Table of contents. d 2 y dx 2, As the equation is linear, these quantities can only be involved in the following manner:

Transcription

1 M ath 0 1 E S 1 W inter Last Updated: January, 01 0 Solving Second Order Linear ODEs Disclaimer: This lecture note tries to provide an alternative approach to the material in Sections and 4. 9 in the textbook. It is not a replacement of the textbook. Table of contents Solving Second Order Linear ODEs Introduction Exploring linearity General solution for the homogeneous problem Finding one particular solution for the nonhomogeneous problem General solution to nd order linear constant-coefficient homogeneous problem r 1 r real r 1 = r real r 1, r complex A theoretical issue checking linear dependence More examples Linear, homogeneous, constant coefficient equations of higher order Nonhomogeneous equations how to find the particular solution The method of undetermined coefficients Variation of parameters Variable-coefficient equations Free mechanical vibrations The undamped, free case System with damping Underdamped motion ( b < 4 m k) Overdamped motion ( b > 4 m k) Critically damped motion ( b = 4 m k) Nonzero external force Examples Introduction. In this note we discuss how to solve general second order linear ODEs. As the equation is nd order, only the following unknown quantities can be involved ( denote the variable by x and the unknown function by y) : d y dx, dy dx, y. ( 1 ) As the equation is linear, these quantities can only be involved in the following manner: a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y ( ) for known functions a, a 1, a 0. Thus the general form of second order linear ODE is as follows: a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x). ( 3) Following the success in studying the first order equations, one may be tempted to try to solve the nd order equations via transformation of the equation into the form d Y = F( x) ( 4) dx

2 for some new unknown function Y and some new variable X. However it turns out this is not possible in most cases. Therefore other ways have to be taken. It turns out that, to solve a second order linear ODE, one only need to find three special solutions, and all other solutions can be obtained from these three.. Exploring linearity. To understand this approach, we have to explore linearity of the equation. Recall that we have mentioned in Introduction: Linear equations are easier than nonlinear equations. Why? The reason lies in the following. Proposition 1. Consider the linear nd order ODEs and a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x) ( 5) a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = g( x). ( 6) If y( x) is a solution to the first equation and z( x) is a solution to the second equation, then y( x) + z( x) solves a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x) + g( x). ( 7) C orollary. The general solution of is given by where y 1 ( x) is problem a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x) ( 8) y( x) = ỹ( x) + y 1 ( x) ( 9) solution of the above equation, and ỹ( x) is the general solution of the homogeneous a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = 0. ( 1 0) In light of the above, we see that to solve the general nd order linear equation we only need to a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x) ( 1 1 ) 1. Find all solutions ( that is, general solution) of the homogeneous problem. Find one solution of the original nonhomogeneous problem a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = 0 ( 1 ) a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x) ( 1 3). 1. General solution for the homogeneous problem. How to find the general solution to the homogeneous problem? We again turn to linearity. Proposition 3. If y 1, y,, y k solves so does c 1 y 1 + c y + + c k y k for any constants c 1,, c k. a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = 0, ( 1 4)

3 Thus there is the possibility of representing a large amount ( actually infinitely many as can be easily seen) of solutions by only a few. Now the questions are, 1. Is it possible to represent all solution by the linear combinations of finitely many?. If yes, how many is needed? The answer to the first question is yes, and the answer to the second is two. More specifically, we have the following theorem. Theorem 4. Let y 1, y be two solutions to a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = 0, ( 1 5) that are linearly independent 1. Then the general solution is given by y = c 1 y 1 + c y ( 1 6) for arbitrary constants c 1, c. Conversely, for any homogeneous nd order linear ODE, there are solutions y 1 ( x), y ( x) such that the general solution can be written y = c 1 y 1 + c y ( 1 7) Remark 5. ( Initial value problem) The above theorem immediately tells us that a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x), y( a) = b ( 1 8) is not the correct initial value problem. The correct way to propose initial value problems is a ( x) d y dx + a 1 ( x) dy dx + a 0( x) y = f( x), y( a) = b, y ( a) = c ( 1 9) as we need two conditions to fix the two arbitrary constants. Remark 6. ( Boundary value problem) Instead of specifying y( a) and y ( a), we can also specify the values of y( a) and y( b). This is called a boundary value problem... Finding one particular solution for the nonhomogeneous problem. We will see that this is actually a quite difficult task. In fact, finding this one particular solution to the nonhomogeneous problem is more difficult than finding all solutions to the homogeneous one. We will discuss useful methods in later lectures. 3. General solution to nd order linear constant-coefficient homogeneous problem. That is, general solution to a y + b y + c y = 0 ( 0) where a, b, c are constants. Recall that all we need to do is to find two solutions that are linearly independent. Note that it doesn t matter how we obtain them. Nevertheless, we have a systematic method of doing so. The idea is to consider solutions of the type y = e rt for some constant r. Substituting this into the equation leads to e rt [ a r + b r + c ] = 0. ( 1 ) Therefore, y = e rt solves the equation if and only if a r + b r + c = 0, or r is either r 1 = b + b 4 a c, or r = b b 4 a c. ( ) a a 1. Meaning: The only pair of constants ( C 1, C ) such that C 1 y 1 + C y 0 is ( 0, 0).

4 Recall from the theory of quadratic equations that there are three cases: Both r 1, r real and r 1 r ; r 1 = r real; r 1, r complex. We discuss them one by one r 1 r real. This is the easiest case. We have two solutions e r1 t and e r t. If they are linearly independent, then all solutions to can be written as a y + b y + c y = 0 ( 3) y = c 1 e r1 t + c e r t ( 4) and the equation is solved. The proof of their linear independence goes as follows. Assume that there are constants C 1, C such that Differentiating with respect to t, we have In other words, the constants C 1, C solve the system 0 = C 1 e r1 t + C e r t. ( 5) 0 = C 1 r 1 e r1 t + C r e r t. ( 6) e r1 t x + e r t y = 0 ( 7) r 1 e r1 t x + r e r t y = 0. ( 8) Solving this system shows that the only solutions are x = y = 0. Therefore C 1 = C = 0. We have shown that If C 1, C are such that 0 = C 1 e r1 t + C e r t, then C 1 = C = 0. ( 9) This is exactly the linearly independence of e r1 t and e r t. Example 7. ( 4.. 3) Solve y + 5 y + 6 y = 0. ( 30) Solution. Substituting y = e rt into the equation, we reach As a consequence, the general solution is given by r + 5 r + 6 = 0 r 1 = 3, r =. ( 31 ) y = c 1 e 3 t + c e t. ( 3) 3.. r 1 = r real. In this case e r1 t = e r t, that is we can only find one solution of the form e rt. We need another solution which is linearly independent. It turns out that this other solution can be taken to be t e rt. It is easy to show that e rt and t e rt are linearly independent. Thus when r 1 = r = r real, the general solution to is given by Example 8. ( 4.. 5) Solve a y + b y + c y = 0 ( 33) y = c 1 e rt + c t e rt. ( 34) y + 8 y y = 0. ( 35) Solution. Substituting y = e rt we easily see that r + 8 r = 0 r 1 = r = 4. ( 36). Note that t e rt = r ( e rt ). This is not a coincidence! Check 6. if you are interested in the underlying reason.

