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# Differential Equations MATH 308

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This 39 page Class Notes was uploaded by Maximilian Turcotte on Monday October 5, 2015. The Class Notes belongs to MATH 308 at Colgate University taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/219591/math-308-colgate-university in Mathematics (M) at Colgate University.

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Solving 21 Linear System Math 308 In this example we see how to solve a linear system of the form x39Ax Suppose and we want to solve where x The differential equation written in matrix form above is really two rst order equations x39 Ax X39O Xt 2W Y39G 4X0 5Yt 7 We de ne the problem in Maple by entering these equations individually 7 gt eql diffxt t xt 2yt E eq1gxtxt 2yt gt eq2 diffytt 4xt5yt E eq2 gyt 4 xt 5 yt We can use the dsolve function to solve the system As usual the rst argument to dsolve is the problem to be solved and the second argument holds the functions to be found In this case the problem is a system so we put both equations in brackets Also there are two functions to be found so we also put them in brackets gt sol dsolve eqleq2 xt yt 3 301yt7C1e 7C2e t 3 if 1 xtEiC1e 7C2elt 7 Note that Maple has returned the two functions in a list enclosed in curly brackets To access the elements of the list we use the notation soli where i is an integer It is important to also notice that Maple did not put xt rst the order tends to be random It might even be different the next time we 7 enter the exact same command gt sol 2 3 t 1 it XI57C1e 7C2e I gt soll 73 it yt7C1 e 7C2e The vector form of the solution would be xt 730 1 it yt 7C1 e 7C2 1e Check thisyou should understand that this is the same solution as the individual functions in sol l NIH Ifwe have an initial value problem to solve we must include the initial conditions as part of the problem to be solved in the dsolve function Suppse we want to solve the above system with the 7 initial conditions X0l y0l Here is how we can use dsolve to do this gt 3011 dsolve eqleq2x0 ly0 1 I xt Yt i 7 73 7 73 3011yt 3et4e xt 3e 2e 0 7 Here is one way to pull apart Maple39s solution and plot the results Note that when I executed these commands Maple put yt first and Xt second in soll Ifthey are in a different order you will have to change the indices in the following commands Also note that I use the rhs function to get the righthand side of the expression returned by Maple gt solx rhs sell 2 if 3 solx 3e 2e gt soly rhssolll 3 soly 3e t4e t gt plotsolxsolytO253 f 4 3 You can tell which curve is which by the initial conditions gt The MassSpring System Math 308 This Maple session uses the massspring system to demonstrate the phase plane direction elds solution curves trajectories the extended phase space and more The differential equation is 32 a m y b a kyt 0 I39ve used b instead of the usual 7 for the coef cient of friction We convert this to a system by de ning E Vrgyr Then 2 a kyt blgym EVI7 m or B kyt bVt VI77 This equation along with the equation that de nes Vt make up a system of two equations gt withDEtools We rst de ne the two equations that make up the system gt eql diffytt Vt Iquot3 t 1 eq aty V gt eq2 diffvtt kmytbmVt E k t b t eq2vmLL at m m Now de ne a variable that holds the system gt sys eqleq2 a a k n bvuq s 3 tv t vt y 394 at m m We substitute in some numerical values for the parameters so we can plot some solutions The quotsubsquot command makes the substitutions given by the rst several arguments in the expression given as the last argument 7 gt sysl subsbl3mlklsys 3 2 l i m1 7 arm v0 avr yr 3 v0 Let39s rst look at the direction eld This is the same as the vector eld but all the vectors are plotted with the same length We use the DEplot command gt DEplot sysl yt vt t0 lyl lvl lscalingconstrained colorblacktitlequotMassSpring Direction Field mlklbl3 II MassSpring Direction Field m1k1b13 A 39 AN Jamp N ieiii iiiiiiiwo iiiiiii 0 5 iiifZii N a x 39 w N N Let39s add a solution curve also called a trajectory We will nd the curve with initial condition y0l v00 The ics argument to DEplot is a list of triples of the form t0y0v0 Here we only put in one initial condition 010 gt ics 010 ics0l0 7 gt DEplot sysl yt vt t5 15 icsyl lvl l scalingconst rainedcolorblacktitlequotMassSpring Direction Field ml klbl3 quot linecolorblue stepsize l MassSpring Direction Field m1k1b13 A 2g AN N f 4amp W mm i Mi E W 5 iii f MM N m N Let39s plot yt vs t for the same solution The option quotsceneVarlVar2quot let s us choose how the i solution is plotted Ifwe say quotscenetyquot we get a plot of yt vs t 7 gt DEplot sysl yt vt t5 15 icsy2 5 l 5vl lcolorbla