Linear Second Order ODEs

Chapter 3 Linear Second Order ODEs In this chapter we study ODEs of the form (3.1) y + p(t)y + q(t)y = f(t), where p, q, and f are given functions. Since there are two derivatives, we might expect that two conditions are required to uniquely determine the solution. Thus we might expect initial conditions like y(0) = y 0, y (0) = y 0, where y 0 and y 0 are given numbers. Equation (3.1) is called a second-order linear ODE. It is time to explain the term linear ODE. Consider the second-order ODE (3.) y + p(t)y + q(t)y = 0. Note the zero on the right side of the equation. This ODE is called a second order homogeneous linear ODE. The term linear means the following. Suppose y 1 and y solve (3.). That is, y i + p(t)y i + q(t)y i = 0 for i = 1,. We say an ODE is linear if c 1 y 1 + c y also solves the homogeneous equation for all constants c 1 and c. In this case it is not difficult to check that (c 1 y 1 + c y ) = c 1 y 1 + c y, (c 1 y 1 + c y ) = c 1 y 1 + c y. Thus plugging c 1 y 1 + c y into (3.), we find (c 1 y 1 + c y ) + p(t)(c 1 y 1 + c y ) + q(t)(c 1 y 1 + c y ) = c 1 y 1 + c y + p(t)(c 1 y 1 + c y ) + q(t)(c 1 y 1 + c y ) = c 1 y 1 + p(t)y 1 + q(t)y 1 + c y + p(t)y + q(t)y = 0 + 0 = 0. Thus (3.) is linear. 17

18 3. Linear Second Order ODEs Unlike first order linear ODEs we studied in Chapter one, there are no known formulas for the solution. The best we can do is describe the form of the solutions. Let s look at some examples. Example 3.1. Consider the equation y = 0. Of course solving this equation is easy - just integrate it twice. Integrating once we find y = c 1, Integrating again, y(t) = c 1 t + c. Notice the solution is in the form y(t) = c 1 y 1 + c y, where y 1 = t and y = 1, and c 1, c are arbitrary constants. Moreover, y 1 and y solve the homogeneous equation y = 0. The following may seem reasonable. Conjecture. All solutions to y +p(t)y +q(t)y = 0 have the form y(t) = c 1 y 1 +c y, where c 1, c are arbitrary constants and y 1, y solve y + p(t)y + q(t)y = 0. Example 3.. Find the solution to y 4y + 3y = 0, y(0) = 1, y (0) = 1. Solution. The form of the ODEs suggests that y = e rt might work. Putting this guess in the ODE, we find (r 4r + 3)e rt = 0. Since exponents are never zero, r 4r + 3 = 0. That is, (r 1)(r 3) = 0. The roots are r = 1, 3. Thus y 1 = e t and y = e 3t are both solutions to the ODE. Since the equation is linear y(t) = c 1 e t + c e 3t, is also a solution for any c 1, c. Once again the solution has the form our conjecture suggests. To solve for the constants we apply the initial data. Here y(0) = 1 = c 1 + c y (0) = 1 = c 1 + 3c. Solving we find c 1 = 1 and c = 0. The solution is (check this!!!). Example 3.3. Solve y(t) = e t y y = 0. y(0) = 1, y (0) = 0. Solution. Note that both y 1 = e t+1 and y = e t solve the ODE. Our conjecture indicates all solutions have the form y(t) = c 1 y 1 + c y = c 1 e t+1 + c e t.

