Name: November 8, 011 Worksheet # : Higher Order Linear ODEs (SOLUTIONS) 1. A set of n-functions f 1, f,..., f n are linearly independent on an interval I if the only way that c 1 f 1 (t) + c f (t) +... + c n f n (t) = 0 holds for all t in I is if c 1 = c =... = c n = 0. Otherwise, they are linearly dependent. (a) Prove that y 1 (t) = t, y (t) = 3t + t + 1, y 3 (t) = t + 1, y 4 = sin t are linearly dependent on (, ) by finding a combination with c 1,..., c 4 not all zero so that for all t. c 1 y 1 (t) + c y (t) + c 3 y 3 (t) + c 4 y(t) = 0 Solution. Set c 1 = 3, c = 1, c 3 = 1 and c 4 = 0, since 3(t ) + (3t + t + 1) (t + 1) + 0(sin t) = 0 for all t. Note there are other possible choices. (b) There is a Wronskian test for verifying that a set of functions is linearly independent. For 3 twice-differentiable functions, the Wronskian is y 1 (t) y (t) y 3 (t) W (y 1, y, y 3 )(t) = y 1(t) y (t) y 3(t) y 1 (t) y (t) y 3 (t) (The Wronskian for 4 or more functions is defined in a similar way (see page 1) but requires computation of determinant larger than 3 3). If W (y 1, y, y 3 )(t) 0 for some point t in I, then the functions must be linearly independent. If, on the other hand, the functions are linearly dependent then the Wronskian must be 0 for all t in I. i. Use the Wronskian to verify that y 1 = x, y = x and y 3 = x 3 are linearly independent on (, ). Solution. x x x 3 W (y 1, y, y 3 )(x) = 1 x 3x 0 6x = x(1x 6x ) x (6x) + x 3 () = x 3. Since this expression is not zero for example at x = 1 the functions are linearly independent. ii. The functions y 1 (t) = t + t, y (t) = t t, and y 3 (t) = t are linearly dependent on (, ). Find c 1, c, c 3 not all zero so that c 1 y 1 (t) + c y (t) + c 3 y 3 (t) = 0 holds for all t. Verify that the Wronskian is 0 for all t. Solution. Let c 1 =, c = 1, and c 3 = 3/. We have (t + t) + 1(t t) + 3 (t) = t t + t t + 3t = 0.
Therefore, the functions are linearly dependent. The Wronskian must be zero. We check that t + t t t t W (y 1, y, y 3 )(t) = t + 1 4t 1 4 0 = (t + t)( 8) (t t)( 4) + t(4(t + 1) (4t 1)) = 8t 8t + 8t 4t + t(8t + 4 8t + ) = 1t + 1t = 0. Consider the homogeneous third order differential equation L[y] = y (3) + p(t)y + q(t)y + r(t)y = 0 Assume that p, q, r are all continuous functions on an open interval I. The general solution to this equation is y = c 1 y 1 + c y + c 3 y 3 where y 1, y, and y 3 are three times differentiable, linearly independent functions which are also solutions to the ODE. The solutions y 1, y, y 3 form a fundamental set. There is again a Wronskian test for linearly independence of solutions: y 1, y, y 3 are linearly independent solutions to the ODE if and only if y 1 (t) y (t) y 3 (t) W (y 1, y, y 3 )(t) = y 1(t) y (t) y 3(t) y 1 (t) y (t) y 3 (t) 0 for some point t in I. (a) Let p(t) = 6, q(t) = 11, and r(t) = 6. Verify that y 1 (t) = e t, y (t) = e t, y 3 (t) = e 3t are linearly independent solutions to the ODE. Construct the general solution and find the constants c 1, c, c 3 so that the initial conditions hold. Solution. The differential equation is Note that y(0) = 1, y (0) = 1, y (0) = 1 L[y] = y 6y + 11y 6y = 0 L[e rt ] = (r 3 6r + 11r 6)e rt so that if y = e rt is a solution r must be a root of the cubic equation This equation factors as r 3 6r + 11r 6 = 0 (r 1)(r )(r 3) = 0 There are three, distinct real roots r = 1,, 3. Each of these gives a solution to the differential equation: y 1 (t) = e t, y (t) = e t, and y 3 (t) = e 3t. We must verify that these are linearly independent. Compute the Wronskian: e t e t e 3t W (y 1, y, y 3 )(t) = e t e t 3e 3t e t 4e t 9e 3t = e t (18e 5t 1e 5t ) e t (9e 4t 3e 4t ) + e 3t (4e 3t e 3t ) = 6e 6t 6e 6t + e 6t = e 6t 0
