4 Differential Equations
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1 Advanced Calculus Chapter 4 Differential Equations 65 4 Differential Equations 4.1 Terminology Let U R n, and let y : U R. A differential equation in y is an equation involving y and its (partial) derivatives. When n = 1 the differential equation is called an ordinary differential equation; when n 2 the differential equation is called a partial differential equation. The order of a differential equation is the highest order of derivative appearing in the equation. Example 1 The following are all examples of differential equations. Equations (a) (e) are ordinary differential equations and (f) (h) are partial differential equations. Equations (a), (b) and (f) have order one, (c), (d), (g) and (h) have order 2, and (e) has order three. (a) = 2x. (b) = ex y + 2. (c) d2 y y = 4 sin x. 2 (d) d2 y 2 = ω2 y. This is the equation for simple harmonic motion. It arises naturally in the solution of many problems in mechanics. (e) d3 y + y 3 xd2 + x2 2 + x3 y = 0. (f) z x + z y = 3xy. (g) 2 z y x = x2 + y 2. (h) U t = U α 2. (The heat equation.) x2 A solution of a differential equation in y is a relation between y and its variables, free from the derivatives and higher derivatives of y, that is consistent with the equation. Example 2 It is easy to check that y = x 2 is a solution of the differential equation = 2x. However, there are other solutions.
2 Advanced Calculus Chapter 4 Differential Equations 66 For example, y = x 2 + 5, y = x 2 7 and y = x are all solutions of this differential equation. In fact, y = x 2 + c is a solution of this differential equation for any constant c. This last solution is called the general solution of the differential equation. For most ordinary differential equations of order one, the general solution involves an arbitrary constant. Given the value of y for a particular value of x we can determine the value of c. Such values of y and x are called boundary conditions for the differential equation. For an ordinary differential equation of order k, the general solution normally involves k arbitrary constants. In this case, k independent boundary conditions are required to find a unique solution. The solutions of the differential equation in Example 2 gave y explicitly in terms of x. However, solutions of differential equations often express y implicitly in terms of its variables, as the following example illustrates. Example 3 The general solution of the ordinary differential equation = ex y + 2 is 1 2 y2 + 2y = e x + c, (1) where c is a constant. We can check this by differentiating expression (1) implicitly with respect to x to get y + 2 = ex. Thus (y + 2) = ex, and dividing throughout by y + 2 gives as required. = ex y + 2, Note that starting from the given solution we get y 2 + 4y = 2e x + d, where d = 2c. Thus, completing the square of the left hand side of the above solution gives (y + 2) 2 4 = 2e x + d,
3 Advanced Calculus Chapter 4 Differential Equations 67 and so (y + 2) 2 = 2e x + f, where f = d + 4. Taking square roots of both sides of this solution, and then subtracting 2 from both sides gives y = 2e x + f 2. So in this case it is also possible to give the solution for y explicitly in terms x. However, this is not always possible (see the next example), and it is debatable whether it is worth the effort to find such a solution in this case. Example 4 The general solution of the ordinary differential equation y(x 1) = x(y + 1) is y + ln y x + ln x = c. This can easily be verified by differentiating the solution with respect to x and rearranging the result to get. However, it is not possible to rearrange this solution to express y explicitly in terms of x. 4.2 Ordinary differential equations of order one In this section we consider the problem of finding the general solution of an ordinary differential equation of order one. This is not always possible, but there are many methods for finding the general solution, depending on the nature of the differential equation, and we consider some of these here. We also consider some special methods for certain families of differential equations. Variables separable differential equations Let M and N be functions of one variable, and consider the differential equation = M(x) N(y). Then N(y) = M(x). Integrating both side with respect to x gives N(y) = M(x),
4 Advanced Calculus Chapter 4 Differential Equations 68 and so, by the chain rule, N(y) = M(x). Thus, if we can evaluate the two integrals, we can find the general solution of the differential equation. Differential equations of this nature are called variables separable, and the method outlined for solving them is called the separation of variables. Example 5 Consider the differential equation 9x = x y Separating the variables gives x y = 9x. Thus 2y = 1 9 ( x + 1 ). x Evaluating the two integrals gives y 2 = 1 ( ) x ln x + c, where c is a constant, and so y = 1 3 x ln x + c. Example 6 Consider the differential equation where a is a constant. = ay, Separating the variables gives y = Evaluating the two integrals gives where c is a constant, and so where A = e c. a. ln y = ax + c, y = e ax+c = Ae ax,
5 Advanced Calculus Chapter 4 Differential Equations 69 Example 7 Consider the differential equation (3y + 2) 3 cos 2 x = 1. Separating the variables gives (3y + 2) 3 = Thus (3y + 2) 3 = Evaluating the two integrals gives 1(3y + 2)4 3 4 where c is a constant, and so where d = 12c. cos 2 x. sec 2 x. = tan x + c, (3y + 2) 4 = 12 tan x + d, Note that, with a bit more effort, we can write the solution of this differential equation as y = tan x + d 2, but it is a matter 3 of mathematical taste as to whether this is better way of expressing the solution than the one given. Exact differential equations Let M and N be functions of two variables. The differential equation M(x, y) + N(x, y) = 0 is exact if M y = N x. When a differential equation of the given form is exact then the general solution has the form f(x, y) = c, for some constant c and some function f, of two variables, satisfying f x = M and f y = N. To see this, we differentiate f(x, y) = c with respect to x. This gives f x + f y = 0. Since as required. = 1, f x = M and f y = N we get M + N = 0,