cktitlequotMassSpring ml klbl3 quot linecolorblue stepsize 1 scene ty MassSpring m1k1b13 5t to 15 V quotlt Now let39s see this solution in the extended phase space Here we use the command DEp10t3d Don39t forget that you can clickanddrag in the 3D plot to see different Views gt DEplot3dsyslytvt 515icsy25l5vllcolorb lacktitle MassSpring Extend Phase Space mlklbl3 quot linecolorbluestepsizelscene tyv axesnor mal MassSpring Extend Phase Space m1k1b13 Math 308 Differential Equations Second Order Nonhomogeneous Differential Equations Three Solved Examples 1 Find the general solution to dtz 9y 5 sin 2t dzy Solution First find yh The homogeneous equation is W 9y 0 so when we guess yh equot we find r2 9 0 Thus r i3z39 and we have yh k1 cos 3t k2 sin 3t Now we find the particular solution Yp The first guess is Yp Acos2t Bsin2t7 and neither term in Yp solves the homogeneous equation so this will work We find Yp 72A sin 2t 2B cos 2t Yp 74A cos 2t 7 4B sin 2t Now put these into the original equation 74A cos 2t 7 4B sin 2t 9A cos 2t 9B sin 2t 5 sin 2t 5A cos 2t 5B sin 2t 5 sin 2t and this implies A 0 and B 1 Thus Yp sin 2t The general solution is yt yh Yp k1 cos 3t k2 sin 3t sin 2t 2 Solve the initial value problem is div dtz 4y 300s 2t y0 07 yO 1 Solution First find yh following the usual procedure we find yh k1 cos 2t k2 sin 2t Now we find the particular solution Yp Our first guess is Yp Acos 2t B sin 2t This will not work because each term in this guess also solves the homogeneous problem So we multiply by t to get the next guess Yp Atcos 2t Btsin 2t Neither term solves the homogeneous problem so this will work We find Yp B 7 2At sin 2t A 2Bt cos 2t Yp 4B cos 2t 7 4At cos 2t 7 4A sin 2t 7 4Bt sin 2t Put these into the original equation 4B cos 2t 7 4At cos 2t 7 4A sin 2t 7 4Btsin 2t 4At cos 2t 4Btsin 2t 300s 2t 4B cos 2t 7 4Asin 2t 300s 2t and we have A 0 and B 34 Thus Yp itsin 2t The general solution is then 3 yt yh Yp k1 cos 2t k2 sin 2t ltsin 2t Now choose k1 and kg to satisfy the initial conditions y0 0 gt k100 0 gt k1 0 Now y t 2kg cos 2t 32tcos 2t 34 sin 2t and y 0 1 gt 2kg 0 0 1 gt k2 12 The solution to the initial value problem is 1 3 W g sin2t it sin 2t 3 Solve the initial value problem dzy dy w 2 By cos3t y0 1 yO 0 Solution The characteristic polynomial for the homogeneous equation is r2 27quot 5 so we find r 72 i M4 7 202 71i 2239 Thus y 016 j cos2t 026 sin2t To find Yp we could use the formula given in class or in the text but it is not hard to simply derive it from scratch by using the method of undetermined coefficients Our first guess is Yp A cos3t B sin3t 2 and this will work because neither term in Yp also solves the homogeneous problem We find Yp 73A sin3t 3B cos3t Yp 79A cos3t 7 9B sin3t Substituting Yp into the differential equation gives 79A cos3t 7 9B sin3t 2 73A sin3t 3B cos3t 5A cos3t B sin3t cos3t7 or 74A 63 cos3t 74B 6A sin3t cos3t This implies 74A 6B 1 and 74B 6A 0 solving these for A and B gives A 7113 and B 326 The general solution is therefore 4 t 1 3 yt yh Yp 616 cos2t 626 Sln2t 7 E cos3t s1n3t Now we must determine 01 and 02 to satisfy the initial conditions First we39ll need y t 7t v 7t it t 3 9 y t 01726 Sln2t 7 c cos2t 0226 cos2t 7 c s1n2t E Sln3t cos3t The given initial conditions y0 1 and y 0 0 imply 11 Cl 13 70 20 9 0 1 2 and we find 01 1413 and 02 1952 Thus the solution to the initial value problem is 14 19 1 3 yt EETt cos2t 56 sin2t 7 E cos3t sin3t Notes on the Periodically Forced Harmonic Oscillator Warren Weckesser Math 308 Differential Equations 1 The Periodically Forced Harmonic Oscillator By periodically forced harmonic oscillator we mean the linear second order nonhomogeneous dif ferential equation my my kg Fcoswt 1 where m gt O 39y 2 O and k gt 0 We can solve this problem completely the goal of these notes is to study the behavior of the solutions and to point out some special cases The parameter 39y is the damping coe icz39ent also known as the coe icz39ent of friction We consider the cases 39y 0 undamped and 39y gt 0 damped separately 2 Undamped 7 0 When 39y O we have the equation my kg Fcoswt 2 k w0 7 m This is the natuml frequency of the undamped unforced harmonic oscillator To solve 2 we must consider two cases to 7 we and w 4 For convenience de ne 21 to 31 Loo By using the method of undetermined coe icients we nd the solution of 2 to be yt c1 cosw0t 02 sinw0t coswt 3 771M 7 w where as usual cl and 62 are arbitrary constants Note that the amplitude of yp becomes larger as w approaches 4 This suggests that something other than a purely sinusoidal function may result when w we Beats Consider the initial conditions y0 0 and y 0 0 Solving for 01 and 02 gives 7 d 0 mltwg 7 LUZ an 32 and thus W By using some