So the solutions are linearly independent. The general solution is y(t) = c 1 e t + c e t + c 3 e 3t The first and second derivatives of this solution are y (t) = c 1 e t + c e t + 3c 3 e 3t y (t) = c 1 e t + 4c e t + 9c 3 e 3t To satisfy the initial conditions at t = 0 we need: 1 = c 1 + c + c 3 1 = c 1 + c + 3c 3 1 = c 1 + 4c + 9c 3 There are many ways to solve this system of algebraic equations. One easy way is to subtract the first equation from the third: 0 = 3c + 8c 3 So c = 8 3 c 3. We can then subtract the second equation from the first to get = c c 3 Substituting the c = 8 3 c 3: = 3 c 3 c 3 = 3 Then and plugging into the first equation c = 8 1 = c 1 8 + 3 c 1 = 6. Therefore, the unique solution satisfying these initial conditions is y = 6e t 8e t + 3e 3t 4.1: #11 Verify that 1, cos t, sin t are solutions to y + y = 0 and compute their Wronskian. Solution. By substituting these functions into the equation, it is clear that each is a solution. We compute the Wronskian 1 cos t sin t 0 sin t cos t 0 cos t sin t = 1(sin t + cos t ) = 1. 4.1: # 15 Verify that 1, x, x 3 are solutions to and compute their Wronskian. xy xy = 0
Solution. By substituting these functions into the equation, it is clear that each is a solution. For example x(x 3 ) x(x 3 ) = 3x(x ) 3x(x ) = 6x(x) 6x = 0. We compute the Wronskian 1 x x 3 0 1 3x 0 0 6x = 6x. 3. Consider the constant coefficient homogeneous, 3rd order ODE ay + by + cy + dy = 0. (a) Substitute y = e rt into the ODE to obtain a characteristic equation. You can always factor this equation into the form a(r r 1 )(r r )(r r 3 ) = 0. List all possible cases for the roots of r. For example, can this equation have three complex roots? One complex and two real? Repeated roots? Answer. The important fact is that complex roots of polynomial equations (with real coefficients) can only occur in complex conjugate pairs. So you cannot have three complex roots or one complex root. Since the roots are conjugates, complex roots in this case can t be repeated. The cases are I. Three real and distinct roots. II. Three real roots, but only two distinct. That is r 1, r, r 3 real with, say r = r 3. III. Three real roots, all repeated. IV. One real root and a complex conjugate pair of roots. (b) If the roots are all real and different, prove that y 1 = e r1t, y = e rt, y 3 = e r3t form a fundamental set. Solution. Use the operator notation If y = e rt, then it is clear that L[y] = ay + by + cy + dy. L[y] = (ar 3 + br + cr + d)e rt and so L[y] = 0 if and only if r is one of r 1, r, r 3. Thus y 1, y, y 3 are solutions. The Wronskian is e r 1t e r t e r 3t r 1 e r1t r e rt r 3 e r3t r 1e r 1t re r t r3e r 3t = (r r3 rr 3 )e (r1+r+r3)t (r 1 r3 r1r 3 )e (r1+r+r3)t + (r 1 r r 1r )e (r 1+r +r 3 )t = (r r 3 r r 3 r 1 r 3 r 1r 3 + r 1 r r 1r )e (r1+r+r3)t = (r r 1 )(r 3 r 1 )(r 3 r )e (r 1+r +r 3 )t This expression is clearly not zero since all of r 1, r and r 3 are distinct. Therefore, the solutions are linearly independent. (c) If the equation has one repeated real root r, prove that y 1 = e rt, y = te rt, y 3 = t e rt form a fundamental set.