5 Thus the general solution is given by y = c 1 e 4t + c t e 4t. ( 37) r 1, r complex. As r 1 = b + b 4 a c a, r = b b 4 a c a with a, b, c real, the only possibility for r 1, r to be complex is that ( 38) r 1 = α + i β, r = α i β ( 39) where α = b, β = 4 a c b are both real. a a In this case, what we have are two linearly independent complex solutions e ( α + i β) t, e ( α iβ) t. ( 40) The task now is the produce two linearly independent real solutions out of them. The idea is the use the properties of complex exponentials together with Euler s formula: to obtain e iθ = cos θ + i sin θ ( 41 ) e ( α + iβ) t = e α t cos β t + i e α t sin β t, e ( α iβ) t = e α t cos β t i e α t sin β t. ( 4) Recall that due to the linearity of the equation, if y 1, y are solutions, so is c 1 y 1 + c y for any constants c 1, c. Inspecting the formulas for e ( α ± i β) t we easily figure out that and e α t cos β t = 1 e( α + i β) t + 1 e( α iβ) t ( 43) e α t sin β t = 1 e( α + iβ) t 1 e( α iβ) t ( 44) are also solutions. Now we show that they are linearly independent. Assume that C 1, C are constants such that This gives C 1 + C C 1 e α t cos β t + C e α t sin β t = 0. ( 45) e ( α + iβ) t + C 1 C As e ( α ± i β) t are linearly independent, necessarily C 1 + C from which C 1 = C = 0 immediately follows. Example 9. ( ) Solve = C 1 C Solution. Substituting y = e rt into the equation, we reach Thus we obtain two complex solutions From this we obtain two real solutions e ( α iβ) t = 0. ( 46) = 0 ( 47) y + 9 y = 0. ( 48) r + 9 = 0 r 1 = 3 i, r = 3 i. ( 49) e 3 it = cos 3 t + i sin 3 t and e 3 it = cos 3 t i sin 3 t. ( 50) cos 3 t, sin3 t. ( 51 )

6 The general solution is then given by y = c 1 cos 3 t + c sin 3 t. ( 5) A theoretical issue checking linear dependence. Lemma 1 0. ( Checking linear independence) If y 1, y are two solutions to on (, ) and if the equality holds at any point τ (only ). a y + b y + c y = 0 ( 53) y 1 ( τ) y ( τ) y 1 ( τ) y ( τ) = 0 ( 54) is needed), then y 1 and y are linearly dependent on (, More examples. We first recall the procedure of solving linear nd order constant-coefficient equations: 1. Write down and solve the auxiliary equation Or equivalently, substitute y = e rt into the equation.. Three cases. a. r 1 r, both real. General solution is given by b. r 1 = r real. General solution is given by a y + b y + c y = 0. ( 55) a r + b r + c = 0 r 1, r. ( 56) y = c 1 e r1 t + c e r t. ( 57) y = c 1 e r1 t + c t e r1 t. ( 58) c. r 1, r complex. In this case r 1 = α + i β, r = α i β and the general solution is given by Now we look at more examples. y = c 1 e α t cos β t + c e α t sin β t. ( 59) Example 1 1. ( 4.. 9) Solve Solution. Try y = e rt. We have r r 1 1 = 0 r 1 = As a consequence, the general solution is given by Example 1. ( ) Solve y y 1 1 y = 0. ( 60) y = c 1 e = , r = ( 61 ) t + c e t. ( 6) y + y 8 y = 0; y( 0) = 3, y ( 0) = 1 ( 63) Solution. First we find the general solution. Substituting y = e rt leads to So the general solution is given by r + r 8 = 0 r 1 = 4, r =. ( 64) y( t) = c 1 e 4t + c e t. ( 65)

7 To determine the constants we use the initial values: 3 = y( 0) = c 1 + c ; 1 = y ( 0) = 4 c 1 + c. ( 66) Solving this linear system ( with c 1, c as unknowns) gives c 1 = 3, c = 0. ( 67) So finally the solution to the initial value problem is given by y( t) = 3 e 4t. ( 68) Example 1 3. ( ) Solve 4 w + 0 w + 5 w = 0. ( 69) Solution. Substituting w = e rt leads to 4 r + 0 r + 5 = 0 r 1 = r = 5. ( 70) Thus the two linearly independent solutions are A a consequence, the general solution is e 5 t and t e 5 t. ( 71 ) w( t) = c 1 e 5 t + c t e 5 t. ( 7) Example 1 4. ( ) Solve y + y + y = 0; y( 0) = 1, y ( 0) = 3. ( 73) Solution. First we find the general solution. Substituting y = e rt leads to r + r + 1 = 0 r 1 = r = 1. ( 74) Thus the general solution is given by y( t) = c 1 e t + c t e t. ( 75) Using the initial values: 1 = y( 0) = c 1 ; 3 = y ( 0) = c 1 + c. ( 76) This gives c 1 = 1, c =. ( 77) Therefore the solution to the initial value problem is given by y( t) = 1 e t t e t. ( 78) Example 1 5. ( ) Solve Solution. Substititing y = e rt leads to y + y + 5 y = 0. ( 79) r + r + 5 = 0 r 1 = 1 + i, r = 1 i. ( 80) The two linearly independent complex solutions are then given by e ( 1 + i) t = e t [ cos t + i sin t], e ( 1 i) t = e t [ cos t i sin t]. ( 81 ) From this we can obtain the real solutions: e t cos t, e t sin t. ( 8) Finally the general solution is given by y( t) = c 1 e t cos t + c e t sin t. ( 83)

8 + = Example 1 6. ( ) Solve y + y + y = 0; y( 0) =, y ( 0) = 1. ( 84) Solution. First we find the general solution. Substituting y = e rt gives r + r + = 0 r 1 = 1 + i, r = 1 i. ( 85) The complex solutions are then which yield the real solutions So the general solution is given by e ( 1 + i) t, e ( 1 i) t ( 86) e t cos t, e t sin t. ( 87) Now we use the initial values: Solving this we obtain y( t) = c 1 e t cos t + c e t sin t. ( 88) = y( 0) = c 1, 1 = y ( 0) = c 1 + c. ( 89) c 1 =, c = 3. ( 90) The solution to the initial value problem is then given by Two examples in each case. One general solution, one initial value. y( t) = e t cos t + 3 e t sin t. ( 91 ) Linear, homogeneous, constant coefficient equations of higher order. The above theory and method naturally extends to linear homogeneous equations of higher orders: We have a n y ( n) + + a 1 y + a 0 y = 0 ( 9) If y 1,, y n are solutions that are linearly independent 3, then the general solution is given by To find y 1,, y n, we try y = e rt to reach y = c 1 y c n y n. ( 93) a n r n + + a 1 r + a 0 = 0 r = r 1,, r n ( 94) Then if r 1,, r n are all different and real, are the n solutions we want. e r1 t,, e rn t ( 95) If some r i s coincide, or some are complex, we use similar tricks. If r is a root of multiplicity m, then we put e rt, r e rt,, r m 1 e rt ( 96) in our list of linearly independent solutions. If α ± i β are complex roots, then we put into our list of linearly independent solutions. 4 e α t cos β t, e α t sin β t ( 97) Example 1 7. ( ) Solve y + y 6 y + 4 y = 0. ( 98) 3. Meaning: If constants C 1,, C n makes C 1 y 1 + C n y n = 0, then C 1 = C = C n = It can be shown that if α + i β is a root, so is α i β. That is, complex roots have to appear in conjugate pairs ( otherwise we are in trouble! ).