6 Advanced Calculus Chapter 4 Differential Equations 70 Example 8 Consider the differential equation 2xy + (x 2 + y 2 ) = 0. Here M = 2xy and N = x 2 +y 2, and since M M y y N = 2x and = 2x = x, the differential equation is exact. Thus we look for a function f of two variables such that f x = M and f y = N. That is, f x f y = 2xy; (2) = x 2 + y 2. (3) Integrating both sides of equation (2) with respect to x gives f = x 2 y + g(y), (4) where g is a function of y. To find g we differentiate equation (4) with respect to y to get Comparing this expression for f y f y = x2 + g (y). x 2 + g (y) = x 2 + y 2. with the one in equation (3) gives Thus g (y) = y 2 and so g(y) = 1 3 y3. Hence f(x, y) = x 2 y y3, and so the solution of the original differential equation is for some constant c. x 2 y y3 = c, Example 9 Consider the differential equation ( ) x ln x + 2x ln y + y 2y = 0. Here M = 1 + ln x + 2x ln y and N = x2 y M 2y, and since = 2x y y, the differential equation is exact. Thus we look and N = 2x = M x y y for a function f of two variables such that f That is, x = M and f y = N. f x f y = 1 + ln x + 2x ln y; (5) = x2 y 2y. (6)
7 Advanced Calculus Chapter 4 Differential Equations 71 Integrating both sides of equation (5) with respect to x gives f = x ln x + x 2 ln y + g(y), (7) where g is a function of y. To find g we differentiate equation (7) with respect to y to get To evaluate ln x, use integration by parts with u = ln x and dv = 1. Comparing this expression for f y f y = x2 y + g (y). with the one in equation (6) gives x 2 y + g (y) = x2 y 2y. Thus g (y) = 2y and so g(y) = y 2. Hence f(x, y) = x ln x + x 2 ln y y 2, and so the solution of the original differential equation is for some constant c. x ln x + x 2 ln y y 2 = c, Integrating factors Let M and N be functions of two variables. In general the differential equation M(x, y) + N(x, y) = 0 will not be exact. However, suppose we can find a function µ of two variables such that the differential equation µ(x, y)m(x, y) + µ(x, y)n(x, y) = 0 is exact. Then we can solve the resulting differential equation using the method described, and hence find the general solution of the original differential equation. The function µ is called an integrating factor for the original differential equation. Unfortunately, in general, it is difficult to find an integrating factor for most differential equations. However, there are certain classes of differential equations for which there is a method for finding one. In the next subsection we consider one important class of differential equations that can be solved using an integrating factor. Linear differential equations A ordinary differential equation of order one is linear if it can be written in the form + P (x)y = Q(x),
8 Advanced Calculus Chapter 4 Differential Equations 72 where P and Q are functions of one variable. Given a linear differential equation of this form, let µ(x) = e P (x). We claim that µ is an integrating factor for the linear differential equation. First note that µ (x) = P (x)e P (x) = P (x)µ(x). To differentiate µ(x), we use the chain ( rule and) the fact that P (x) = P (x). d Now rearrange the linear differential equation in the form P (x)y Q(x) + = 0, and then multiply both sides by µ(x) to get µ(x)(p (x)y Q(x)) + µ(x) = 0. We now need to show that this differential equation is exact. Let M = µ(x)(p (x)y Q(x)) and N = µ(x). Then M y = µ(x)p (x); N x = µ (x), = µ(x)p (x). Thus M = N, and so the differential equation is exact. Thus µ(x) y x is an integrating factor for the linear differential equation. Note that in this case we do not have to reduce the linear differential equation to the exact form since (using the Product Rule), d (µ(x)y) = µ (x)y + µ(x), = (µ(x)p (x))y + µ(x), ( = µ(x) P (x)y + ), = µ(x)q(x). d Thus (µ(x)y) = µ(x)q(x), and so integrating both sides with respect to x gives µ(x)y = µ(x)q(x).
9 Advanced Calculus Chapter 4 Differential Equations 73 Thus y = 1 µ(x) µ(x)q(x). (8) So to solve the linear differential equation + P (x)y = Q(x), first find the integrating factor µ(x) = e P (x), and then use identity (8) to find y. Example 10 Consider the linear differential equation where A, a and b are constants. + ay = Aebx, Here P (x) = a and Q(x) = Ae bx. Let Then µ(x) = e a = e ax. 1 y = µ(x)q(x), µ(x) = e ax (e ax )(Ae bx ), = e ax Ae (a+b)x, ( ) Ae = e ax (a+b)x + c, a + b A = a + b ebx + ce ax. So the general solution of the differential equation is where c is a constant. y = A a + b ebx + ce ax, Example 11 Consider the linear differential equation Then (x 2 + 1) + xy = x(x2 + 1). + x (x 2 + 1) y = x.
10 Advanced Calculus Chapter 4 Differential Equations 74 Thus the differential equation is linear with P (x) = Q(x) = x. Let µ(x) = e = e 1 2 x (x 2 +1), 2x (x 2 +1), = e 1 2 ln(x2 +1), ( ) 1 = e ln(x2 +1) 2, x and (x 2 +1) = (x 2 + 1) 1 2. Then 1 y = µ(x)q(x), µ(x) = (x 2 + 1) 1 2 ((x 2 + 1) 1 2 )x, ( ) = (x 2 + 1) (x2 + 1) c), = 1 3 (x2 + 1) + c(x 2 + 1) 1 2, = 1 3 (x2 + 1) + c x So the general solution of the differential equation is where c is a constant. y = 1 3 (x2 + 1) + c x2 + 1, Homogeneous differential equations Recall that a function f of two variables is homogeneous of degree n if f(λx, λy) = λ n f(x, y). Let M and N be functions of two variables. Consider the differential equation M(x, y) + N(x, y) = 0, where M and N are both homogeneous of degree n. (Such a differential equation is called homogeneous.) Then we put y = vx, where v is a function of x. By the product rule = xdv + v.