3. Linear Second Order ODEs 19 Applying the initial data we find y(0) = 1 = c 1 e + c y (0) = 0 = c 1 e + c. Solving by subtracting the two, we find 1 = 0! What went wrong? Well, notice that y 1 and y are multiples. Indeed, y 1 = ey. So there is really only one solution. We must modify our conjecture. Theorem 3.4. All solutions to y + p(t)y + q(t)y = 0 have the form y(t) = c 1 y 1 (t) + c y (t) where c 1, c are arbitrary constants, and y 1, y are two independent (not multiples) solutions to the ODE. This is the best we can do - we know the form of the solutions. Example 3.5. Find the general solution to y y = 0. Solution. We are looking for a function whose second derivative is itself. This suggests trying y = e rt. Plugging this in the ODE reveals, (r 1)e rt = 0. This implies r 1 = 0 or r = ±1. Thus y 1 = e t and y = e t. These two functions are not multiples of one another since their quotient is not constant. Thus by Theorem 3.4 all solutions are in the form y(t) = c 1 e t + c e t. Note that both y 1 = sinh t and y cosh t solve the ODE and are not multiples. Therefore, all solutions look like y(t) = c 1 sinh t + c cosh t. Let us check this. Note y = e t solves the ODE. If our theorem is correct, constants c 1 and c exist so that and indeed c 1 = c = 1 works. e t = c 1 sinh t + c cosh t = c 1 e t e t Example 3.6. Consider the ODE y + 4y = 0. e t + e t + c, Show that y 1 = sin t and y = cos t are two independent solutions. Note that y = sin t cos t is also a solution. Hence Theorem 3.4 implies constants c 1, c exist so that y = c 1 y 1 + c y. Find c 1 and c. Solution. It is easy to check y 1, y, and y are solutions. Since the quotient of y 1, y is tan t, which is not constant, y 1 and y 1 are not multiples. Theorem 3.4 applies and all solution to the ODE are in the form y = c 1 y 1 + c y. Since y is also a solution and all solutions are of this form, constants exist so that y = sin t cos t = c 1 sin t + c cos t. Since sin t cos t = 1 sin t, we find c 1 = 1/ and c = 0.

0 3. Linear Second Order ODEs Homework 3.0 1. Consider the ODE y y = 0. Check that y 1 = sinh t, y = cosh t, and y 3 = e t 6e t solve the ODE. According to Theorem 3.4 all solutions may be written as y = c 1 y 1 + c y. Since y 3 is a solution, it can be written this way too. Find the constants which make y 3 = c 1 y 1 + c y.. Consider the ODE (1 + t )y 4ty + 6y = 0. Check that y 1 = 1 3t, y = t t 3 /3, and y 3 = 1 + 3t 3t t 3 solve the ODE. According to Theorem 3.4 all solutions may be written as y = c 1 y 1 + c y. Since y 3 is a solution, it can be written this way too. Find the constants which make y 3 = c 1 y 1 + c y. 1. c 1 = 7, c = 5. c 1 = 1, c = 3. Answers 3.1. Homogeneous Linear ODEs with Constant Coefficients In the case the functions p(t) and q(t) are constant in (3.), exact solutions can be found. We study ODEs in the form (3.3) ay + by + cy = 0, where a 6= 0, b, and c are real numbers. There are three possibilities. 3.1.1. Real Distinct Roots. In this section we consider the case when b 4ac > 0. Example 3.7. Solve y + y y = 0, y(0) = 4, y (0) =. Solution. We are looking for a function whose derivatives are multiples of itself. This suggests to guess y = e rt. Doing so gives (r + r )e rt = 0. The polynomial is called a characteristic polynomial. Note its relation to the ODE. Here r + r = (r 1)(r + ) = 0 and the roots are r = 1,. Thus y 1 = e t and y = e t are two independent solutions. Theorem 3.4 implies all solutions can be expressed as y(t) = c 1 e t + c e t. To find the constants, apply the initial data. We find 4 = y(0) = 4 = c 1 + c = y (0) = = c 1 c. Solving, we find c 1 = and c =, and the solution is y(t) = e t + e t.

