Key Point. The nth order linear homogeneous equation with constant coefficients

General Solutions of Higher-Order Linear Equations In section 3.1, we saw the following fact: Key Point. The nth order linear homogeneous equation with constant coefficients a n y (n) +... + a 2 y + a 1 y + a 0 y = 0 has n essentially different solutions y 1, y 2,...,y n, and Y = c 1 y 1 + c 2 y 2 +... + c n y n, where each c i is an arbitrary constant, is the most general solution to the equation. In section 3.3, we learned how to find these n essentially different solutions to an nth order LHCC equation, and were thus able to build the general solution. In this section, we will discuss the phrase essentially different in detail; in addition, we will learn how solutions to LHCC equations can be used to help solve the associated linear nonhomogeneous equations with constant coefficients. We will need to recall the special forms in which higher-order equations can appear; an equation is linear if it can be written in the form A n (x)y (n) +... + A 2 (x)y + A 1 (x)y + A 0 (x)y = F (x); a linear equation is homogeneous if the only term not containing y or one of its derivatives is 0, i.e. the equation has the form A n (x)y (n) +... + A 2 (x)y + A 1 (x)y + A 0 (x)y = 0; a linear equation has constant coefficients if it can be written in the form a n y (n) +... + a 2 y + a 1 y + a 0 y = F (x), where each a 1 is a constant; and a linear homogeneous equation with constant coefficients (LHCC equation) has form a n y (n) +... + a 2 y + a 1 y + a 0 y = 0. One final note: many of the theorems in this section apply to all linear equations, or to all linear homogeneous equations, and not just to LHCC equations. Particular Solutions We have seen that we can use known solutions to an equation to build new ones. In section 3.1, we saw that v 1 = sin x and v 2 = cos x are solutions to y + y = 0; but so are q = 3 sin x cos x, r = 5 sin x 3 + π cos x, and s = cos x. The idea of using a few solution functions (like v 1 and v 2 ) to build many more (like q, r, and s) is an extremely important one, so we will give this idea a name: 1

Definition. A linear combination of functions y 1, y 2,..., y n, is any sum of constant multiples of the y i, and has the form c 1 y 1 + c 2 y 2 +... + c n y n, where each c i is a constant. In the language of this new definition, each of q, r, and s is a linear combination of v 1 and v 2. It is easy to see that q = 3v 1 v 2, r = 5v 1 3 + πv 2, and s = v 2. It turns out that any linear combinations of solutions to a linear homogeneous equation (with or without constant coefficients) is also a solution, as indicated by the following theorem: Theorem 1. Let y 1,..., y n be n (particular) solutions of an nth order linear homogeneous equation on an interval I. Then the linear combination y = c 1 y 1 +... + c n y n is also a (particular) solution of the equation on the interval I for any constants c 1,..., c n. In section 3.1, we saw the following point: Key Point. We must have n initial conditions in order to find a particular solution to an nth order LHCC equation. We will state this theorem precisely in a moment; but before we do, we should think back to the analogous idea from our study of first-order equations. When we worked with first-order equations, we were concerned that an initial value problem (an equation along with one initial condition) might have no solution, or might have more than one solution; but in chapter 1, we saw that well-behaved equations will have precisely one solution corresponding to the initial condition. We may have some of the same concerns about finding particular solutions to higher-order equations; given a linear nth order equation and n initial conditions, can we be certain that there will be a solution? Can we be sure that there is no more than one solution? The theorem below answers that question; as with first-order equations, if our linear nth order equation is reasonably well-behaved, we can be certain that the initial value problem will have precisely one solution: Theorem 2. If p 1 (x), p 2 (x),..., p n (x) and f(x) are continuous on an open interval I, and if a is in the interval I, then the nth order linear equation y (n) + p 1 (x)y (n 1) +... + p n 1 (x)y + p n (x)y = f(x) has precisely one solution on the entire interval I satisfying the conditions y(a) = b 0, y (a) = b 1,..., y (n 1) (a) = b n 1. 2