9 Solution. Substituting y = e rt gives r 3 + r 6 r + 4 = 0. ( 99) Inspection suggests r = 1 is a solution. Thus r 3 + r 6 r + 4 = ( r 1 ) [ ]. Assuming to be a r + b r + c and then solving a, b, c, we obtain r 3 + r 6 r + 4 = ( r 1 ) [ r + r 4 ] ( ( ) ) ( ( ) ) = ( r 1 ) r r 1 5. ( 1 00) Thus we have three different roots: r 1 = 1, r = 1 + 5, r3 = 1 5. ( 1 01 ) The general solution is then given by Example 1 8. ( ) Solve y( t) = c 1 e t + c e Solution. Substituting z = e rt leads to Inspecting for a while we realize that Thus the roots are and the general solution is given by ( ) ( ) t 1 5 t + c3 e. ( 1 0) z + z 4 z 8 z = 0. ( 1 03) r 3 + r 4 r 8 = 0. ( 1 04) r 3 + r 4 r 8 = ( r + ) [ r 4 ] = ( r + ) ( r ). ( 1 05) r 1 = r =, r 3 = ( 1 06) z( t) = c 1 e t + c t e t + c 3 e t. ( 1 07) 4. Nonhomogeneous equations how to find the particular solution. Having understood linear nd order constant-coefficient homogeneous equations, we turn to equations with nonzero right hand side: a y + b y + c y = f( t). ( 1 08) Recall that all we need to do is to find one particular solution y p ( t) to this equation, then the general solution is given by y( t) = ỹ ( t) + y p ( t) ( 1 09) where ỹ( t) is the general solution to the corresponding homogeneous problem a y + b y + c y = 0. ( 1 1 0) There is a general method called variation of parameters. However the calculation involved may get a bit messy in practice in particular, it involves integration of possibly complicated functions. On the other hand, for several class of special f( t) s, one can take short-cuts by cleverly guess the form of the solution. In the following we will first discuss these special cases, and then turn to the general method The method of undetermined coefficients. The basic idea is as follows. Since functions like e a t, { sin β t, cos β t}, { t n } are closed under differentiation, when such functions appear on the right hand side, we guess that there are solutions of similar form. Example 1 9. ( ) Find a particular solution of z + z = 9 e t. ( )

10 Solution. We guess a solution of the form z = A e t. Substituting into the equation we have It is clear that taking A = 9/5 leads to a particular solution Example 0. ( ) Find a particular solution of z + z = 8 A e t + A e t = 9 A e t. ( 1 1 ) z( t) = e t. ( 1 1 3) x + x = 3 t. ( 1 1 4) Solution. As ( t n ) = n t n 1, we would like to guess that there is a particular solution of the form A t n. Substituting into the left hand side, we find x + x = A n t n 1 + A t n. ( 1 1 5) This clearly cannot be equal to the right hand side. But this first try tells us two things: 1. To obtain 3 t, we cannot use a single power t n. We have to use a polynomial A n t n + A n 1 t n The highest power on the left hand side is t n, coming from x. It cannot be cancelled by any term coming from x. As a consequence, n =. Therefore we would like to try Substituting into the equation, we have z = A t + B t + C. ( 1 1 6) x + x = [ A t + B] + A t + B t + C = A t + ( 4 A + B) t + ( B + C ). ( 1 1 7) Equating with the right hand side: gives the following system: A t + ( 4 A + B) t + ( B + C ) = 3 t ( 1 1 8) A = 3 ( 1 1 9) 4 A + B = 0 ( 1 0) B + C = 0 ( 1 1 ) whose solution is A = 3, B = 1, C = 4. ( 1 ) As a consequence, the particular solution can be taken as x p ( t) = 3 t 1 t + 4. ( 1 3) Example 1. ( ) Solve y y + 9 y = 3 sin 3 t. ( 1 4) Solution. Recall that differentiating sin one obtains cos, and differentiating cos one obtains sin. Thus the set { A sin 3 t + B cos 3 t} is closed under differentiations. Inspired by this, we guess Substituting into the left hand side, we have y = A sin 3 t + B cos 3 t. ( 1 5) y y + 9 y = [ 9 A sin 3 t 9 B cos 3 t] [ 3 A cos 3 t 3 B sin 3 t] + 9 [ A sin 3 t + B cos 3 t] = 3 B sin 3 t 3 A cos 3 t. ( 1 6) Equating with the right hand side: 3 B sin 3 t 3 A cos 3 t = 3 sin 3 t ( 1 7)

11 we obtain Thus the particular solution is given by B = 1, A = 0. ( 1 8) y p ( t) = cos 3 t. ( 1 9) Example. ( ) Find a particular solution for Solution. Writing x = e ( ln ) x, we naturally guess y ( x) + y( x) = x. ( 1 30) y( x) = A e ( ln ) x. ( 1 31 ) Substituting into the equation, we reach [ ] y + y = A ( ln ) e ( ln ) x + A e ( ln ) x = A ( ln ) + 1 x. ( 1 3) Equating with the right hand side, we have [ ] A = ( ln ) ( 1 33) which leads to [ ] y p ( x) = ( ln ) 1 [ ] + 1 e ( ln ) x = ( ln ) x. ( 1 34) Thus we can find the general solution to equations with right hand side of the above forms. Example. ( ) Find the general solution to y y = t. ( 1 35) Solution. Recall that all we need to do is to find one particular solution and the general solution to the corresponding homogeneous equation y y = 0. ( 1 36) Finding one particular solution. Try y = A t + B. Substituting into the left hand side, we have y y = 0 ( A t + B). ( 1 37) Equating with the right hand side, we find out A = 1, B = 0. Thus y p = t is a particular solution. Finding general solution to the homogeneous equation. Substituting y = e rt gives Thus the general solution is given by r 1 = 0 r 1 = 1, r = 1. ( 1 38) ỹ = c 1 e t + c e t. ( 1 39) Putting everything together, we see that the general solution to the non-homogeneous equation is Example 3. ( ) Find the general solution to y( t) = c 1 e t + c e t t. ( 1 40) y + 3 y 4 y = t + sin t + 3. ( 1 41 ) Solution. The idea now is to use superposition principle, that is, if y i solves a y + b y + c y = f i, ( 1 4)