11 Advanced Calculus Chapter 4 Differential Equations 75 Also, since M is homogeneous of degree n, M(x, y) = M(x, vx), = M(x.1, x.v), = x n M(1, v), = x n ˆM(v), for some function ˆM of one variable. Similarly N(x, y) = x n ˆN(v), for some function ˆN of one variable. So, with y = vx the original differential equation becomes ( x n ˆM(v) + x n ˆN(v) x dv ) + v = 0. Since this is valid for all x, we can cancel the common factor of x n to get ( ˆM(v) + ˆN(v) x dv ) + v = 0. This final differential equation is variables separable. Thus, using the method given earlier, we can find v, and consequently y. Note When solving a homogeneous differential equation, it is not necessary to identify the functions ˆM and ˆN. You just have to remember to eliminate y by putting y = vx. If the resulting differential equation is not variables separable then you have either made a mistake, or the original differential equation is not homogeneous. Example 12 Consider the differential equation x 3 + y 3 + 2xy 2 = 0. It is easy to verify that M = x 3 + y 3 and N = 2xy 2 are both homogeneous of degree 3. Putting y = vx in the differential equation gives ( x 3 + v 3 x 3 + 2xv 2 x 2 x dv ) + v = 0. Cancelling the common factor of x 3, and rearranging the resulting differential equation gives Separating the variables gives 2v 2 2xv 2 dv = (1 + 3v3 ) v 3 dv = x.
12 Advanced Calculus Chapter 4 Differential Equations 76 Evaluating the two integrals we get 2 9 ln(1 + 3v3 ) = ln x + c, where c is a constant. Since y = vx then v = y, and so substituting x for v in the above solution, we get the general solution of the original differential equation as follows: ) 2 (1 9 ln + 3y3 + ln x = c. x 3 Example 13 Consider the differential equation ( y y + x sec x x) = 0. It is easy to verify that M = y +x sec ( y x) and N = x are both homogeneous of degree 1. Putting y = vx in the differential equation gives ( vx ) ( vx + x sec x x dv ) x + v = 0. Cancelling the common factor of x, and rearranging the resulting differential equation gives x dv = sec v. Separating the variables gives dv sec v = x, and, since sec v = 1, we get cos v cos v dv = x. Evaluating the two integrals, we get sin v = ln x + c, where c is a constant. Since y = vx then v = y, and so substituting x for v in the above solution, we get the general solution of the original differential equation as follows: ( y sin = ln x + c. x)
13 Advanced Calculus Chapter 4 Differential Equations 77 Change of variable When solving a homogeneous differential equation we changed the variables with the substitution y = vx. Often other first order differential equations can be solved by changing the variables. For some differential equations it is clear what the change of variables should be, as illustrated in the following example. Example 14 Consider the differential equation = (x + 4y + 3)2. Here we put z = x + 4y + 3. Then dz = 1 + 4, = 1 + 4(x + 4y + 3) 2, = 1 + 4z 2, and so we get the variables separable equation dz = 1 + 4z2. Separating the variables gives dz 1 + 4z = 2 and evaluating the two integrals we get for some constant c. Substituting for z gives, 1 arctan(2z) = x + c, 2 1 arctan(2(x + 4y + 3)) = x + c, 2 which is the general solution of the original equation. The general solution can be tidied up (with a little effort) to get where d = 2c. y = 1 (tan(2x + d) 2x 6), 8 In general, it is not always obvious what the change of variables should be. However, there are some classes of differential equations that can be solved with a standard change of variable. We give two such special cases in the next two subsections.
14 Advanced Calculus Chapter 4 Differential Equations 78 Special case 1 Let a, b, c, d, e, f R with at least one of d, e or f nonzero. Consider the differential equation = ax + by + c + ey + f. The differential equations in this class can be solved using a change of variables, depending on whether the lines ax + by + c = 0 and + ey + f = 0 are parallel. Suppose first that ax + by + c = 0 and + ey + f = 0 are not parallel. Then let (α, β) be the point where these two lines intersect, and put x = X + α, y = Y + β. We claim that dy =. To see this, suppose that y = g(x) for dx some function g. Then = g (x). Also Y + β = g(x + α) and so, by the chain rule, Thus dy = dx dy dx = g (X + α) = g (x) =., as claimed. Since (α, β) is on the line ax+by+c = 0, it follows that aα+bβ+c = 0. Thus Similarly ax + by + c = a(x + α) + b(y + β) + c, = ax + by + aα + Bβ + c, = ax + by. + ey + f = dx + ey. Using the above, after the change of variables, the differential equation becomes dy ax + by = dx dx + ey. This differential equation is homogeneous, and so can be solved using the method given earlier. Thus we can find the general solution of the original equation. Suppose now that ax + by + c = 0 and + ey + f = 0 are parallel. Then there are λ, g R such that + ey + f = λ(ax + by + g). Thus = ax + by + c λ(ax + by + g).