Applying this theorem to LHCC equations gives us a particularly nice result; since an LHCC equation has form a n y (n) +... + a 2 y + a 1 y + a 0 y = 0, each of the functions p i (x) from Theorem 2 are just constant functions. Of course, constant functions are continuous on the entire real line, so when we study LHCC equations, we can be certain that their solutions are unique for all real numbers. General Solutions Notice that Theorem 1 above specifically deals with particular solutions, not with finding a general solution to the given linear homogeneous equation. In terms of general solutions, recall that we have seen the following point several times: Key Point. The nth order linear homogeneous equation with constant coefficients a n y (n) +... + a 2 y + a 1 y + a 0 y = 0 has n essentially different solutions y 1, y 2,...,y n, and Y = c 1 y 1 + c 2 y 2 +... + c n y n, where each c i is an arbitrary constant, is the most general solution to the equation. We are now ready to be more precise about what we mean by the phrase essentially different. We ll start with a couple of examples; in the first, we will look at three functions that I claim are not essentially different. In the second example, the three functions presented are essentially different. Example. Let f 1 = x 1, f 2 = 3x, and f 3 = 2. In a sense, these functions are not essentially different because it is possible to build one of them from the other two by using linear combinations. For example, we can view f 1 as a linear combination of f 2 and f 3 by writing In a sense, f 1 depends on f 2 and f 3. 1 3 f 2 1 2 f 3 = 1 3 (3x) 1 2 (2) = x 1 = f 1. Example. On the other hand, g 1 = 1, g 2 = x, and g 3 = x 2 are essentially different: it is impossible to build any one of them from the other two using linear combinations. For example, the linear combination c 1 g 1 + c 2 g 2 = c 1 + c 2 x can never produce g 3 = x 2, no matter how well we pick the constants c 1 and c 2. In a sense, these three functions are independent of each other. 3

In the first example above, we wrote alternatively, 1 3 f 2 1 2 f 3 = f 1 ; f 1 + 1 3 f 2 1 2 f 3 = 0. In the second example, we tried to write or c 1 g 1 + c 2 g 2 = g 3, c 1 g 1 + c 2 g 2 g3 = 0, but saw that this was impossible. We give a name to each type of behavior in the definition below: Definition. The functions f 1,..., f n are linearly dependent if there are constants c 1,..., c n (at least one of which is non-zero) so that c 1 f 1 +... + c n f n = 0. If there is no way to find constants c 1,..., c n (with at least one c i 0) so that c 1 f 1 +... + c n f n = 0, then the functions f 1,..., f n are linearly independent. Using this language, we would describe the three functions f 1 = x 1, f 2 = 3x, and f 3 = 2 as linearly dependent, while we would say that the functions are linearly independent. g 1 = 1, g 2 = x, and g 3 = x 2 Saying that a collection of functions is linearly independent is a precise way to say that they are essentially different. With this in mind, we are now ready to discuss the general solutions of linear homogeneous equations. Notice that the theorem below applies to any linear homogeneous equation, not just to LHCC equations. Theorem 4. If y 1, y 2,..., and y n are linearly independent solutions to a linear homogeneous nth order equation, then the general solution to the equation is Y = c 1 y 1 + c 2 y 2 +... + c n y n, where each c i is an arbitrary constant, and any solution to the equation is of this form. 4

The most important point to note here is this: any particular solution to a linear homogeneous nth order equation is just a linear combination of the n linearly independent solutions y 1,...,y n. We can apply Theorem 4 specifically to LHCC equations; from Section 3.3, we know how to find n solutions to any nth order LHCC equation, and we will soon be able to show that these n solutions are linearly independent. Thus the technique we learned in Section 3.3 does actually produce the general solution. Because of Theorem 4, it will be extremely important to be able to tell if a system of n equations is actually linearly independent. For example, suppose that we know that a particular third order linear homogeneous equation has solutions y 1 = 2x 3, y 2 = 2x 2 + 1, y 3 = x 2 + 1, y 4 = 2x 2 x, and y 5 = 3x 2 + x, and we would like to be able to write a general solution. Since the original equation is third order, we simply need to find three linearly independent solutions so we hope that three of the solutions above are linearly independent. Unfortunately, it is quite difficult to do this by inspection; for example, if we think that y 1, y 2, and y 3 are linearly independent, then we have to show that there are no constants c 1, c 2, and c 3 (at least one of which is nonzero) so that c 1 y 1 + c 2 y 2 + c 3 y 3 = 0. Fortunately, there is a tool that we can use to check a set of functions for independence or dependence. This tool is called the Wronskian, and is defined below: Definition. Assume that the functions f 1,...,f n are n times differentiable (i.e. f i, f i (n),...,f i all exist for each of the functions). The Wronskian of the functions f 1,...,f n is the determinant W of the n n matrix f 1 f 2... f n f 1 f 2... f n.... f (n) 1 f (n) 2... f n (n) The utility of the Wronskian in determining if n functions are independent or dependent is apparent due to the following theorem: Theorem 3. Suppose that y 1,..., y n are n solutions of the linear homogeneous nth order equation y (n) + p 1 (x)y (n 1) +... + p n 1 (x)y + p n (x)y = 0 on an open interval I, and let W be the Wronskian of the functions. If each p i is continuous, then: (a) If y 1,...,y n are linearly dependent, then W = 0 on each point of I. (b) If y 1,...,y n are linearly independent, then W 0 on each point of I. The theorem is extremely helpful: since a collection of functions must be either dependent or independent, then their Wronskian will either be identically 0 (if they are dependent), or never 0 (if they are independent). Putting Theorems 3 and 4 together gives us a method for finding the general solution to an nth order linear homogeneous equation: 5