12 then c 1 y c k y k solves a y + b y + c y = c 1 f 1 + c f + + c k f k. ( 1 43) Guided by this, what we need to do ( to solve the problem) are the following: First we simplify the right hand side using the trignometric formula Thus the right hand side becomes We need to Find a particular solution y p1 for Find a particular solution y p for Find a particular solution y p3 for Find the general solution ỹ for cos t = 1 sin t sin t = 1 1 cos t. ( 1 44) Then the general solution to the nonhomogeneous equation is given by Now we carry this plan out. Find a particular solution y p1 for t 1 cos t + 7. ( 1 45) y + 3 y 4 y = t; ( 1 46) y + 3 y 4 y = 1 cos t; ( 1 47) y + 3 y 4 y = 7 ; ( 1 48) y + 3 y 4 y = 0. ( 1 49) y = ỹ + y p1 + y p + y p3. ( 1 50) y + 3 y 4 y = t; ( 1 51 ) Based on the right hand side, we guess y = A t + B. Substituting into the equation we reach Thus 4 A t + 3 A 4 B = t. ( 1 5) and A = 1, B = 3 8 ( 1 53) y p1 = 1 t 3 8. ( 1 54) Find a particular solution y p for We use the ansatz Then y + 3 y 4 y = 1 cos t; ( 1 55) y = A cos t + B sin t. ( 1 56) y + 3 y 4 y = [ 4 A cos t 4 B sin t] + 3 [ A sin t + B cos t] Equating with the right hand side 4 [ A cos t + B sin t] = ( 1 A + 6 B) cos t + ( 1 B 6 A) sin t. ( 1 57) 1 A + 6 B = 1, 1 B 6 A = 0. ( 1 58)

13 This gives Therefore we can take Find a particular solution y p3 for As the right hand side is a constant, we try This quickly leads to Find the general solution ỹ for Substituting y = e rt leads to r + 3 r 4 = 0 r 1 = A = 1 30, B = ( 1 59) y p = 1 30 cos t 1 sin t. ( 1 60) 60 y + 3 y 4 y = 7 ; ( 1 61 ) y = A. ( 1 6) y p3 = 7 8. ( 1 63) y + 3 y 4 y = 0. ( 1 64), r = We have two distinct real roots, thus the general solution is given by ỹ = c 1 e Putting everything together, the solution to the original problem is y = ỹ + y p1 + y p + y p3 = c 1 e ( 1 65) t + c e ( 1 66) t + c e t cos t 1 60 sin t 5 4. ( 1 67) From the above experience we see that, basically, when the right hand side is of the form e a t, we guess the solution to be of the form A e a t ( Note that constant right hand side belongs to this case) ; When the RHS is sin β t or cos β t, we guess the solution to be A cos β t + B sin β t; When the RHS is of the form t m, we guess the solution to be a polynomial of degree m, that is A m t m + A m 1 t m A 1 t + A 0. However life is not so simple, as is shown by the following example. Example 4. ( ) Solve Solution. As usual we try y y + y = 8 e t. ( 1 68) y = A e t. ( 1 69) Substituting into the equation, we have [ ( y y + y = A ) ] e t e t + e t = 0 ( 1 70) which clearly cannot be made 8 e t! Inspecting the above, we see that the problem is that e t is a solution to the corresponding homogeneous equation, thus everything cancelled out and we have nothing left to adjust. The idea now is to put factor of the form t k in front of e t. Let s see what is the appropriate k. Substituting y = A t k e t we have y y + y = A [ t k e t + k t k 1 e t + k ( k 1 ) t k e t ] A [ t k e t + k t k 1 e t ] + A t k e t = A t k [ e t e t + e t ] + A t k 1 [ k e t k e t ] + A k ( k 1 ) t k e t = A k ( k 1 ) t k e t. ( 1 71 )

14 As the right hand side is 8 e t, we see that the appropriate k is. Substituting y = A t e t into the equation we reach So the desired particular solution is A e t = 8 e t A = 4. ( 1 7) y = 4 t e t. ( 1 73) Looking at the above example again, we see that e t corresponds to r = 1, which is a double root of the auxiliary equation of the homogeneous equation r r + 1 = 0 ( 1 74) y y + y = 0. ( 1 75) That is, the multiplicity of the root r = 1 is. We notice that this is exactly the value of k we need to take in the ansatz t k e t. This is not coincidence! Method of undetermined coefficients To find a particular solution to the differential equation where P m ( t) is a polynomial of degree m, use the form where a y + b y + c y = P m ( t) e rt, ( 1 76) y p ( t) = t s ( A m t m + + A 1 t + A 0 ) e rt ; ( 1 77) r is not a root of the associated auxiliary equation s = 0; r is a simple root s = 1 ; r is a double root s =. To find a particular solution to the differential equation a y + b y + c y = P m ( t) e αt cos β t + Q n ( t) e α t sin β t ( 1 78) where P m ( t) is a polynomial of degree m and Q n ( t) is a polynomial of degree n, use the form y p ( t) = t s( A k t k + + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + + B 1 t + B 0 ) e α t sin β t. ( 1 79) where k is the larger of m and n ( that is k = max { m, n} ), and s is determined by α + β i is not a root of the associated auxiliary equation s = 0; α + β i is a root s = 1. 5 Armed with this general procedure, we now study a few more examples. Example 5. ( ) Find a particular solution to y ( x) + y( x) = 4 x cos x. ( 1 80) Solution. First we need to figure out the correct form of the solution. Comparing with the general procedure above, we write the right hand side as 4 x e 0 x cos x. ( 1 81 ) 5. Think: What happens to the double root case?