15 Advanced Calculus Chapter 4 Differential Equations 79 Put z = ax + by. Then dz = a + b, and so ( ) / dz = a b. If λ < 1 it is slightly easier to put z = + ey. Thus, changing the variables in the differential equation we get which can be rearranged to get ( ) / dz a b = z + c λ(z + g), dz = b(z + c) λ(z + g) + a. This differential equation is variables separable, and so can be solved using the method given earlier. Thus we can find the general solution of the original equation. Example 15 Consider the differential equation = x + y 6 x y 2. The lines x + y 6 = 0 and x y 2 = 0 are not parallel and intersect at the point (4, 2). Putting x = X + 4 and y = Y + 2 into the differential equation gives dy dx = X Y X + 4 Y 2 2 = X + Y X Y. This differential equation is homogeneous, so putting Y = XV we get dy = X dv + V, and the differential equation becomes dx dx Thus X dv dx + V = X + XV X XV = 1 + V 1 V. X dv dx = 1 + V 1 V V = 1 + V V (1 V ) 1 V = 1 + V 2 1 V. The last differential equation is variables separable and so 1 V 1 + V dv = 2 dx X. To find the point where the lines intersect solve the equations x + y = 6, x y = 2. Hence ( V V ) dx dv = V 2 X. Evaluating the two integrals gives arctan V 1 2 ln(1 + V 2 ) = ln X + c,
16 Advanced Calculus Chapter 4 Differential Equations 80 where c is a constant. Multiplying throughout by 2, and rearranging the solution gives 2 arctan V (ln(1 + V 2 ) + 2 ln X) = d, where d = 2c. Now 2 ln X = ln X 2, and so ln(1 + V 2 ) + 2 ln X = ln(1 + V 2 ) + ln X 2 = ln(x 2 + X 2 V 2 ). Thus Since V = Y X 2 arctan V ln(x 2 + X 2 V 2 ) = d. the solution becomes ( ) Y 2 arctan ln(x 2 + Y 2 ) = d. X Finally, putting X = x 4 and Y = y 2 we get ( ) y 2 2 arctan ln((x 4) 2 + (y 2) 2 ) = d, x 4 which is the general solution of the original differential equation. Example 16 Consider the differential equation = x + 2y + 5 2x + 4y 3. The lines x + 2y + 5 = 0 and 2x + 4y 3 = 0 are parallel so we put z = x + 2y. Then dz = 1 + 2, and so = 1 ( ) dz 2 1. Thus, changing the variable, the differential equation becomes ( ) 1 dz 2 1 = z + 5 2z 3. Hence dz = 2z z = 4z + 7 2z 3. This differential equation is variables separable, and so separating the variables gives 2z 3 4z + 7 dz =. (9)
17 Advanced Calculus Chapter 4 Differential Equations 81 Now 2z 3 4z + 7 dz = 1 4z 6 2 4z + 7 dz, = 1 4z dz, 2 4z + 7 = z + 7 dz, = 1 ( z 13 ) ln(4z + 7). 2 4 Thus evaluating the two integrals in equality (9) gives 1 (z 134 ) 2 ln(4z + 7) = x + c, where c is a constant. Multiplying both sides of the last equality by 8 we get 4z 13 ln(4z + 7) = 8x + d, where d = 8c. Substituting z = x + 2y in the last expression and simplifying the result gives 8y 4x 13 ln(4x + 8y + 7) = d, which is the general solution of the original differential equation. Special case 2 Let P and Q be functions of one variable, and consider the differential equation + P (x)y = Q(x)yn. To solve this differential equation first divide both sides by y n to get 1 y n + P (x) y = Q(x). n 1 Now put u = 1 y n 1 du = y 1 n. For this change of variable = (1 n)y n = 1 n y n, and so = yn du 1 n. After the change of variable, the differential equation becomes 1 y n ( ) y n du + up (x) = Q(x), 1 n
18 Advanced Calculus Chapter 4 Differential Equations 82 and so 1 du + up (x) = Q(x). 1 n Multiplying both sides by 1 n yields the differential equation du + (1 n)up (x) = (1 n)q(x), which is linear, so it can be solved by finding an integrating factor and using the method given earlier. Example 17 Consider the differential equation + y x = x3 y 6. Dividing both sides by y 6 gives 1 y xy = 5 x3. Putting u = 1, we get du = 6, and so = y6 du. Changing y 6 y 6 5 the variable in the differential equation gives ) 1 ( y6 du + u y 6 5 x = x3. Simplifying this differential equation, and then multiplying both sides by 5 gives du 5 x u = 5x3. The preceding differential equation is linear so we first find the integrating factor µ(x) = e 5 x, = e 5 ln x, = e ln x 5, = x 5 = 1 x 5. Then 1 u = µ(x)( 5x 3 ), µ(x) = x 5 5 x, ( 2 ) 5 = x 5 x + c, where c is a constant. Thus u = x 4 (5 + cx). Since u = 1 y 5, the general solution of the original differential equation is 1 y 5 = x4 (5 + cx), or x 4 y 5 (5 + cx) = 1.
19 Advanced Calculus Chapter 4 Differential Equations Ordinary differential equations of order two To find the general solution of a differential equation of order two is more difficult than for order one. In this section we only consider linear differential equations. That is, differential equations of the form P (x) d2 y + Q(x) + R(x)y = S(x), 2 where P, Q, R and S are functions of one variable. The homogeneous case with constant coefficients We start with the simplest case where P, Q and R are constant functions and S is identically zero. That is, the differential equation has the form P d2 y + Q + Ry = 0, (10) 2 where P, Q, R R. Such a differential equation is called homogeneous with constant coefficients. In looking for the general solution of equation (10) we note first that if y is a solution of the differential equation then so is Ay, for a constant A. Also, since the differential equation is of order two, we expect the general solution to involve two arbitrary constants. Finally, in Example 6 we found that the general solution of a homogeneous linear differential equation of order one was an exponential function. Inspired by this, we look for solutions of equation (10) having the form y = e tx, for some constant t. For this y, = tetx and d2 y = t 2 e tx. Substituting for y, and d2 y in equation (10) 2 2 gives P t 2 e tx + Qte tx + Re tx = 0. Since e tx 0, we can cancel this factor from the previous equation to get P t 2 + Qt + R = 0. This is a quadratic equation in t, and is called the auxiliary equation of the differential equation given in (10). Let α and β be the roots of the auxiliary equation. Then the general solution of (10) depends on the nature of α and β. There are three cases to consider. Case 1: α, β R with α β. This is the easiest case; since we know that e αx and e βx are distinct solutions of equation (10), the general solution is y = Ae αx + Be βx, where A and B are (real) constants.