3.1. Homogeneous Linear ODEs with Constant Coefficients 1 3.1.. Complex Roots. In this section we consider the case when b 4ac < 0. In this case the roots of the characteristic polynomial are complex. Example 3.8. Find the general solution to y + y = 0. Solution. The coefficients in the ODE are constant, so the guess e rt will work. The characteristic polynomial is r + 1 = 0, and the roots are r = ±i. Thus, in accordance with Theorem 3.4, all solutions have the form (3.4) y(t) = c 1 e it + c e it. While this is the correct general solution, its form is not pleasing-the original ODE had no complex numbers, and we should expect the solution to be free of complex numbers. Physically the ODE y + y = 0 is F = ma for a spring without friction. We expect oscillatory behavior. This observation allows us to guess that y 1 = sin t and y = cos t are two independent solutions. Therefore, all solutions can be written as (3.5) y(t) = c 1 cos t + c sin t. Here is where the importance of Theorem 3.4 cannot be understated. Since all solutions to the ODE can either be written as (3.4) or (3.5), it must the that constants exist so that e it = c 1 cos t + c sin t. This equation holds for all t. To find the constants we first set t = 0. We find e i0 = 1 = c 1 cos 0 + c sin 0 = c 1. That is, c 1 = 1. To find c we take the derivative again set t = 0. We find ie i0 = c 1 sin 0 + c cos 0, and c = i. We have now discovered an important formula: (3.6) e it = cos t + i sin t. This is called Euler s formula. To see how this solves our problems, we return to (3.4) an use Euler s formula y(t) = c 1 e it + c e it = c 1 (cos t + i sin t) + c (cos( t) + i sin( t) = (c 1 + c ) cos t + i(c 1 c ) sin t. Recall that cos( t) = cos t and sin( t) = sin t. Since the constants are arbitrary, we set C 1 = c 1 + c and C = i(c 1 c ) and we recover (3.5) from (3.4). That is, Example 3.9. Find the solution to y(t) = C 1 cos t + C sin t. y + y + y = 0 y(0) = 1 y (0) = 1.

3. Linear Second Order ODEs Solution. The characteristic equation is r +r+ = 0, and the roots are r = 1±i. Thus the general solution is y(t) = c 1 e ( 1+i)t + c e ( 1 i)t. Again we would like no complex numbers in the solution. We use Euler s formula to remove them. y(t) = c 1 e t e it + c e t e it = c 1 e t (cos t + i sin t) + c e t (cos( t) + i sin( t) = (c 1 + c )e t cos t + i(c 1 c )e t sin t. As in the previous example, we set C 1 = c 1 + c and C = i(c 1 c ) and the general solution is y(t) = C 1 e t cos t + C e t sin t. To find the constants we employ the initial data. Here 1 = y(0) = C 1 1 = y (0) = C 1 + C. Solving, we find C 1 = 1 and C =. The solution is y(t) = e t cos t + e t sin t. In general if the root of the characteristic polynomial are r = λ ± iµ, then the general solution is (3.7) y(t) = c 1 e λt cos µt + c e λt sin µt Example 3.10. Find the general solution to y 4y + 13y = 0. Solution. The characteristic equation is r 4r + 13 = 0, and the roots are r = ± 3i. Thus the general solution is y(t) = c 1 e t cos 3t + c e t sin 3t. Homework 3.1. (just for fun) Use Euler s formula to express the following as a complex number (a + ib). 1. i i 3. e iπ. π i 1. e π. cos(ln π) + i sin(ln π) 3. 1 4. cos(ln ) i sin(ln ) Answers 4. 1 i