15 We see that the polynomial involved is x, which is of degree one. This suggests that a factor ( A 1 x + A 0 ) is in the solution. Thus the solution is of the form x s ( A 1 x + A 0 ) cos x + t s ( B 1 x + B 0 ) sin x. ( 1 8) To fix the power s, we need to check whether 0 + i is a root to the auxiliary equation Clearly the answer is yes. Therefore we need to take s = 1. Substituting into the equation, we have Thus r + 1 = 0. ( 1 83) y = x [ ( A 1 x + A 0 ) cos x + ( B 1 x + B 0 ) sin x] ( 1 84) 4 x cos x = y + y = [ ( A 1 x + A 0 x ) cos x + ( B 1 x + B 0 x ) sin x ] Solving this gives + ( A 1 x + A 0 x ) cos x + ( B 1 x + B 0 x ) sin x = [ A 1 cos x ( A 1 x + A 0 ) sin x ( A 1 x + A 0 x ) cos x ] + [ B 1 sin x + ( B 1 x + B 0 ) cos x ( B 1 x + B 0 x ) sin x ] + ( A 1 x + A 0 x ) cos x + ( B 1 x + B 0 x ) sin x = 4 B 1 x cos x + ( 4 A 1 ) x sin x + ( A 1 + B 0 ) cos x + ( B 1 A 0 ) sin x. ( 1 85) So the particular solution is given by Example 6. ( ) Find a particular solution to 4 B 1 = 4 ( 1 86) 4 A 1 = 0 ( 1 87) A 1 + B 0 = 0 ( 1 88) B 1 A 0 = 0. ( 1 89) B 1 = 1, A 1 = 0, B 0 = 0, A 0 = 1. ( 1 90) y p = x [ cos x + x sin x]. ( 1 91 ) Solution. First recall that y y + 3 y = cosh t. ( 1 9) cosh t = 1 [ e t + e ] t. ( 1 93) Thus to find the particular solution, we need to find a particular solution for y y + 3 y = 1 et and y y + 3 y = 1 e t ( 1 94) respectively. Particular solution for y y + 3 y = 1 et. ( 1 95) Guided by the method of undetermined coefficients, we look for solutions of the form y = A t s e t. ( 1 96) To determine s, we need to check the relation between r = 1 and the auxiliary equation r r + 3 = 0. ( 1 97)

16 We see that r = 1 is not a solution. Thus s = 0. Substituting y = A e t into the equation we reach 1 et = y y + 3 y = A e t A = 1 4 y p1 = 1 4 et. ( 1 98) Particular solution for We look for solutions of the form y y + 3 y = 1 e t. ( 1 99) y = A t s e t. ( 00) This time r = 1 is also not a solution to the auxiliary equation r r + 3 = 0 ( 01 ) and consequently s = 0. Substituting y = A e t into the equation leads to 1 e t = y y + 3 y = 6 A e t A = 1 1 y p = 1 1 e t. ( 0) Putting everything together, the particular solution is given by The same idea works for higher order equations. Example 7. ( ) Find a particular solution to Solution. We look for solutions of the form y p = 1 4 et e t. ( 03) y + y y = t e t. ( 04) y = t s [ A 1 t + A 0 ] e t. ( 05) To determine s we need to check the relation between r = 1 and the auxiliary equation r 3 + r = 0 ( r 1 ) ( r + r + ) = 0. ( 06) We see that r = 1 is a simple root. Therefore s = 1. Substituting y = t ( A 1 t + A 0 ) e t into the equation we have Thus t e t = y + y y = [ ( A 1 t + A 0 t ) e t ] + [ ( A1 t + A 0 t ) e t ] ( A1 t + A 0 t ) e t = [ A 1 t + ( 6 A 1 + A 0 ) t + 6 A A 0 ] e t + [ A 1 t + ( 4 A 1 + A 0 ) t + A 1 + A 0 ] e t ( A 1 t + A 0 t ) e t = 1 0 A 1 t e t + ( 8 A A 0 ) e t. ( 07) The special solution is then given by [ y p ( t) = t A 1 = 1 1 0, A 0 = 4 5. ( 08) t ] ( t e t = t ) e t. ( 09) Variation of parameters. What if the right hand side is not a combination of powers, exponential, sin s and cos s? In this section we present a more general method: variation of parameters. Consider the nonhomogeneous problem a y + b y + c y = g( t). ( 1 0)

17 As we can easily solve the homogeneous equation a y + b y + c y = 0, ( 1 1 ) we would like to check the possibility of obtaining the particular solution from modifying the general solutions of the homogeneous problem. More specifically, let y 1, y be two linearly independent solutions to the homogeneous problem. We know that the general solution of the homogeneous problem is given by y = c 1 y 1 + c y. ( 1 ) Now the idea is to replace the constants c 1, by two functions and try to find particular solution of the form Substituting this into the nonhomogeneous equation, we obtain g( t) = a y + b y + c y Thus as long as we can find v 1, v such that y p ( t) = v 1 ( t) y 1 ( t) + v ( t) y ( t). ( 1 3) = a [ v 1 y 1 + v y ] + b [ v 1 y 1 + v y ] + c [ v 1 y 1 + v y ] = a [ v 1 y 1 + v 1 y 1 + v 1 y 1 + v y + v y + v y ] + b [ v 1 y 1 + v 1 y 1 + v y + v y ] + c [ v 1 y 1 + v y ] = v 1 [ a y1 + b y 1 + c y 1 ] + v [ a y + b y + c y ] + a [ v 1 y 1 + v 1 y 1 + v y + v y ] + b [ v 1 y 1 + v y ] = a [ v 1 y 1 + v 1 y 1 + v y + v y ] + b [ v 1 y 1 + v y ]. ( 1 4) a [ v 1 y 1 + v 1 y 1 + v y + v y ] + b [ v 1 y 1 + v y ] = g( t), ( 1 5) we are done. This equation is quite complicated. We need to simplify it. Note that we have two unknowns v 1, v and just one equation, it is likely that we can require v 1, v to satisfy another equation without losing existence of solutions. One good choice is requiring v 1 y 1 + v y = 0. ( 1 6) Note that this not only makes the term containing b vanish, but also greatly simplify the first term through Thus our equations for v 1, v are Rewrite this system to 0 = ( v 1 y 1 + v y ) = v1 y 1 + v 1 y 1 + v y + v y. ( 1 7) a [ v 1 y 1 + v y ] = g( t) ( 1 8) v 1 y 1 + v y = 0. ( 1 9) y 1 v 1 + y v = g( t) /a ( 0) y 1 v + y v = 0. ( 1 ) This system admits a unique solution when y 1, y are linearly independent. 6 The solution is given by 6. Recall from linear algebra that the system v 1 = g( t) y ( t) a [ y 1 y y 1 y ], v g( t) y 1 ( t) = a [ y 1 y y 1 y ]. ( 6) a x + b y = e ( ) c x + d y = f ( 3)

18 Then v 1, v can be obtained from integrating the above. Example 8. ( ) Find the general solution to the differential equation y + 4 y = tan t. ( 7) Solution. We need to do two things: finding general solution to the homogeneous equation, and finding one particular solution to the nonhomogeneous equation. General solution to the homogeneous problem. The auxiliary equation is r + 4 = 0 r 1 = i, r = i z 1 = e i t = cos t + i sin t, z = cos t i sin t. ( 8) Thus the general solution to the homogeneous equation is given by ỹ = c 1 cos t + c sin t. ( 9) Particular solution to the nonhomogeneous problem. As the right hand side is tan t, it is not possible to use the method of undetermined coefficients. We use variation of parameters instead. We have y 1 = cos t, y = sin t, g( t) = tan t = sin t cos t, a = 1. ( 30) We compute Thus the equations for v 1, v are v 1 = g( t) y ( t) a [ y 1 y y = 1 1 y ] y 1 y y 1 y =. ( 31 ) sin t cos t, v = g( t) y 1 ( t) a [ y 1 y y = 1 sin t. ( 3) 1 y ] Integrating, we have We evaluate v 1 ( t) = 1 sin t cos t dt = 1 1 cos t dt cos t = 1 1 cos t dt + 1 cos t dt. ( 33) 1 cos t dt = 1 sin t. ( 34) cos t dt = 1 cos t cos t dt = 1 d sin t 4 1 sin t = 1 [ ] dsin t d sin t 8 1 sin t sin t = 1 [ ln 1 + sin t ln 1 sin t ]. ( 35) 8 has a unique solution for all e, f if and only if ( ) a b det = a d b c c d 0. ( 4) When this is the case, the unique solution is given by Cramer s rule ( ) ( ) det e b det a e f d c f x = ( ), y = ( ). ( 5) det a b det a b c d c d