20 Advanced Calculus Chapter 4 Differential Equations 84 Case 2: α, β R with α = β. In this case, since α = β, e αx = e βx and so we only have one solution of equation (10). However, it is straightforward to check that xe αx is also a solution of equation (10), and so the general solution is y = (A + Bx)e αx, where A and B are (real) constants. Case 3: α, β R. We first recall that it is not possible to have one of α or β in R and the other not in R, since the non-real roots of a polynomial equation with coefficients in R occur in conjugate pairs. Thus, in this case, α = a + bi and β = a ib where a, b R, Recall that i = 1. with b 0. Doing the same as in case 1 we could write the general solution of equation (10) in the form y = Âe(a+bi)x + ˆBe (a ib)x, where  and ˆB are complex constants. However, it is unsatisfactory to have the solution of a differential equation involving a real function expressed in terms of complex functions, so we aim to find a better way of expressing the solution. Now Similarly, e (a+bi)x = e ax+bxi = e ax e bxi = e ax (cos(bx) + i sin(bx)). e (a bi)x = e ax bxi = e ax e bxi = e ax (cos(bx) i sin(bx)). Using these equivalent expressions for e (a+bi)x and e (a bi)x we get Âe (a+bi)x + ˆBe (a ib)x = Âeax (cos(bx) + i sin(bx)) + ˆBe ax (cos(bx) i sin(bx)), = e ax (( + ˆB) cos(bx) + i(â ˆB) sin(bx)), = e ax (A cos(bx) + B sin(bx)), where A =  + ˆB and B = i(â ˆB). In summary, we have the following method for finding the general solution of equation (10). First find the solutions α and β of the auxiliary equation P t 2 + Qt + R = 0. Then the following table gives the general solution of equation (10), where A and B are constants. Nature of the roots of General solution the auxiliary equation of equation (10) α, β R, α β y = Ae αx + Be βx α, β R, α = β y = (A + Bx)e αx α = a + ib, β = a ib, a, b R, b 0 y = e ax (A cos(bx) + B sin(bx)) Here we use the standard rules for exponentials together with the result that for all θ R. e iθ = cos θ + i sin θ, Note, since A, B R, it follows that ˆB is the complex conjugate of Â.
21 Advanced Calculus Chapter 4 Differential Equations 85 Example 18 Consider the differential equation d 2 y 5 + 6y = 0. 2 The auxiliary equation of this differential equation is t 2 5t + 6 = 0. Thus (t 2)(t 3) = 0, and so t = 2, 3. Hence the general solution of the differential equation is where A and B are constants. y = Ae 2x + Be 3x, Example 19 Consider the differential equation 4 d2 y y = 0. The auxiliary equation of this differential equation is 4t 2 4t + 1 = 0. Thus (2t 1) 2 = 0, and so t = 1 2 of the differential equation is (twice). Hence the general solution where A and B are constants. y = (Ax + B)e 1 2 x, Example 20 Consider the differential equation d 2 y y = 0. 2 The auxiliary equation of this differential equation is Thus t 2 10t + 29 = 0. t = 10 ± , 2 = 10 ± 16, 2 = 5 ± 2i. Hence the general solution of the differential equation is where A and B are constants. y = e 5x (A cos(2x) + B sin(2x)),
22 Advanced Calculus Chapter 4 Differential Equations 86 The nonhomogeneous case with constant coefficients We now consider the case where P, Q and R are constant functions and S is a function of x. That is, the differential equation has the form P d2 y + Q + Ry = S(x), (11) 2 where P, Q, R R. Such a differential equation is called nonhomogeneous with constant coefficients. To find the general solution of equation (11) we have to do two things. First we find the general solution C(x)of the homogeneous part P d2 y + Q + Ry = 0. 2 This is called the complementary function of equation (11). Next we find any solution p(x) of equation (11). This is called a particular integral of the differential equation. Then the general solution of equation (11) is y = C(x) + p(x). Finally, if there are any boundary conditions we use them to determine the arbitrary constants in the general solution. Finding a particular integral In the above method, we know how to find the complementary function. However, finding a particular integral is very hard for most functions S(x), but there are some families of functions for which it is possible to find a particular integral. Here we give three such families of functions. Family 1: S(x) = Me γx, where M, γ R. for a particular integral of the form p(x) = Gx j e γx, In this case we look where j is the number of times γ is a root of the auxiliary equation, and G is a constant. Family 2: S(x) = M cos(cx) + N sin(cx), where M, N, c R. this case we look for a particular integral of the form p(x) = G cos(cx) + H sin(cx), where G and H are constants, unless the roots of the auxiliary equation are ±ci, in which case we look for a particular integral of the form p(x) = x(g cos(cx) + H sin(cx)). In
