Higher-Order Linear Differential Equations
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Higher-Order Linear Differential Equations Sean Z. Roberson March 29, 2014 In Chapter 2, we examined a linear differential equation of the form y0 + P y = Q, where P and Q were functions of x (or whatever independent variable 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 function 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 set 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 linear equation. 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 linear algebra 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 integer, 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 calculus 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. p p −1 ± −3 −1 ± i 3 The auxiliary equation is r2 + r + 1 = 0. Using the quadratic formula gives r = = . 2 2 p ! p !! − 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 parameters. 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.