Solution. You can verify exactly as above that y 1 (t) = e rt is a solution. Compute L[y ] = a(te rt ) + b(te rt ) + c(te rt ) + dte rt = a(e rt + rte tr ) + b(e rt + rte tr ) + c(e rt + rte tr ) + dte rt = a(re rt + re tr + r te rt ) + b(re rt + re tr + r te rt ) + c(e rt + rte rt ) + dte rt = a(3r e rt + r 3 te rt ) + b(r + r t)e rt + c(1 + rt)e rt + dte rt = (ar 3 + br + cr + d)te rt + (3ar + br + c)e rt Now clearly ar 3 + br + cr + d = 0. I claim the second term is also zero. To see why, notice that ax 3 + bx + cx + d = a(x r) 3 Expanding the right hand side we see that ax 3 + bx + cx + d = a x 3 3 a r x + 3 a r x a r 3. Matching coefficients, we have that b = 3ar, c = 3ar, d = ar 3. Therefore and it follows that 3ar + br + c = 0 L[y ] = 0. The same trick will work to show that L[y 3 ] = 0 and y 1, y, y 3 are all solutions. We compute the Wronskian and carefully simplify to obtain e rt te rt t e rt re rt (1 + rt)e rt (t + rt )e rt r e rt r( + rt))e rt ( + 4rt + r t )e rt = e3rt 0 So the solutions y 1, y, y 3 are linearly independent on (, ) and hence form a fundamental set for the ODE. (d) Complex roots always occur in conjugate pairs λ ± µi. A complex conjugate pair gives rise to solutions y 1 (t) = e λt cos µt and y (t) = e λt sin µt. What is the fundamental solution set in this case? Be sure to verify that the solutions in your fundamental set are linearly independent. Solution. The characteristic equation has one real root, call it r 1, and a complex conjugate pair of roots r,3 = µ ± λt. The fundamental set in this case is y 1 (t) = e rt, y (t) = e λt cos µt, y 3 (t) = e λt sin µt. It is clear that y 1 (t) is a solution by the arguments of the previous problem(s). You do not have to check that y (t), y 3 (t) are solutions, since it is a given in the statement of the problem.
If you want to see why they are though, substitute y (t) into the equation and simplify to obtain: L[y ] = ( (a µ 3 + ( 3 a λ b λ c ) µ ) sin (µ t) + ( ( 3 a λ b) µ + a λ 3 + b λ + c λ + d ) cos (µ t) ) e λ t On the other hand, you can plug r back in to the characteristic equation and collect real and imaginary parts to see that 0 = ar 3 + br + cr + d = ( ( 3 a λ b) µ + a λ 3 + b λ + c λ + d ) + (( 3 a λ + b λ + c ) µ a µ 3) i. For this to hold, it must follow that ( 3 a λ + b λ + c ) µ a µ 3 = 0 ( 3 a λ b) µ + a λ 3 + b λ + c λ + d = 0 and y must be solution. A similar argument works for y 3. The Wronskian to expand is: e r t cos (µ t) e t λ sin (µ t) e t λ r e r t ( (cos (µ t) λ µ sin (µ t)) e t λ (sin (µ t) λ + µ cos (µ t)) e t λ r e r t cos (µ t) λ µ sin (µ t) λ µ cos (µ t) ) ( e t λ sin (µ t) λ + µ cos (µ t) λ µ sin (µ t) ) e t λ If you compute things correctly, you should find that W (y 1, y, y 3 )(t) = µ(λ rλ + r + µ )e tλ+rt Can this quantity ever be 0? Well, the exponential part is never 0 and µ 0, since r, r 3 must have an imaginary part. This leaves the possibility that λ r λ + r + µ = 0. But this is a quadratic equation in r and if you solve it, you will find that the roots are precisely r = µ ± λi. This cannot occur since by assumption r must be the real root. We therefore may conclude that W (y 1, y, y 3 )(t) 0 and so the solutions are linearly independent on (, ). Therefore, y 1, y, y 3 form a fundamental set. (e) Now provide the fundamental solution sets for all other cases you described in part (a). Solutions. The only remaining case is (II). Here we have the root r 1 and the twice repeated root r and these lead to the fundamental set y 1 (t) = e r 1t, y (t) = te r t, and y 3 (t) = te r 3t. You can verify that this is a fundamental set as we did above but I do not require you to do this. (f) 4.: # 9: Solve y + y = 0 y(0) = 0 y (0) = 1 y (0) =