23 Advanced Calculus Chapter 4 Differential Equations 87 Family 3: S(x) = M k x k + M k 1 x k M 1 x + M 0, where M k, M k 1,..., M 1, M 0 R. In this case we look for a particular integral of the form p(x) = x j (m k x k + m k 1 x k m 1 x + m 0 ), where j is the number of times 0 is a root of the auxiliary equation, and m k, m k 1,..., m 1, m 0 are constants. Observe that, if 0 is a root of the auxiliary equation then the differential equation P d2 y + Q + Ry = S(x) 2 must have R = 0. That is the differential equation is P d2 y 2 + Q = S(x). In this case, it is easier to change the variable by putting u = (so du = d2 y ) to get 2 P du + Qu = S(x), and then solve the resulting first order differential equation. If we do this, then whenever S(x) is a polynomial of degree k the particular integral is also a polynomial of degree k. In general, we note that for all three families of functions considered, the particular integral has the same form as S(x), except when the roots of the auxiliary equation take certain values. In the latter case we multiple the form of the particular integral by a suitable power of x; this is required because, in this case, S(x) is part of the complementary function and so cannot be a particular integral of the nonhomogeneous differential equation. Finally, we observe that if S(x) is a sum of functions from the three families considered then the particular integral will be a sum of the corresponding particular integrals. Example 21 Consider the differential equation d 2 y y = 20e 3x, with boundary conditions y = 5, = 3 when x = 0. From Example 18, the auxiliary equation of the homogeneous part of this differential equation is t 2 5t + 6 = 0, with roots t = 2, 3. Hence the complementary function of the differential equation is C = Ae 2x + Be 3x,
24 Advanced Calculus Chapter 4 Differential Equations 88 where A and B are constants. Here S = 20e 3x, so we look for a particular integral of the form p = Ge 3x. Then dp = 3Ge 3x and d2 p = 9Ge 3x. Now, since p is 2 a particular integral of the differential equation it must satisfy the differential equation. So putting y = p in the differential equation gives 9Ge 3x 5( 3Ge 3x ) + 6Ge 3x = 20e 3x. Cancelling the common factor of e 3x from this expression, and simplifying the result gives 30G = 20, and so G = 2. Thus the 3 particular integral is p = 2 3 e 3x. From the above, the general solution of the differential equation is y = C + p = Ae 2x + Be 3x e 3x. We now use the boundary conditions to determine A and B. Differentiating the general solution with respect to x gives = 2Ae2x + 3Be 3x 2e 3x. Now, when x = 0, y = 5 and = 3. Thus, putting x = 0 in the general solution and its derivative, and then comparing them with the corresponding boundary condition gives A + B = 5, 2A + 3B 2 = 3. Simplifying these equations we get 3A + 3B = 13, (12) 2A + 3B = 5. (13) Subtracting equation (13) from equation (12) gives A = 8; substituting A = 8 in equation (13) gives B = 5, which gives B = 11. Thus the solution of the differential equation with the 3 given boundary conditions is y = 16e 2x 11 3 e3x e 3x. Example 22 Consider the differential equation d 2 y y = 3e2x.
25 Advanced Calculus Chapter 4 Differential Equations 89 As in the previous example, the roots of the auxiliary equation are t = 2, 3 and the complementary function of the differential equation is C = Ae 2x + Be 3x, where A and B are constants. Here S = 3e 2x. Since 2 is a single root of the auxiliary equation, we look for a particular integral of the form p = Gxe 2x. Then dp = Ge2x + 2Gxe 2x, = G(1 + 2x)e 2x ; d 2 p 2 = 2Ge 2x + 2G(1 + 2x)e 2x, = 4G(1 + x)e 2x. So putting y = p in the differential equation gives 4G(1 + x)e 2x 5(G(1 + 2x)e 2x ) + 6Gxe 2x = 3e 2x. Cancelling the common factor of e 2x from this expression gives G(4 + 4x 5 10x + 6x) = 3. Thus G = 3, and so G = 3. Thus the particular integral is p = 3xe 2x. Note that all of the x terms cancel. If this does not happen then we have done something wrong. From the above, the general solution of the differential equation is y = C + p = Ae 2x + Be 3x 3xe 2x. Example 23 Consider the differential equation d 2 y y = 29x2 + 38x 18. From Example 20, the auxiliary equation of the homogeneous part of this differential equation is t 2 10t+29 = 0, with roots t = 5±2i. Hence the complementary function of the differential equation is where A and B are constants. C = e 5x (A cos(2x) + B sin(2x)), Here S = 29x x 18, so we look for a particular integral of the form p = ax 2 + bx + c. Then dp = 2ax + b and d2 p = 2a. So 2 putting y = p in the differential equation gives 2a 10(2ax + b) + 29(ax 2 + bx + c) = 29x x 18.
