to blindly create a general solution for the DE, we would write

2x 2x y(x) = c1e + c2e .

But this solution is partial nonsense, since one of our fundamental solutions is repeated. To solve (no pun intended) this problem, we must append x (or the appropriate independent variable) to the appended solution, once for every repetition of the solution. So in our example, we write

2x 2x y(x) = c1e + c2xe

Note the extra x on the term beginning with c2. If we have an auxiliary equation that factors as, say (x − r)n, where n is an , then we write the solution to its associated differential equation as

rx rx 2 rx n rx y(x) = c1e + c2xe + c3x e + ... + cnx e .

2.2 Complex Roots What if our roots of the auxiliary equation look like a + bi? We would still exponentiate like normal, but it would be silly to leave a solution in the form e(a+bi)x. We create a simpler way to express this solution.

It is well-known that for any angle θ, eiθ = cos θ+i sin θ. This is called Euler’s Formula, and a derivation can be found in a text on a section on power series. We use Euler’s Formula to find an alternative form for solutions corresponding to complex roots. To that end, consider e(a+bi)x. This can be rewritten as eaxebix. By Euler’s Formula, we have

eaxebix = eax(cos bx + i sin bx)

2 which is the form for complex solutions. (We drop the i in solutions.) Note that to have a general ax solution, we write y(x) = e (c1 cos bx + c2 sin bx). Note that for the root a − bi, the solution is still written ax as e (c1 cos bx + c2 sin bx). We are only concerned about the real and imaginary parts to build our solution. Try to see that if b is replaced with −b, the solution is still the same.

We continue with examples, one of each type of solution. Example 2. Solve y00 − 12y0 + 28y = 0.

Solution 2. We begin by writing the auxiliary equation associated for this DE, which is m2 − 12m + 28 = 0. Fac- toring gives (m − 14)(m + 2) = 0, so m = 14 and m = −2. It follows that our solution to the DE is 14x −2x y(x) = c1e + c2e .

Example 3. Solve y00 − 2y0 + y = 0 subject to y(0) = 1, y0(0) = 1. Solution 3.

Again, we build the auxiliary equation, which is m2 − 2m + 1 = 0. This factors as (m − 1)2 = 0, so m = 1 x x is a repeated root. Now, our two fundamental solutions are y1(x) = e and y2(x) = xe . So our general x x solution is y(x) = c1e + c2xe . Note that we have initial conditions, so we must evaluate both y(x) and y0(x) at the indicated point and solve for constants.

For the first condition, we have 1 = c1 + 0c2, so c1 = 1. To solve for c2, we must take derivatives and 0 x x x repeat the process. We then have y (x) = c1e + c2xe + c2e . Using our conditions gives 1 = c1 + c2 + c2. We already know that c1 = 1, so this equation simplifies to 1 = 1 + 2c2. Solving for c2 gives c2 = 0. So, our specific solution is y(x) = ex .

Example 4.

Solve y00 + y0 + y = 0. Solution 4. √ √ −1 ± −3 −1 ± i 3 The auxiliary equation is r2 + r + 1 = 0. Using the gives r = = . 2 2 √ ! √ !! − x x 3 x 3 Since we have complex roots, our solution will be e 2 c cos + c sin . 1 2 2 2

3 Non-Homogeneous Linear DEs

So, far we have only considered DEs of the form L(y) = 0 (where L is a differential operator; this is commonly used to abbreviate DEs). What if the right-hand side is nonzero? For this, we consider DEs of the form L(y) = g(x), where g is some function. There are two methods to solve non-homogeneous linear DEs - undetermined coefficients and variation of . The method of undetermined coefficients has two methods associated with it - the first is a weaker form involving a guess of a particular solution, while the second is stronger and uses the annihilator operator. The solution to any non-homogeneous DE will have the form y = yc + yp, where yc is the solution obtained from solving L(y) = 0, the associated homogeneous DE, and yp is found by one of the two methods previously mentioned.

3 3.1 Undetermined Coefficients Suppose one has the DE L(y) = g(x). The first step in solving this equation is to solve the associated complementary equation L(y) = 0 by the method previously disucssed. After this solution yc is found, one must now find the solution yp. The form of yp is determined by looking at g(x) in the original differential equation. The form of g(x) will suggest the form of yp. This will be illustrated in the next examples, step by step. Example 5. Solve y00 + y0 − 6y = 2x. Solution 5. Step 1. We must form the associated homogeneous DE, which just means to replace the right-hand side of the original equation with 0. Step 2. Next, we build the auxiliary equation and solve it. Here, the auxiliary equation is m2 +m−6 = 0. 3x −2x The roots are m = 3 and m = −2, which means our solution yc is yc = c1e + c2e . Step 3. The third step in the process involves making a guess as to what the form of yp should be. To do that, we refer to the right-hand side of the original DE. We see that the function on the right-hand side is g(x) = 2x, so we may make a guess and assume that our solution yp = Ax + B, where A and B are constants to be determined. Now, why do we have a constant when the right-hand side is purely linear? We include the constant term B because we must assume that our particular solution matches the form of the right side of the DE. In this case, it must be a linear function. Step 4. After we have our guess, we must take derivatives of our guess function, substitute it in our DE, 0 00 and solve for A and B. Here, we have yp = A and yp = 0. Placing each part in the differential equation gives A − 6(Ax + B) = 2x. We now equate coefficients and solve the equations A − 6B = 0 and −6A = 2. 1 1 Solving gives A = − and B = − . 3 18 x 1 Step 5. We build our solution y = y + y . Our final answer is y = c e3x + c e−2x − − . c p 1 2 3 18 The next example will use a different right-hand side. The procedure remains the same. Example 6. Solve y00 + 6y0 + 5y = e−2x. Solution 6. The steps will not be explicitly listed. We first form the auxiliary equation after replacing the RHS by −5x −x zero. Solving this equation gives m = −5 and m = −1. So, we have yc = c1e + c2e . Now we must make a guess as to what yp will look like. This guess comes from the RHS of the original equation, so we −2x assume that yp = Ae , where A is a constant to be determined. After taking derivatives, we find that 1 y0 = −2Ae−2x and y00 = 4Ae−2x. Substituting and solving for A in the DE gives A = − . We then find p p 3 1 1 that y = − e−2x, so y = c e−5x + c e−x − e−2x . p 3 1 2 3