Key Point. To find the general solution of an nth order linear homogeneous equation, find n functions y 1,...,y n, whose Wronskian is nonzero. Since these n functions are then linearly independent, Theorem 4 says that y = c 1 y 1 + c 2 y 2 +... + c n y n is the general solution to the equation, and any solution is of this form. We see from the remark above that finding the general solution to an equation will come down to evaluating the Wronskian, which is just a matrix determinant. We should recall how to make the calculation for 2 2 and 3 3 determinants: The determinant of a 2 2 matrix ( ) a b, c d denoted by is a b c d The determinant of a 3 3 matrix denoted by is which can be expanded as So a b c d e f g h j a e h a c b d, = ad bc. a b c d e f, g h j a b c d e f g h j f j b d g, f j + c d g e h a(ej fh) b(dj fg) + c(dh eg). = a(ej fh) b(dj fg) + c(dh eg). There is an alternate algorithm for finding determinants of 3 3 matrices that makes the computation a bit quicker. As an example, consider the matrix 3 0 1 1 2 1 2. 4 2 1 Start by putting the first two columns at the end of the matrix, so that we know have a 3 5 array: 6

The lines drawn below in the array indicate lines along which we must multiply; blue lines indicate that we leave the sign of the product the same, while red lines indicate that we must change the sign of the product: Finally, we add up the products of the lines; in this case, the determinant of the matrix is 3 1 1 + 0 2 4 + 1 1 2 2 0 1 1 3 2 2 1 1 4 = 3 + 1 12 4 = 12. 2 Example. A particular third order linear homogeneous equation has solutions y 1 = 2x 3, y 2 = 2x 2 + 1, y 3 = x 2 + 1, y 4 = 2x 2 x, and y 5 = 3x 2 + x. Find a general solution to the equation. We need to find three functions from the list above whose Wronskian is nonzero; we may have to try several combinations to find such a triplet. Let s start with y 1, y 2, and y 5. Their Wronskian is the determinant of the matrix y 1 y 2 y 5 2x 3 2x 2 + 1 3x 2 + x y 1 y 2 y 5 = 2 4x 6x + 1. y 1 y 2 y 5 0 4 6 Rewriting the matrix as 2x 3 2x 2 + 1 3x 2 + x 2x 3 2x 2 + 1 2 4x 6x + 1 2 4x 0 4 6 0 4 7