3.1. Homogeneous Linear ODEs with Constant Coefficients 3 3.1.3. Repeated Roots. We start with an example. Example 3.11. Find the general solution to y y + y = 0. Solution. The characteristic equation is r r + 1 = 0, and the roots are r = 1, 1. One solution is clear; y 1 = e t. Clearly,...,clearly y = e t will not work. How to find another independent solution? Let s work backwards. If the roots are repeated, the characteristic polynomial must be (r r 1 )(r r 1 ) = 0 = r rr 1 + r 1. This suggests that the ODE must have the form (3.8) y r 1 y + r 1y = 0. One solution is easy to fine, y 1 = e r1t. How to find the other?? Here is a clever trick. We take the one solution we have, y 1 = e r1t, and we try to modify it to produce another independent solution. Here is one possible modification y(t) = c(t)y 1 (t) = c(t)e r1t. That is, we let the constant in front float, change, and we try to force this guess to solve (3.8). We first have to compute the derivative of our guess. Here y (t) = c e r1t + cr 1 e r1t Now we plug all this in (3.8) and find y (t) = c e r1t + c r 1 e r1t + cr 1e r1t c (t)e r1t = 0. So we require c = 0 and c(t) = c 1 + c t. There one possible new solution is y (t) = (c 1 + tc )e r1t. Now we check your understanding of Theorem 3.4. We know y solves (3.8) for any constants c 1 and c. All we need is another solution independent of y 1 = e r1t. The simplest choice for y would be y = ty 1 = te r1t. That is, we simply multiply the one solution we guessed by t and move on. Example 3.1. Find the solution to y + 6y + 9y = 0, y(0) = 1, y (0) =. Solution. The characteristic equation is r + 6r + 9 = 0, and the roots are r = 3, 3. Hence, y 1 = e 3t and y = te 3t. The general solution is Using the initial data, we find y(t) = c 1 e 3t + c te 3t. 1 = y(0) = c 1 = y (0) = 3c 1 + c. Solving, we find c 1 = 1 and c = 5. The solution is y(t) = e 3t + 5te 3t.

4 3. Linear Second Order ODEs Summary Suppose the first order, linear, homogeneous, constant coefficient ODE ay + by + cy = 0 characteristic equation, ar + br + c = 0 has roots r 1 and r. If the roots are real and distinct, the general solution is y(t) = c 1 e r1t + c e rt. If the roots are complex with the form λ ± iµ, the general solution is y(t) = c 1 e λt cos µt + c e λt sin µt. If the roots are repeated, r = r 1 = r the general solution is y(t) = c 1 e rt + c te rt. Homework 3.1 (Constant Coefficient ODEs) Find either the general solution or the solution as indicated. 1. y + y 3y = 0. y + y y = 0 3. y + 3y = 0 4. y + y y = 0, y(0) = 1, y (0) = 1 5. y + 4y + 3y = 0, y(0) =, y (0) = 1 6. y + 3y = 0, y(0) =, y (0) = 3 7. Find a differential equation whose general solution is y = c 1 e t/ + c e t/3 8. y y + y = 0 9. y y + 4y = 0 10. 4y + 8y + 5y = 0 11. y + 4y = 0, y(0) = 0, y (0) = 1 1. y + 4y + 5y = 0, y(0) = 1, y (0) = 1 13. y y + 5y = 0, y(π/) = 1, y (π/) = 1 14. 9y + 6y + y = 0 15. y 6y + 9y = 0 16. y 4y + 4y = 0, y(0) = 1, y (0) = 1 17. y 4y + 4y = 0, y(1) = 0, y (1) = 1 Answers 1. y = c 1 e t + c e 3t. y = c 1 e t/ + c e t 3. y = c 1 + c e 3t 4. y = e t 5. y = 5 e t 1 e 3t 6. y = 1 e 3t 7. 6y + 5y + y = 0 8. y = c 1 e t cos t + c e t sin t 9. y = c 1 e t cos 3t + c e t sin 3t 10. y = c 1 e t cos t/ + c e t sin t/ 11. y = 1 sin t 1. y = e t cos t + 3e t sin t 13. y = e t π/ cos t 14. y = c 1 e t/3 + c te t/3 15. y = c 1 e 3t + c te 3t 16. y = e t te t 17. y = e (t 1) (t 1)e (t 1)