26 Advanced Calculus Chapter 4 Differential Equations 90 Comparing the coefficients of x 2 and x, and the constant term, respectively, for both sides of the previous expression gives 29a = 29, 20a + 29b = 38, 2a 10b + 29c = 18. Solving the system of linear equations gives a = 1, b = 2 and c = 0. Thus the particular integral is p = x 2 + 2x. From the above, the general solution of the differential equation is y = C + p = e 5x (A cos(2x) + B sin(2x)) + x 2 + 2x. Example 24 Consider the differential equation d 2 y + y = 4 sin x + 3 cos x. 2 The auxiliary equation of the homogeneous part of this differential equation is t = 0, and so t = ±i. Hence the complementary function of the differential equation is where A and B are constants. C = A cos x + B sin x, Here S = 4 sin x + 3 cos x. Since ±i are roots of the auxiliary equation, we look for a particular integral of the form p = x(g sin x + H cos x). Then dp = G sin x + H cos x + x(g cos x H sin x); d 2 p 2 = G cos x H sin x + G cos x H sin x + x( G sin x H cos x), = 2(G cos x H sin x) x(g sin x + H cos x). Putting y = p in the differential equation gives 2(G cos x H sin x) x(g sin x+h cos x)+x(g sin x+h cos x) = 4 sin x+3 cos x. Thus 2(G cos x H sin x) = 4 sin x + 3 cos x. Thus G = 3 and 2 H = 2, and so the particular integral is ( ) 3 p = x sin x 2 cos x. 2 From the above, the general solution of the differential equation is ( ) 3 y = C + p = A cos x + B sin x + x sin x 2 cos x. 2
27 Advanced Calculus Chapter 4 Differential Equations 91 Example 25 Consider the differential equation d 2 y y = 2ex + 3x + 4. The auxiliary equation of the homogeneous part of this differential equation is t 2 + t + 1 = 0, and so t = 1 ± 1 4, 2 = ± 2 i. Hence the complementary function of the differential equation is ( ) ( )) C = e (A x cos 2 x + B sin 2 x, where A and B are constants. Here S = 2e x + 3x + 4. This is the sum of functions from two of the families considered. Hence we look for a particular integral of the form p = Ge x + Hx + K. Then dp = Gex + H; d 2 p 2 = Ge x. Putting y = p in the differential equation gives Ge x + Ge x + H + Ge x + Hx + K = 2e x + 3x + 4. Thus 3Ge x + Hx + H + K = 2e x + 3x + 4. Thus G = 2 3, H = 3 and H + K = 4, so K = 1. Thus the particular integral is p = 2 3 ex + 3x + 1. From the above, the general solution of the differential equation is ( ) ( )) y = C +p = e (A x cos 2 x + B sin 2 x ex +3x+1. Order two linear differential equations We now consider general order two linear differential equations. That is, differential equations of the form P (x) d2 y + Q(x) + R(x)y = S(x), (14) 2
28 Advanced Calculus Chapter 4 Differential Equations 92 where P, Q, R and S are functions of one variable. For such a differential equation, the homogeneous part is the differential equation P (x) d2 y + Q(x) + R(x)y = 0. 2 As in the case with constant coefficients, finding the general solution of equation (14) depends upon being able to find a solution of the homogeneous part. To see this, suppose y = g is a solution of the homogeneous part, for some function g of x. Then P (x) d2 g dg + Q(x) + R(x)g = 0. 2 Now we change the variable in equation (14) by putting y = gv for some function v of x. Then, using the product rule, we get = g dv + v dg ; d 2 y = g d2 v dg dv 2 + v d2 g. 2 Thus, with the change the variable, equation (14) becomes ( P (x) g d2 v + 2 dg ) ( dv 2 + v d2 g +Q(x) g dv 2 + v dg ) +R(x)gv = S(x), which can be rewritten as ( ) v P (x) d2 g + Q(x) dg + R(x)g + 2 ( ) P (x) g d2 v + 2 dg dv + Q(x)g dv = S(x). (15) 2 Now, since g is a solution of the homogeneous part of equation (14), the first term of equation (15) is zero. Thus equation (15) becomes ( P (x) g d2 v + 2 dg ) dv + Q(x)g dv 2 = S(x), which can be rewritten as P (x)g d2 v 2 + ( 2P (x) dg + Q(x)g ) dv = S(x). (16) Now we put w = dv P (x)g dw + in equation (16) to get ( 2P (x) dg ) + Q(x)g w = S(x), which is a first order differential equation for w. Solving this differential equation gives w. We can then find the general solution of equation (14) by integrating w with respect to x to get v and finally getting y = vg.
29 Advanced Calculus Chapter 4 Differential Equations 93 Note You do not have to remember the first order differential equation for w in terms of P, Q, S and g to solve equation (14); just remember the change of variable y = gv, and derive the first order differential equation from scratch. Example 26 Consider the differential equation x 2 (x + 1) d2 y + x(3x 2) 2 (3x 2)y = 10x3. It is easy to verify that y = x is a solution of the homogeneous part of this equation. So putting y = vx we get = xdv + v; d 2 y 2 = x d2 v + 2dv 2. Thus the differential equation becomes ( ) ( x 2 (x+1) x d2 v + 2dv +x(3x 2) x dv ) 2 + v (3x 2)vx = 10x 3. Collecting terms in the previous equation involving d2 v 2 get and dv we x 3 (x + 1) d2 v 2 + ( 2x 2 (x + 1) + x 2 (3x 2) ) dv = 10x3, and simplifying gives x 3 (x + 1) d2 v dv + 5x3 2 = 10x3. Dividing the resulting equation throughout by x 3 (x + 1), and then putting w = dv we get dw + 5w x + 1 = 10 x + 1, which is a first order linear differential equation. To solve this differential equation we first find the integrating factor µ(x) = e 5 x+1, = e 5 ln(x+1), = e ln(x+1)5, = (x + 1) 5.