we can now calculate the determinant by multiplying along diagonals: the determinant is W (y 1, y 2, y 5 ) = (2x 3)(4x)(6) + (2x 2 + 1)(6x + 1)(0) + (3x 2 + x)(2)(4) (2x 2 + 1)(2)(6) (2x 3)(6x + 1)(4) (3x 2 + x)(4x)(0) = 48x 2 72x + 24x 2 + 8x 24x 2 12 (8x 12)(6x + 1) = 48x 2 72x + 24x 2 + 8x 24x 2 12 (48x 2 72x + 8x 12) = 48x 2 72x + 24x 2 + 8x 24x 2 12 48x 2 + 72x 8x + 12 = 24x 2 24x 2 + 48x 2 48x 2 72x + 8x + 72x 8x 12 + 12 = 0. Since W (y 1, y 2, y 5 ) = 0, the three functions are linearly dependent; in other words, while the linear combination c 1 y 1 + c 2 y 2 + c 3 y 5 can be used to describe many solutions of the equation, it will miss some other solutions entirely. We need to try again: let s use y 1, y 3, and y 4. Their Wronskian is the determinant of the matrix y 1 y 3 y 4 2x 3 x 2 + 1 2x 2 x y 1 y 3 y 4 = 2 2x 4x 1. y 1 y 3 y 4 0 2 4 To calculate the determinant, we use 2x 3 x 2 + 1 2x 2 x 2x 3 x 2 + 1 2 2x 4x 1 2 2x 0 2 4 0 2. Multiplying along diagonals, we get W (y 1, y 3, y 4 ) = (2x 3)(2x)(4) + (x 2 + 1)(4x 1)(0) + (2x 2 x)(2)(2) (x 2 + 1)(2)(4) (2x 3)(4x 1)(2) (2x 2 x)(2x)(0) = 16x 2 24x + 8x 2 4x 8x 2 8 (2x 3)(8x 2) = 16x 2 24x + 8x 2 4x 8x 2 8 16x 2 + 24x + 4x 6 = 16x 2 + 8x 2 8x 2 16x 2 24x + 24x 4x + 4x 8 6 = 14. Since W (y 1, y 3, y 4 ) 0, the three functions are linearly independent, and the most general solution to the original equation is y = c 1 y 1 + c 2 y 3 + c 3 y 4 ; any solution to the original equation is of this form. 8

Nonhomogeneous Equations Recall that a linear nonhomogeneous equation has associated homogeneous equation y (n) +... + A 2 (x)y + A 1 (x)y + A 0 (x)y = F (x) y (n) +... + A 2 (x)y + A 1 (x)y + A 0 (x)y = 0. Our original interest in homogeneous equations stemmed from the fact that we could use them to help find general solutions to nonhomogeneous equations. The following theorem tells us how this interaction plays out: Theorem 5. Suppose that y p is a solution of the nth order linear nonhomogeneous equation y (n) + p 1 (x)y (n 1) +... + p n 1 (x)y + p n (x)y = f(x) on an open interval I, and that each p i and f(x) are continuous on I. If y c is a solution of the associated homogeneous equation, then y = y p + y c is also a solution of the nonhomogeneous equation on I. If y 1,...,y n are linearly independent solutions to the associated nth order equation, then the general solution of the nonhomogeneous equation is Y = c 1 y 1 +... + c n y n + y p, and every solution to the nonhomogeneous equation is of this form. We will call a solution y p of the nonhomogeneous equation a particular solution; any solution y c of the associated homogeneous equation is called a complementary solution. The theorem tells us that we can find the general solution to the nonhomogeneous equation by finding any solution to the original equation, then adding it to the general solution of the associated homogeneous equation. Example. Find a general solution to y y = e 2x. We should begin by trying to find a particular solution: since the difference of y and the original function y is e 2x, it seems reasonable to guess that y should be something like e 2x ; perhaps a constant multiple of this would work. If y = ce 2x, then y = 2ce 2x and y = 4ce 2x. So y y = 4ce 2x ce 2x = 3ce 2x. 9

Since we want y y = e 2x, we should choose c = 1 3. It is clear that y p = 1 3 e2x is the particular solution we re looking for. To find a general solution to the equation, we need to find a general solution to the associated homogeneous equation y y = 0. The characteristic equation is r 2 r = 0, and since r 2 r factors as (r 0)(r 1), the roots are r 1 = 0 and r 2 = 1. The general solution to the homogeneous equation is thus and the general solution to y y = e 2x is y c = c 1 e 0x + c 2 e x = c 1 + c 2 e x, Y = c 1 + c 2 e x + 1 3 e2x. Theorem 5 may seem a bit unbelievable why should a solution to a homogeneous equation help us build the general solution to a nonhomogeneous equation? It seems that we lose a lot of information when we switch to the homogeneous version of the equation. Let s think about the theorem using our example above. Since y c = c 1 + c 2 e x is a solution to y y = 0, we know that y c y c = 0. In addition, we know that y p = 1 3 e2x is a solution to y y = e 2x, so that Adding these two equalities gives us y p y p = e 2x. which we can rewrite as y c y c + y p y p = e 2x, (y c + y p ) (y c + y p ) = e 2x, proving that y c + y p is indeed the general solution we were looking for. Finding the particular solution y p to the nonhomogeneous equation may seem a bit daunting; we will study this problem more in section 3.5. 10