3.. Non Homogeneous Linear ODEs 5 3.. Non Homogeneous Linear ODEs In this section we consider ODEs of the form (3.9) y + p(t)y + q(t)y = f(t). The non homogeneous term, f(t), can be thought of as a forcing term. As before, there is no known explicit solution for the general case. However, when the coefficients are constant methods are available to solve (3.9) for general f(t). Moreover, if the forcing term f is not too complicated, we can develop a method to guess the solution. We detail this method first. We consider the simplified form of (3.9) (3.10) ay + by + cy = f(t), where a, b, c are real numbers. In some cases we can guess the solution. Example 3.13. Find the general solution to y + y = e t. Solution. We seek a function y p (where the p stands for particular) so that y p + y p = e t. Since an e t must appear on the right side, an obvious guess would be y p = Ae t. Putting this back in the ODE we find Ae t + Ae t = e t. Solving for A we find A = 1/, and y p = e t /. This is not the whole solution however. If we were given initial data, we would have no freedom to impose the constraints required by y(0) and y (0) (there is no c 1, c to choose). To remedy this we add on the homogeneous solution. That is, the solution to y + y = 0. Here y h = c 1 cos t + c sin t, and y h satisfies y h + y h = 0. Thus, if we add this solution to y p, no harm is done. In particular, y(t) = y h + y p solves y + y = 0 + f. The general solution is therefore y(t) = c 1 cos t + c sin t + et. There is an art to guessing as the next examples show. Example 3.14. Solve y + y y = cos t. Solution. We seek a function y p satisfying y p + y p y p = cos t. We can see the guess y p = A cos t will not work since the y term will introduce a sin t which cannot be balanced with the right side. We need a way of canceling the sin. Thus we try guessing y p = A cos t + B sin t. We need to compute the derivatives and require y p + y p y p = cos t. Here y p = A cos t + B sin t y p = A sin t + B cos t y p = A cos t B sin t.

6 3. Linear Second Order ODEs We can avoid actually plugging these back in the ODE - we just need to balance the elements. In particular, the right side of the ODE has one cos, and zero sin. The left side has to balance. The total number of cos on the left side can be computed by looking where the cos in the derivatives of y p go. By inspection (no writing) the cos balance is A + B A = 1 and the sin balance is B A B = 0. Solving these we find A = 3/10 and B = 1/10. The general solution is y(t) = c 1 e t + c e t 3 10 cos t + 1 sin t. 10 Example 3.15. Find the general solution to y y = e t. Solution. We seek a function y p satisfying y p y p = e t. An obvious guess would be y p = Ae t. However, notice that the homogeneous solution is y h = c 1 e t + c e t. Thus (e t ) e t = 0 6= e t! That is, e t is part of the homogeneous solution - it is annihilated by the ODE. What to do? Based on our previous experience with repeated roots, we might try y p = Ate t. This turns out to work. Indeed, we first take the appropriate derivatives y p = Ate t y p = Ae t + Ate t y p = Ae t + Ate t. Again we do not have to plug these back into the ODE directly and expand the resulting mess. Instead we observe that there are two elements - e t, and te t. We just need to balance them. Balancing the e t s we find A A = 1. Balancing the te t s, we find A A = 0. Thus A = 1, and the general solution is y(t) = y h + y p = c 1 e t + c e t + te t. Guessing y p is not so easy. Here is a no-think algorithm to help. Examples follow.

3.. Non Homogeneous Linear ODEs 7 Guessing Algorithm for y p c (1) List the functions in f(t) without constants. () Differentiate the list. (3) Add any new functions to the list - again ignoring constants in front. (4) Repeat 1-3 until the list is closed - no new functions. (5) If any function in the list is also part of the homogeneous solution, multiply it repeatedly by t until it is out of the homogeneous list and out of the list in step 4. (6) Guess y p = Af 1 +Bf +Cf 3 +..., where f i are the functions in the list. Example 3.16. Find a suitable form for y p if y p solves y y + y = te t + 3t e t. Solution. Before we can apply the algorithm, we need the homogeneous solution. The characteristic polynomial is r r + 1 = (r 1)(r 1) = 0. Thus y h = c 1 e t + c te t, or as a list y h = {e t, te t }. Now we apply the algorithm. (1) The functions in f are {te t, t e t }. () Differentiating the list {e t + te t, te t + t e t }. (3) Adding the new functions {e t, te t, t e t }. (4) The list is closed to differentiation (5) If a function is in y h hit it with t. Since the first function is, we hit is with a t as prescribed and the list becomes {t 3 e t, te t, t e t }. Since the second function is in y h, the same happens and the list becomes {t 3 e t, t 4 e t, t e t }. The last function is not in y h, so nothing happens. (6) Guess y p = At e t + Bt 3 e t + Ct 4 e t. Example 3.17. Find the solution to y + 4y = 8t + 10e t, y(0) = 1 y (0) =. Solution. The characteristic polynomial is r + 4 = 0. Thus the homogeneous solution is y h = c 1 cos t + c sin t. Next we apply the algorithm. (1) The functions in f are {t, e t }. () Differentiating the list {t, e t }. (3) Adding the new functions {t, t, e t }. (4) Repeating, {1, t, t, e t } and the list is closed. Notice what (and why) the algorithm did! (5) If a function is in y h hit it with t. Since the homogeneous list consists of cos and sin, nobody gets hit here. (6) Guess y p = A + Bt + Ct + De t.