19 Thus v 1 ( t) = On the other hand, Putting things together, we have Thus the desired general solution is given by 7 Example 9. ( ) Solve sin t cos t dt = 1 8 [ ln 1 + sin t ln 1 sin t ] + 1 sin t. ( 36) 4 v ( t) = 1 sin t dt = 1 cos t. ( 37) 4 y p ( t) = v 1 y 1 + v y [ = 1 8 [ ln 1 + sin t ln 1 sin t ] + 1 ] 4 sin t cos t 1 cos t sin t 4 = 1 [ ln 1 + sin t ln 1 sin t ] cos t. 8 ( 38) y( t) = c 1 cos t + c sin t 1 [ ln 1 + sin t ln 1 sin t ] cos t. ( 39) 8 Solution. Clearly we should first simplify the equation to General solution to The auxiliary equation is Thus the general solution is given by Finding one particular solution to We use variation of parameters. We have We compute Thus we have Integrating, we have x x 4 x = e 3 t. ( 40) x x x = e 3 t. ( 41 ) x x x = 0. ( 4) r r = 0 r 1 =, r = 1. ( 43) x = c 1 e t + c e t. ( 44) x x x = e 3 t. ( 45) x 1 = e t, x = e t, g = e 3 t, a = 1. ( 46) x 1 x x 1 x = e t e t = 3 e t. ( 47) v 1 = g( t) y ( t) a [ x 1 x x = 1 1 x ] 3 et, v g( t) y 1 ( t) = a [ x 1 x x = 1 1 x ] 3 e4t. ( 48) The particular solution is then given by v 1 ( t) = 1 3 et, v ( t) = 1 1 e4t. ( 49) x p ( t) = v 1 x 1 + v x = 1 4 e3 t. ( 50) 7. This formula looks different from the answer in the book ( B-7). But actually they are the same.

20 Finally, the general solution to the nonhomogeneous problem is x( t) = c 1 e t + c e t e3 t. ( 51 ) Remark 30. Clearly we can also find the particular solution to x x x = e 3 t. ( 5) using the method of undetermined coefficients: Guess x = A e 3 t and sutstitute into the equation: [ 9 A 3 A A] e 3 t = e 3 t A = 1 4 x p = 1 4 e3 t. ( 53) Remark 31. As we can see from the above calculation, when the method of undetermined coefficients can be applied, usually it s a bit simpler than the method of variation of parameters 8. On the other hand, variation of parameters can be applied to any right hand side. More importantly, it s theoretically much cleaner and can be generalized to the case of variable-coefficient equations. 5. Variable-coefficient equations. The theory of variable-coefficient equations parallel that of constant-coefficient ones, although we don t have a general procedure to obtain explicit formulas for the solutions anymore even for homogeneous problems. Consider the following genera linear nd order ODE a ( t) y + a 1 ( t) y + a 0 ( t) y = g( t). ( 54) We first summarize theoretical facts: Note: the following are true on intervals in which a 1 / a, a 0 /a and g/ a are simultaneously continuous. 9 The general solution can be represented as y = c 1 y 1 + c y + y p ( 55) where y 1, y are any two linearly independent solutions of the corresponding homogeneous equation a ( t) y + a 1 ( t) y + a 0 ( t) y = 0 ( 56) and y p is one particular solution to the nonhomogeneous problem. The solution to the initial value problem exists and is unique. y( t 0 ) = Y 0, y ( t 0 ) = Y 1 ( 57) ( Checking linear dependence/independence) If y 1, y are two solutions to the homogeneous equation over an interval I where p, q are continuous. Then y + p( t) y + q( t) y = 0 ( 58) y 1, y linearly independent W[ y 1, y ] ( t) = y 1 y y 1 y 0 ( 59) for any t I. W[ y 1, y ] ( t) is usually referred to as the Wronskian. The method of variation of parameters can be generalized to the variable-coefficient case almost without any modification. 8. The higher the order, the simpler ( compared to variation of parameters). 9. There are other cases where these properties hold. But discussion of this issue is beyond the level of this course.

21 Let y 1, y be two linearly independent solutions to the homogeneous equation, then v 1 y 1 + v y is a particular solution, where v 1 = g y a W[ y 1, y ], v g y = 1 a W[ y 1, y ]. ( 60) As we have mentioned, there is not general method producing solutions to the general nd order linear variable-coefficient equation. In general we have to rely on luck ( combined with educated guesses whose accuracy depends on experience). Example 3. ( , C auchy-euler equaitoin) Solve t y + 7 t y 7 y = 0. ( 61 ) Solution. Guided by the theory, we only need to find two linearly independent solutions. The key now is to realize the following property of t r : ( t r ) ( k) t k = C t r. Substitute y = t r into the equation, we have 0 = t y + 7 t y 7 y = [ r ( r 1 ) + 7 r 7] t r = 0 r + 6 r 7 = 0 r 1 = 7, r = 1. ( 6) Thus the general solution is y = c 1 t 7 + c t. ( 63) One easily sees that this strategy would work for equations of the form a t y + b t y + c y = 0 ( 64) where a, b, c are constants. Such equations, together with the nonhomogeneous version are called Cauchy-Euler, or Equidimensional, equations. Consider the homogeneous Cauchy-Euler equation Substituting y = t r leads to This is called the associated characteristic equation. As the characteristic equation is quadratic, we have three cases: a t y + b t y + c y = g( t) ( 65) a t y + b t y + c y = 0 ( 66) a r ( r 1 ) + b r + c = 0. ( 67) Two distinct real roots r 1, r. In this case the two linearly independent solutions are t r1, t r. ( 68) A double root r = r 1 = r. As a, b, c are real, r is real. In this case t r is one solution and we need another one. It turns out we can take the second one to be A pair of complex roots r 1, = α ± i β. Then we have y = t r ln t. ( 69) t r1 = e ( α + iβ) ln t = e α ln t [ cos ( β ln t) + i sin( β ln t) ], t r = e α ln t [ cos ( β ln t) sin( β ln t) ]. ( 70) Using the linearity of the solution, we see that two linearly independent solutions are given by y 1 = e α ln t cos ( β ln t) = t α cos( β ln t), y = t α sin( β ln t). ( 71 ) Remark 33. The above looks similar to our theory for linear constant-coeffcient equations. This similarity becomes more striking if we introduce a new variable x = ln t: The three cases become Two distinct roots e r1 x, e r x ; One double root e rx, x e rx ;