Solution. The characteristic equation is r 3 + r = r(r + 1) = 0. The roots of this equation are Thus the general solution is r 1 = 0 r,3 = ±i. y(t) = c 1 + c cos t + c 3 sin t. Compute y (t) and y (t): y (t) = c sin t + c 3 cos t y (t) = c cos t c 3 sin t Using the initial conditions, we obtain the equations c 1 + c = 0 c 3 = 1 c = So c 1 =, c =, and c 3 = 1 The solution is y(t) = cos t sin t (g) Problem 4. #3 : Solve y y + y y = 0 y(0) = y (0) = 1 y (0) = Solution. The characteristic equation is r 3 r + r 1 = 0. There are a number of ways you can find the roots of simple polynomial equations like this. One method is to guess a root, e.g. r = 1, verify that its a root, then do polynomial division to break the polynomial into a linear term and a quadratic term. Here r = 1 is a root and doing polynomial division, we factor this equation as (r 1)(r + 1) = 0. So the roots are r 1 = 1 and r,3 = ±i. The general solution is thus The derivatives are y(t) = c 1 e t + c cos t + c 3 sin t. y (t) = c 1 e t c sin t + c 3 cos t y (t) = c 1 e t c cos t c 3 sin t. Using the initial conditions, we obtain the coefficient equations c 1 + c = c 1 + c 3 = 1 c 1 c = If you add the first and last equation, you obtain c 1 = 0 so that c = by the first equation and c 3 = 1 by the second. The solution is y(t) = cos t sin t
4. Consider the the constant coefficient homogeneous n-th order ODE: a 0 y (n) + a 1 y (n 1) +... + a n 1 y + a n y = 0. By substituting in y = e rt you will obtain a characteristic equation a 0 r n + a 1 r n 1 +... + a n 1 r + a n = a 0 (r r 1 )(r r ) (r r n ) = 0. There are many cases for the roots here and it can be extraordinarily difficult to factor higher order polynomial equations 1. You should use software to find roots for higher order cases. By finding all the roots you can generate a fundamental set of solutions exactly as we did in the second order and third order cases. Combinations of distinct real roots and complex roots work the same way as in the previous problem. For higher order equations you can have many repeated roots and also complex repeated roots. Consult the text page 8 to see how to handle these scenarios. Use these ideas to solve the following problems from the text. (a) 4.: # 19: Find the general solution to Solution. The characteristic equation is y (5) 3y (4) + 3y 3y + y = 0. r 5 3r 4 + 3r 3 3r + r = r(r 4 3r 3 + 3r 3r + ) = 0 It is clear that r = 0 is one root. You could use Maxima to further factor the polynomial or guess roots and do polynomial division. Obtain r(r 1)(r )(r + 1) = 0 so that the characteristic equation has roots r = 0, 1,, ±i. The general solution is y(t) = c 1 + c e t + c 3 e t + c 4 cos t + c 5 sin t. (b) 4.: # 1: Find the general solution of Solutions. The characteristic equation is y (8) + 8y (4) + 16y = 0. r 8 + 8r 4 + 16 = 0 Again you might use software to factor this but I indicate a different approach. Let u = r 4. But you may also be able factor this by hand: r 8 + 8r 4 + 16 = (r 4 + 4) = ( (r r + )(r + r + ) ) = 0 The roots of the first quadratic term are r 1, = 1 ± i and the roots of the second are r 3,4 = 1 ± i. Since the quadratic factors are squared, each of these roots is repeated. This gives the general solution y(t) = e t (c 1 cos t + c sin t) + e t (c 3 cos t + c 4 sin t) + e t t (c 5 cos t + c 6 sin t) + e t t (c 7 cos t + c 8 sin t) 1 There are formulas analogous to the quadratic formula which find all the roots of cubic and quartic polynomials. They are very complicated and it is worth a Google search to see just how involved the quartic formula is. An amazing fact is that it is impossible to write down general closed formulas for the roots of 5th and higher degree polynomials. It is not just that none have been found but that it cannot be done!
(c) 4.: # 30: Solve y (4) + y = 0 y(0) = 0 y (0) = 0 y (0) = 1 y (0) = 0 Solution. The characteristic equation is The roots of this equation are r 4 + 1 = 0 r 1, = 1 ± 1 i r 3,4 = 1 ± 1 i The general solution is ( ( y(t) = e t 1 c 1 cos t ) ( )) ( ( ) ( )) 1 + c sin t + e t 1 1 c 3 cos t + c 4 sin t. So y(0) = c 1 + c 3 = 0 Compute the first through third derivative of y and evaluate at t = 0. You can do this by hand or with a computer and use the initial conditions to obtain the equations c 1 + c 3 = 0 c 1 + c c 3 + c 4 = 0 c c + 4 = 1 c 1 + c + c 3 + c 4 = 0 One way to solve this system is to add the second and fourth equations to get c + c 4 = 0 c + c 4 = 0 Adding the third equation to this, you ll find that c = 1 and then that c 4 = 1. The fourth equation is now c 1 + c 3 = 0 and using the first equation it follows that c 1 = c 3 = 0. So the solution is ) (e 1 t y(t) = e 1 t ( ) t sin.