8 3. Linear Second Order ODEs Next we find the constants. As done previously y p = A + Bt + Ct + De t y p = B + Ct + De t y p = C + De t. There are four elements. They must be balanced. The multiples of one must balance: C 4A = 0. The ts must balance: 4B = 0. The t s must balance: 4C = 8. The e t must balance: D + 4D = 10. Solving, we find A = 1, B = 0, C =, and D =. The general solution is y(t) = c 1 cos t + c sin t + 1 + t + e t. Now, and only now, we use the initial data to find c 1, c (we want the entire solution to satisfy the initial data, not half of it!!). Here 1 = y(0) = c 1 + 1 + = y (0) = c +. Solving we find, c 1 =, c = 0. The solution is y(t) = cos t + 1 + t + e t. Homework 3. (Non Homogeneous ODEs) Find either the general solution or the solution as indicated. 1. y y 3y = e t. y y 3y = 3te t 3. y + y = + 8 cos t 4. y + y = cos t + t sin t 5. y + y y = t, y(0) = 0, y (0) = 1 6. y y = 1 + e t, y(0) = 0, y (0) = 1 7. y + y + 5y = 4e t cos t, y(0) = 1, y (0) = 0 Find suitable forms for y p. You need not evaluate the constants. 8. y + y = t(1 + sin t) 9. y y + y = te t + 7 10. y 4y + 4y = t + 4te t + t sin t 11. y IV + 4y + 8y + 8y + 4y = (1 + t )e t cos(t) + (t ). The Characteristic polynomial is (r + r + ) = 0.

3.3. Variations of Parameters 9 1. y = c 1 e 3t + c e t 1 3 et Answers. y = c 1 e 3t + c e t + 3 16 te t + 3 8 t e t 3. y = c 1 + c e t + t + cos t sin t 4. y = c 1 cos t + c sin t+?? 5. y = e t 1 e t t 1 6. y = 1 t + e t + te t 7. y = e t cos t + 1 e t sin t + te t sin t 8. y p = At sin t + Bt sin t + Ct cos t + Dt cos t + E + F t 9. y p = At e t + Bt 3 e t + C 10. y p = A + Bt + Ct + D cos t + E sin t + F t cos t + Gt sin t + Ht e t + It 3 e t 11. y p = At e t cos t + Bt 3 e t cos t + Ct 4 e t cos t + Dt e t sin t + Et 3 e t sin t + F t 4 e t sin t + G + Ht 3.3. Variations of Parameters There is way to find y p without guessing. The method does not require the coefficients to be constant either. Suppose, somehow, two independent solutions can be found to y + p(t)y + q(t)y = 0. Call the two solutions y 1 and y. To solve the inhomogeneous equation (3.11) y + p(t)y + q(t)y = f(t) we employ a similar trick used when we encountered repeated roots. Specifically, we alter the one solution we have to make if fit the new situation. Here the only solution in sight is y h = c 1 y 1 + c y. We alter it by letting the constants float; we try (3.1) y p = u 1 (t)y 1 t) + u (t)y (t), where u 1, u are unknown. We have to put y p back in the nonhomogeneous ODE and find u 1, u. This will require taking derivatives. He have We arbitrarily set y p = u 1y 1 + u y + u 1 y 1 + u y. (3.13) u 1 y 1 + u y = 0 (we can pick the u s anyway we like). Then y p = u 1 y 1 + u y + u 1y 1 + u y. Next we put this back in (3.11) and recall that y i + py i + qy i = 0 (y 1, y are homogeneous solutions). We find (3.14) u 1 y 1 + u y = f(t).