22 Complex roots e α x cos β x, e α x sin β x. This is not coincidence! In fact, setting x = ln t gives y = dy dt = dy dx dx dt = t 1 dy dx, Substituting into the equation y = d y dt = d [ dx ] t 1 dy dx dx dt = t d y dx t dy dx. ( 7) 0 = a t y + b t y + c y = a d y dy + dx ( b a) + c y. ( 73) dx Thus we have transformed the Euler-Cauchy equation into a constant-coefficient equaiton. Furthermore, the auxiliary equation for this equation is a r + ( b a) r + c = a r ( r 1 ) + b r + c ( 74) which is exactly the characteristic equation of the Cauchy-Euler equation! Example 34. ( ) Solve the following initial value problem for the Cauchy-Euler equation Solution. Substituting y = t r gives Thus the general solution is given by Using the initial values, we have Solving this we reach t y 4 t y + 4 y = 0; y( 1 ) =, y ( 1 ) = 1 1. ( 75) r ( r 1 ) 4 r + 4 = 0 r 5 r + 4 = 0 r 1 = 4, r = 1. ( 76) y( t) = c 1 t 4 + c t. ( 77) = y( 1 ) = c 1 + c ; 1 1 = y ( 1 ) = 4 c 1 + c. ( 78) Thus the solution to the initial value problem is given by Example 35. ( ) solve Solution. Substituting y = t r we have c 1 = 3, c = 1. ( 79) y( t) = 3 t 4 + t. ( 80) t y 3 t y + 4 y = 0. ( 81 ) r ( r 1 ) 3 r + 4 = 0 r 4 r + 4 = 0 r 1 = r =. ( 8) This is double root so the general solution is given by Example 36. ( ) Solve Solution. Multiply both sides by t : Substituting y = t r gives y = c 1 t + c t ln t. ( 83) y 1 t y + 5 y = 0. ( 84) t t y t y + 5 y = 0. ( 85) r ( r 1 ) r + 5 = 0 r r + 5 = 0 r 1 = 1 + i, r = 1 i. ( 86) The general solution is then given by y = c 1 t cos ( ln t) + c t sin( ln t). ( 87)

23 Cauchy-Euler equation is just one special class of equations. For most equations, however, we will not be able to obtain two linearly independent solutions from guessing. On the other hand, we may be able to find one single solution via guesswork. Fortunately, we can use this one solution as leverage and pull the following trick to obtain a nd solution which is linearly independent of the guessed one, and thus obtain the general solution. Consider the equation y + p( t) y + q( t) y = 0. ( 88) Let y 1 be a solution. The idea is to find v( t) such that y = v y 1 is also a solution. Note that this clearly requires y 1 0, since otherwhse y = y 1 = 0 no matter what v is. On the other hand, as long as v is not a constant, y 1 and y are linearly independent. 1 0 Substituting y = v y 1 into the equation we obtain 0 = y + p( t) y + q( t) y = ( v y 1 ) + p( t) ( v y 1 ) + q( t) v y 1 = v y 1 + v y 1 + v y 1 + p v y 1 + p v y 1 + q v y 1 = v [ y 1 + p y 1 + q y 1 ] + v y 1 + v y 1 + p v y 1 = v y 1 + ( y 1 + p y 1 ) v. ( 89) Thus all we need to do is to solve y 1 v + ( y 1 + p y 1 ) v = 0. ( 90) This is a linear first order equation ( after setting w = v ). Thus we should multiply both sides by to obtain ( ( e ) y1 + py 1 / y1 = e y 1 y 1 e p = y 1 e p ( 91 ) y 1 e ) p v = 0 y1 e p v = C v = C p e e p( t) dt v = C y 1 y 1 ( t) dt. ( 9) As all we need is v satisfying y 1 v + ( y 1 + p y 1 ) v = 0. ( 93) we have the freedom to take any constant C. For example, we can take C = 1 to obtain the following theorem. Theorem 37. ( Reduction of order) Consider the equation y + p( t) y + q( t) y = 0. ( 94) Let y 1 be a solution, not identically zero. Then e p( t) dt y ( t) = y 1 ( t) y 1 ( t) dt ( 95) is a second, linearly independent solution. Example 38. ( & ) Solve Solution. Based on the above theory, we proceed as follows. t y ( t + 1 ) y + y = t, t > 0. ( 96) 1. Find the general solution to the homogeneous equation t y ( t + 1 ) y + y = 0, t > 0. ( 97) 1 0. Another linear algebra exercise!

24 This step consists of two substeps: a. Find one nonzero solution through guessing. b. Find a second one via reduction of order.. Find a particular solution to the nonhomogeneous equation using variation of parameters. Now we carry this plan out. 1. Find the general solution to the homogeneous equation t y ( t + 1 ) y + y = 0, t > 0. ( 98) a. Find one nonzero solution through guessing. We notice that t ( t + 1 ) + 1 = 0. Thus if we can find y such that y = y = y, this y is a solution. Fortunately such y exists: y = e t. b. Find a second one via reduction of order. We have found y 1 = e t. Rewrite the equation into Thus Calculate y t + 1 y + 1 y = 0. ( 99) t t ( p( t) = ), q( t) = 1 t t. ( 300) e p( t) dt y ( t) = y 1 ( t) y 1 ( t) dt ( / t) e = e t e t e = e t t e ln t e t ( Note that as t > 0, t = t) = e t t e t dt [ ] = e t t de t [ ] = e t t e t + e t dt = ( t + 1 ). ( 301 ) As the above calculation is quite complicated, it is a good idea to substitute y = ( t + 1 ) into the equation and see that it is indeed a solution. As the equation is linear, c ( t + 1 ) can also serve as y for arbitrary constant c 0. We take c = 1 to make things simple. Thus the general solution to the homogeneous equation is given by y = c 1 e t + c ( t + 1 ). ( 30). Now we apply variation of parameters, that is, we try to find a particular solution to the nonhomogeneous equation of the form Here y 1 = e t, y = ( t + 1 ). Compute the Wronskian Then we have ( Here a = t, g = t ) v 1 = t y ( t + 1 ) y + y = t, t > 0. ( 303) y = v 1 y 1 + v y. ( 304) W[ y 1, y ] ( t) = y 1 y y 1 y = e t e t ( t + 1 ) = t e t. ( 305) g y = e t ( t + 1 ), v g y a W[ y 1, y = 1 = 1. ( 306) ] a W[ y 1, y ]