30 Advanced Calculus Chapter 4 Differential Equations 94 Then where C is a constant. Now, dv where D = C 4 1 w = µ(x) 10 µ(x) x + 1, 1 = (x + 1) 5 10 (x + 1) 5 x + 1, 1 = 10(x + 1) 4, (x + 1) 5 1 ( = 2(x + 1) 5 + C ), (x + 1) 5 C = 2 + (x + 1), 5 = w, and so v = w, = ( ) C 2 +, (x + 1) 5 = D 2x + (x + 1) + E, 4 and E is a constant. Finally, y = vx, and so the general solution of the original differential equation is y = 2x 2 + Dx (x + 1) + Ex. 4 Euler equations In general, it is difficult to find a solution of the homogeneous part of an order two linear differential equation. However, there are some families of equations where this is possible. Here we consider one such family, the Euler equations. These are differential equations of the form ax 2 d2 y + bx + cy = S(x), 2 where a, b and c are constants, and S is a function of one variable. There are two possible ways to look for the general solution of an Euler equation. Look for a solution of the homogeneous part of the equation of the form y = x k. If a suitable value of k is found then apply the method described above to find the general solution of the nonhomogeneous equation. (This method does not always work as the values of k found could be non-real.)
31 Advanced Calculus Chapter 4 Differential Equations 95 Change the variable by putting x = e t. This reduces the differential equation to one with constant coefficients. (This method always works, but is probably more difficult.) We now give some examples to illustrate the two methods. Example 27 Consider the differential equation We look for a solution of x 2 d2 y 2 7x + 16y = 16x4. x 2 d2 y 2 7x + 16y = 0 of the form y = x k. Then = kxk 1 and d2 y = k(k 1)x k 2. 2 Substituting for y, and d2 y in the homogeneous part gives 2 and so x 2 k(k 1)x k 2 7xkx k x k = 0, k(k 1)x k 7kx k + 16x k = 0. The last equation is valid for all x and so we can cancel the x k term. Thus k(k 1) 7k + 16 = 0 k 2 8k + 16 = 0, (k 4) 2 = 0, k = 4. So y = x 4 is a solution of the homogeneous part of the differential equation. We now change the variable in the differential equation by putting y = vx 4. Then dv = x4 + 4x3 v, ( = x 3 x dv d 2 y 2 = x 3 ) + 4v ; ( x d2 v + dv 2 + 4dv = x 2 ( x 2 d2 v 2 + 8xdv + 12v ) + 3x 2 ( x dv ). ) + 4v Substituting the expressions found for y, dv and d2 v in the differential equation we get 2 ( ) ( x 4 x 2 d2 v + 8xdv v 7x 4 x dv ) + 4v + 16vx 4 = 16x 4.,
32 Advanced Calculus Chapter 4 Differential Equations 96 Cancelling the common factor of x 4 in the previous equation, and simplifying the result gives x 2 d2 v 2 + xdv = 16. Putting w = dv in this equation, and dividing throughout by x2 we get dw + w x = 16 x, 2 which is a first order linear differential equation. To solve this differential equation we first find the integrating factor µ(x) = e 1 x, = e ln x, = x. Then where C is a constant. Now, dv 1 w = µ(x) 16 µ(x) x, 2 = 1 x 16 x x, 2 = 1 16 x x, = 1 (16 ln x + C), x 16 ln x = + C x x = w, and so v = w, = ( 16 ln x + C ), x x = 8(ln x) 2 + C ln x + D, To evaluate 16 ln x x, put u = ln x. where D is a constant. Finally, y = vx 4, and so the general solution of the original differential equation is y = (8(ln x) 2 + C ln x + D)x 4. Example 28 Consider again the differential equation x 2 d2 y 2 7x + 16y = 16x4.
33 Advanced Calculus Chapter 4 Differential Equations 97 We now consider the second way of solving this equation using the change of variable x = e t. The dt = dt, = et, = x ; ( ), dt = d ( x ), dt = ( d x ), dt ( ) = e t + y xd2, 2 d 2 y dt 2 = d dt = x ( + xd2 y 2 = x + x2 d2 y 2. ), Using the final expressions for and d2 y dt dt 2 x = dt ; x 2 d2 y 2 = d2 y dt 2 x, = d2 y dt 2 dt. we get Replacing there terms x and x2 d2 y 2 the above equivalent terms gives of the original equation with d 2 y dt 2 dt 7 dt + 16y = 16x4. Simplifying the previous equation, and noting that x 4 = (e t ) 4 = e 4t, we get d 2 y dt 8 2 dt + 16y = 16e4t. This is a linear differential equation with constant coefficients, so we can solve it using the method given earlier. The auxiliary equation of this equation is s 2 8s + 16 = 0. Thus (s 4) 2 = 0, and so s = 4 (twice). Hence the complementary function of the differential equation is C(t) = (At + B)e 4t,
34 Advanced Calculus Chapter 4 Differential Equations 98 where A and B are constants. Since 4 is a double root of the auxiliary equation, we look for a particular integral of the form p = Gt 2 e 4t. Then dp dt = 2Gte 4t + 4Gt 2 e 4t, = 2Gt(1 + 2t)e 4t ; d 2 p dt 2 = 2G ( (1 + 4t)e 4t + 4t(1 + 2t)e 4t), = 2G(1 + 8t + 8t 2 )e 4t. So putting y = p in the differential equation gives 2G(1 + 8t + 8t 2 )e 4t 16Gt(1 + 2t)e 4t + 16Gt 2 e 4t = 16e 4t. Cancelling the common factor of e 4t from this expression gives G(2 + 16t + 16t 2 16t 32t t 2 ) = 16. Thus 2G = 16, and so G = 8. Thus the particular integral is p = 8t 2 e 4t. From the above, the general solution of the differential equation is y = C + p, = (At + B)e 4t + 8t 2 e 4t, = (At + B + 8t 2 )e 4t. Now e t = x, and so e 4t = x 4 and t = ln x. Thus y = (A ln x + B + 8(ln x) 2 )x 4.
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