30 3. Linear Second Order ODEs Equations (3.13) and (3.14) must be solved simultaneously for u 1 and u. We set w(y 1, y ) = y 1 y y 1 y. This is called the Wronskian. We could, without much effort, show that w(y 1, y ) 6= 0 when y 1 and y are independent homogeneous solutions (maybe in an appendix later). Solving for u 1 and u, we find u 1 = y (t)f(t) w(y 1, y )(t), u = y 1(t)f(t) w(y 1, y )(t). We may integrate to find y p. Indeed, returning to (3.1), Z Z y (t)f(t) (3.15) y p (t) = y 1 (t) w(y 1, y )(t) dt + y y 1 (t)f(t) (t) w(y 1, y )(t) dt Example 3.18. Find the general solution to t y y = 3t 1 given y 1 = t and y = 1/t are two independent solutions to the homogeneous equation t y y = 0. Solution. We need to apply (3.15). Here the Wronskian is w(y 1, y )(t) = y 1 y y 1 y = t 1 t (t) 1 t = 3. Notice here that f(t) = (3t 1)/t (the formula for y p assumed a one in front of y ). Thus Z t y p (t) = t 1 (3t Z 1) t 3t t 1 (3t 1) 3t Z = t 3 3 t 1 Z t dt t 1 (3t 1) dt 3 = t ln t + 1. There is no reason to add a constant (why?? understand!). The general solution is y(t) = c 1 t + c t 1 t + t ln t + 1. You can see why we would have never guessed y p. Find the general solution. Homework 3.3 (Variation of Parameters) 1. y + 9y = 9 sec 3t. y y + y = e t /(1 + t ) 3. y + 4y = g(t), (arbitrary g)

3.4. Dynamics of Second-order ODEs 31 Answers 1. y = c 1 cos 3t + c sin 3t + sin(3t) ln tan 3t + sec 3t 1. y = c 1 e t + c te t 1 et ln(1 + t ) + te t tan t t R 3. y = c 1 cos t + c sin t + 1 g(s) sin((t s)) dt 3.4. Dynamics of Second-order ODEs The roots of the characteristic polynomial often have physical meaning. The physics of a problem suggests a particular behavior and understanding this behavior suggests the nature of the roots of the characteristic polynomial. To illustrate the connection between roots of the characteristic polynomial and the physics of a system, consider a spring attached to a wall and a mass which can move on a table. We allow for friction. The equations of motion (F=ma), are my = γy ky + f(t), where γ > 0 is a constant related to the friction in the system, and f(t) is an external force. The initial data in this case represents the initial position, y(0) and the initial velocity, y (0). The ODE also describes an electrical circuit comprised of a capacitor, resistor, inductor, and a voltage source. Here m would be the inductance, γ the resistance, 1/k the capacitance, and f the derivative of the voltage source. High Friction Let s take m = k = 1, and assume γ > with no external force. Thus the ODE is y + γy + y = 0. The roots of the characteristic equation are given by r = γ ± p γ 4. The goal here is to be able to graph the solution given only the roots, or at least understand qualitatively the behavior of solutions given only the roots. To do this we need to know the sign of the roots. Notice that p γ 4 < p γ. Thus r 1 = γ + p γ 4 < γ + p γ = γ + γ = 0. Thus, r 1 < 0. The other root is clearly negative. So the solution is in the form y = c 1 e r1t + c e rt with r 1, r both negative. Figure 1 below shows plots with various initial data. Given the roots, the behavior of the plots in the figure should be clear. The spring is said to be over damped. Critical Friction Suppose γ =. In this case the roots are r = γ ± p γ 4 = γ,