25 Integrating, we obtain v 1 = ( t + 1 ) e t dt = ( t + 1 ) de t = ( t + 1 ) e t + e t d( t + 1 ) = ( t + 1 ) e t + e t dt = ( t + ) e t. ( 307) v = 1 = t. ( 308) Thus the particular solution is given by y p = v 1 y 1 + v y = ( t + ) + ( t) ( t + 1 ) = t ( t + 1 ). ( 309) As ( t + 1 ) is a solution to the homogeneous equation, we can simply take y p = t. Summarizing, we have the general solution to be y = c 1 e t + c ( t + 1 ) t. ( 31 0) Note that even if we have used y p = t ( t + 1 ), the term ( t + 1 ) would be absorbed into c ( t + 1 ) here. 6. Free mechanical vibrations. Consider the mass-spring system whose governing equation is or ( written in symbols) [ intertia] d y dx + [ damping] dy + [ stiffness] y = [ External force]. ( 31 1 ) dx m y + b y + k y = f. ( 31 ) We consider the case in which m, b, k are independent of the solution y. In this case the equation is a linear nd order homogeneous differential equation. We would like to understand how the coefficients m, b, k, f affects the properties of the solution. The natural approach is divide and conquer. More specifically, we first consider the undamped case b = 0. Then we add the damping term b y and study its effect on the dynamics of the system. Finally we add external force which leads to important effects such as resonance The undamped, free case. We consider Dividing both sides by m, we reach where The auxiliary equation is Consequently and the general solution is given by m y + k y = 0. ( 31 3) y + ω y = 0 ( 31 4) ω = k/m. ( 31 5) r + ω = 0 r 1 = ω i, r = ω i. ( 31 6) y 1 = cos ω t, y = sin ω t ( 31 7) y = c 1 cos ω t + c sin ω t. ( 31 8)

26 To further understand this formula, it is beneficial to introduce A = c 1 + c and sin φ = c 1 A, cos φ = c A and φ such that ( 31 9) y = c 1 cos ω t + c sin ω t = A [ cos ω t sin φ + sin ω t cos φ] = A sin( ω t + φ). ( 30) From this we see that the mass is simply oscillating, with starting location y 0 = A sinφ, amplitude A, period π/ ω. Such a motion is called simple harmonic motion. 6.. S ystem with damping. Next we add the damping term b y. The equation is then whose associated auxiliary equation reads m r + b r + k = 0 r 1 = b + b 4 m k m Clearly we need to discuss three cases: b < 4 m k, corresponding to two complex roots, b > 4 m k, corresponding to two distinct real roots, b = 4 m k, corresponding to one double root. m y + b y + k y = 0 ( 31 ), r = b b 4 m k. ( 3) m Underdamped motion ( b < 4 m k). Let α = b m, β = 4 m k b. ( 33) m Then we have and the general solution is given by Setting A = c 1 + c we reach and φ such that We see that the mass is still oscillating, with Starting location y 0 = A sin φ, Amplitude A e α t, 1 3 period π/ β. 1 4 r 1 = α + i β, r = α i β ( 34) y = c 1 e α t cosβ t + c e α t sinβ t. ( 35) sin φ = c 1 A, cos φ = c A ( 36) y = e α t [ c 1 cos β t + c sin β t] = A e α t sin( β t + φ). ( 37) 1 1. When c 1 = c = 0, there is no need to introduce A, φ as the solution is identically Note that, these two requirements are not the same as one single requirement tan φ = c 1 /c Recall that α = b < 0, the amplitude is decreasing. Note that it takes infinitely long time for the amplitude to m decay to 0, that is the mass-spring system will stop moving at time t =.

27 Overdamped motion ( b > 4 m k). In this case both roots are real: The general solution is As r 1 = b + b 4 m k m, r = b b 4 m k. ( 38) m y = c 1 e r1 t + c e r t. ( 39) b 4 m k < b, ( 330) both r 1, r are negative. As a consequence both terms c 1 e r1 t and c e r t converges to 0 as t. We see that the mass is not oscillating anymore. It starts at location y 0 = c 1 + c, approaches the origin as t. Note that, in general the amplitude y is not decreasing all the time! Although once it starts decreasing, it will keep decreasing. 1 5 Critically damped motion ( b = 4 m k). The remaining case is a double root The general solution is then As r < 0, we still have r = b m. ( 331 ) y = c 1 e rt + c t e rt = ( c 1 + c t) e rt. ( 33) y( t) 0 as t. ( 333) Qualitatively there is no difference between this case and the overdamped case. It s called critically... just because if b is any smaller, oscillation would kick in Nonzero external force. This is covered in Section which is not required Examples. Example 39. ( ) The motion of a mass-spring system with damping is governed by y + b y y = 0, y( 0) = 1, y ( 0) = 0 ( 334) Find the equation of motion and sketch its graph for b = 0, 6, 8, 1 0. Solution. This equation modeling a mass starting from location 1 with initial velocity 0. finding a general solution and then substituting b = 0, 6, 8, 1 0, we solve each case separately. b = 0. The equation is whose auxiliary equation reads so the general solution is Using the initial values we have Instead of y y = 0 ( 335) r = 0 r 1, = ± 4 i ( 336) y = c 1 cos 4 t + c sin 4 t. ( 337) 1 = y( 0) = c 1, 0 = y ( 0) = 4 c y = cos 4 t. ( 338) 1 4. C alled quasiperiod in the book. As the motion is not exactly periodic Those who would like to understand this should consider: Suppose y( 0) > 0, y ( 0) = 0, then for what c 1, c, r 1, r will y ( t) > 0 for t close to 0.

28 b = 6. In this case the auxiliary equation becomes r + 6 r = 0 r 1, = 3 ± 7 i ( 339) Thus the general solution is y = e 3 t ( ) c 1 cos 7 t + c sin 7 t. ( 340) Using initial values we obtain 1 = y( 0) = c 1, 0 = y ( 0) = 3 c c c 1 = 1, c = 3/ 7. ( 341 ) Thus ( ) y = e 3 t 3 cos 7 t + sin 7 t 7 = 4 7 where φ is such that sin φ = 7 4, cos φ = 3 4 ( ) which ( in this case) is the same as arctan 7 /3. ( ) To plot it, we can first plot sin 7 t + φ b = 8. The auxiliary equation is The general solution is then Using initial values, The solution is then b = 1 0. The auxiliary equation is The general solution is then Initial conditions lead to The solution is given by ( ) e 3 t sin 7 t + φ, then stretch/squeeze it using 4 e 3 t. 7 ( 34) ( 343) r + 8 r = 0 r 1 = r = 4. ( 344) y = ( c 1 + c t) e 4t. ( 345) 1 = y( 0) = c 1, 0 = y ( 0) = 4 c 1 + c c 1 = 1, c = 4. ( 346) y = ( t) e 4t. ( 347) r r = 0 r 1 =, r = 8. ( 348) y = c 1 e t + c e 8 t. ( 349) 1 = c 1 + c, 0 = c 1 8 c c 1 = 4/3, c = 1 /3. ( 350) We can see that its plot looks quite similar to the b = 8 case. y = 4 3 e t 1 3 e 8 t. ( 351 )