This next example describes what to do if there is repitition in yc and the guess for yp. That is, what do we do if our right-hand side already appears in yc? Example 7. Solve y00 + 9y = 2 sin 3x. Solution 7. The auxiliarry equation for this DE is m2 + 9 = 0. Clearly, we will have complex roots, namely m = ±3i. This means that yc = c1 cos 3x + c2 sin 3x. The RHS of the original equation suggests that yp = A sin 3x. But, this leads to

−9A sin 3x + 9A sin 3x = 2 sin 3x =⇒ 0 = 2 sin 3x

4 an absurd statement, since we eliminated the coefficient A accidentally. To remedy this, our guess for yp must include an x after the undetermined constant. So, the appropriate guess for yp is Ax sin 3x + Bx cos 3x. The cosine term is added to ensure particular terms cancel. So, after taking derivatives, we must solve 1 x 6A cos 3x − 6B sin 3x = 2 sin 3x. Equating coefficients gives A = 0 and B = − . So, y = − cos 3x and 3 p 3 x y = c cos 3x + c sin 3x − cos 3x . 1 2 3

3.2 Annihilator Operators The annihilator operator is a linear operator that forces a function to be zero by repeated differentiation. d This operator is D, and is precisely the differentiation operator dx . For second derivatives, one may write 2 d2 D , which corresponds to dx2 . We provide some examples of annihilators. Example 8. What annihilates the function y = x3 + 6x?

Solution 8. Observe that this function is a . One derivative will reduce the largest power to 2, a second derivative will reduce it to 1, and a third will reduce the power to zero, leaving a constant. So, it takes four derivatives to annihilate the function, and the appropriate annihilator is D4(y) .

Example 9.

What annihilates y = 4e−x? Solution 9. If we take two derivatives, we are have 4e−x. We need zero to annihilate the function. But, note that this second derivative is equal to the function itself. Subtracting the two gives zero. So, the annihilator is (D2 − 1)(y) . The annihilator, in some cases, may provide a stronger guess as to what the particular solution may be. The trick is to find the annihilator operator for the right side of the differential equation. Afterwards, proceed with undetermined coefficients. We do not provide examples of using the annihilator in a differential equation, since it still results in undetermined coefficients.

4 Variation of Parameters

Sometimes we have differential equations with unconvential right-hand sides. Instead of typical , sines, cosines, and exponentials, we may have , tangents, and other functions that make undeter- mined coefficients difficult. To remedy this, the method of variation of parameters allows us to solve a DE by changing the way we solve for our particular solution.

4.1 Description of Method Describing variation of parameters is tough. The method relies on the use of reduction of order. As in the method of reduction of order, we assume our particular solution is of the form

yp = u1y1 + u2y2

where y1, y2 are the fundamental solutions built from solving the associated homogeneous equation (re- member, that solution is yc = c1y1 + c2y2). The text will show the work, but to avoid extra problems, we 0 0 0 0 0 0 further assume that y1u1 + y2u2 = 0 so that we may also have y1u1 + y2u2 = f(x), where f(x) is the RHS of the DE to be solved. These two foregoing equations form a system of two equations in two unknowns, namely, we must solve

5 ( 0 0 y1u1 + y2u2 = 0 0 0 0 0 y1u1 + y2u2 = f(x)

0 0 for the functions u1 and u2. We can do this by means of Cramer’s Rule (consult any text on linear algebra) and express the solutions in terms of Wronskians. Recall that the Wronskian is a determinant whose columns are derivatives. Namely,

y1 y2 W (y1, y2) = 0 0 y1 y2

0 To solve for u1, we replace the first column of the Wronskian with the entries 0 and f(x), going from the top down. We then take this new determinant W1 and divide by W , the original Wronskian. This gives us 0 0 the function u1, and we integrate to solve for u1. The process is similar for u2, but the second column is instead replaced. We use an example. Example 10. Solve y00 + y = sec t tan t. Solution 10.

Before we begin, observe that the RHS is not a good candidate for undetermined coefficients, since it is not one of our basic functions. The upside to variation of parameters is that it always works.

Now, we begin by solving the auxiliary equation and build our complementary solution, which is yc = c1 cos t + c2 sin t. We next build the Wronskian W . By making the identifications y1 = cos t and y2 = sin t, the Wronskian W is

cos t sin t W = = 1 − sin t cos t

0 0 and the functions u1 and u2 are

0 sin t

W1 sec t tan t cos t u0 = = = − tan2 t 1 W 1

cos t 0

W2 sin t sec t tan t u0 = = = tan t 2 W 1

And, after integrating, we find that u1 = t − tan t and u2 = − ln(cos t). Putting all the pieces together, we find that our solution is y = c1 cos t + c2 sin t + (t − tan t) sin t − cos t ln(cos t) .

Variation of parameters may take some time to complete, but it is a method that will always work.

Examples and explanations are taken from Dennis Zill’s A First Course in Differential Equations with Modeling.

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