3 3. Linear Second Order ODEs 1.5 1 Displacement 0.5 0 0.5 1 0 5 10 15 Time Figure 1. Displacement of a spring for various starting initial data. damped with γ = 4. Over and the roots are repeated. The general solution is y(t) = c 1 e γ t + c te γ t. Again the goal is to understand the relationship between the qualitative behavior of the solutions and the roots of the characteristic polynomial. The spring is said to be critically damped. Figure shows plots of the solution for various initial data. 1 0.8 0.6 0.4 Displacement 0. 0 0. 0.4 0.6 0.8 1 0 5 10 15 Time Figure. Displacement of a spring for various starting initial data. Critically damped with γ =. Notice the spring only crosses the axis once. Low Friction Suppose 0 < γ <. In this case the roots are r = γ ± p γ 4 = γ ± ip 4 γ and the roots are complex with negative real part. That is, they look like λ ± iµ with λ < 0. The general solution is p! p! y(t) = c 1 e γ 4 γ t cos t + c e γ 4 γ t sin t.

3.4. Dynamics of Second-order ODEs 33 Once again the qualitative behavior should be clear from only the knowledge of the roots. 1 0.8 0.6 0.4 Displacement 0. 0 0. 0.4 0.6 0.8 1 0 5 10 15 Time Figure 3. Displacement of a spring for various starting initial data. Under damped with γ =.65. No Friction Suppose γ = 0. In this case the roots are r = ±i, and the general solution is y(t) = c 1 cos t + c sin t. So purely complex roots indicate oscillatory behavior. 1 0.5 Displacement 0 0.5 1 0 5 10 15 Time Figure 4. Displacement of a spring for various starting initial data. No friction (γ = 0). Forced Vibrations Let s consider the effects of adding a forcing term. In particular, an understanding of the physics of multiplying by t in Step 5 of the algorithm for y p is desirable.

34 3. Linear Second Order ODEs Consider the spring mass system, initially at rest, and without friction given by y + ω0 y = f 0 cos ωt y(0) = 0, y (0) = 0. Using the methods of the previous section, the general solution, for ω 0 6= ω, is y(t) = c 1 cos ω o t + c sin ω o t + ω ω0 cos ωt. y(t) = f 0 ω ω0 cos ω 0 + ω ω0 cos ωt. Applying the initial data, we find We can clean this solution up a bit by using the identity f 0 f 0 cos(a B) cos(a + B) = sin A sin B. With A = (ω + ω 0 )/ and B = (ω ω 0 )/, we find Amplitude z } { f 0 ω0 ω ω0 + ω y(t) = sin t sin t. ω ω 0 For ω 0 close to ω the first two terms may be thought of a the amplitude of the displacement (it varies slowly compared with the frequency of the remaining term). The figure below depicts a typical case. 8 6 4 Displacement 0 4 6 8 0 50 100 150 Time Figure 5. Forced spring with ω 0 =.75 and ω = 1.

3.4. Dynamics of Second-order ODEs 35 Homework 3.4 (Dynamics of ODEs) Have fun with the following questions. 1. The behavior of a physical system is believed to be described by the ODE for α > 0. Show that for all initial data y + αy + y = 0, lim t y(t) = 0.. Suppose that the ODE y + γy = 0 describes an electrical gadget. The engineers of this gadget hope that the device produces oscillatory behavior. How should they choose γ? Why? 3. Suppose that oscillations in a bridge are modeled by the ODE y + ω y = sin(ω 0 t). How should the parameter ω 0 be chosen so that the solutions remain bounded for all time? 4. A system with friction may be modeled by the ODE y + γy + λ y = 0, where γ 0 is the friction coefficient. How small must γ be chosen so that the system exhibits oscillatory behavior? 5. The manufacture of bathroom scales wants the dial that indicates the weight of a person not to oscillate. Suppose that the position of the dial satisfies the ODE my + 5y + y = 0, where mg is the person s weight (g = 3.ft/sec ). How heavy can a person be and not cause the dial to oscillate? 6. Suppose that some interesting process may be modeled by the ODE y y = 0. How should the initial data be chosen so that the solution remains bounded as t approaches infinity?