trigonometric identities we may rewrite this as 2F mg 7 wt mg wt yt 37 LUZ s1n lt 2 sin 2 4 Consider the two factors mw l wz sin LO Wt and sin L V Suppose w x we then lwo 7 w is small and the second expression has a much higher frequency than the rst We see that the solution given in 4 is a high frequency oscillation with an amplitude that is modulated by a low frequency oscillation In Figure 1 we consider an example where F 1 m 1 and we 3 In the rst three graphs the solid lines are yt given by 4 and the dashed lines show the envelope or modulation of the amplitude of the solution Note that the vertical scale is different in each graph the amplitude increases as w approaches we Remember that the solutions in Figure 1 are for the special initial conditions y0 0 and y 0 O Beats occur with other initial conditions but the function that modulates the amplitude of the high frequency oscillation will not be a simple sine function Figure 2 shows a solution to 2 with w 35 but with y0 1 and y 0 72 Wcoswt 7 cosw0t 22 w woz Resonance When w we in 2 the method of undetermined coefficients tells us that ypt Acosw0t Bsinw0t can not be the particular solution because each term also solves the homogeneous equation In this case we multiply the guess by t to obtain ypt At cosw0t Bt sinw0t This means that the particular solution is an oscillation whose amplitude grows linearly with t This phenomenon is known as resonance When we substitute yp into 2 and simplify we obtain 72mw0A sinw0t 2mw0B cosw0t Fcosw0t The terms containing a factor of t automatically cancel For this equation to hold for all t we must have A 0 and B F2mw0 Thus the general solution to 2 when w we is yt 01 cosw0t 02 sinw0t tsinw0t 2mw0 Special Case For the special initial conditions y0 0 and y 0 O we obtain 01 02 0 and so yt Ft sinw0t2w0 This solution is the fourth graph shown in Figure 1 with F 1 m1andw03 Beats and Resonan ce i i i Figure 1 Solutions to 2 for several values of w The other parameters are F 17 m 1 and we 3 The initial conditions are y0 0 and y 0 O The solid curves are the actual solutions7 while the dashed lines show the envelope or modulation of the amplitude Beats with Nonzero Initial Conditions i i i i i 1 0 1 n i i i i i i i i i 0 5 10 15 20 25 30 35 40 45 50 Figure 2 Solution to 2 for w 357 with the initial conditions y0 1 and y 0 72 3 Damped 7 gt 0 To solve 1 with the method of undetermined coe icients we rst guess Ygt Acoswt B sinwt We then check that neither term in Y1 solves the homogeneous equation When 39y gt 0 there are three possible forms for the homogeneous solution underdamped critically damped and overdamped but in all cases the homogeneous solutions decay to zero as t increases so neither term in Y can be a solution to the homogeneous equation So our rst guess for Yp will work Substituting this guess into 1 and choosing the coef cients to make Y1 a solutions yields mg 7 w2Fm w yF Ypt W coswt sinwt 5 m2wg 7 w22 39y LUZ Recall that an expression of the form Acoswt Bsinwt may be written as Rcoswt 7 6 where R VA2 1 B2 and tan6 BA R is the amplitude and 6 is the phase angle With this we have Y t 7 4mm 7 6 6 1 mzwg w22 yzwz 7 7 WV 6 7 arctan The general solution to 1 is the usual W 24W 33275 As mentioned earlier when 39y gt O the homogeneous solution will decay to zero as t increases For this reason the homogeneous solution is sometimes called the transient solution and Y1 is called the steady state response where Example Consider the initial value problem 24 2210y 0080737 240 127 20 4 We have m 1 39y 2 k 10 F 1 and w 2 We also nd wo wkm V10 The homogeneous solution is yht cleft cos3t cgeit sin3t and using 5 we nd the particular solution to be 3 1 Yt cos2t E sin2t The solution to the initial value problem is yt 7e t cos3t e t sin3t cos2t 1713 sin2t Figure 3 shows yh Y1 and y y Yp W 0 o5 4 10 05 n A lh t 05 0 1 2 3 4 5 6 7 8 9 10 05 n I 05 0 1 2 3 4 5 6 7 8 9 10 Figure 3 Graphs of yh Y and the full solution to 1 with parameter values m 1 39y 2 k 10 F 1 and with initial conditions y0 712 and y 0 4 31 Amplitude of the steady state Let us consider the particular solution given by The amplitude of the steady state is 4 mzwg w22 Yzwz39 We see that the amplitude depends on all the parameters in the differential equation recall that we wkm so the amplitude depends on k through we In particular the amplitude of the steady state solution depends on the frequency w of the forcing Let s take F 1 m 1 we 3 so we have 1 w2 g2 Yzwz39 The amplitude as a function of w for several values of 39y is shown in Figure 4 Note that as 39y gets smaller the peak in the response becomes larger and the location of the peak approaches we The graph of the amplitude when 39y O for which the response blows up77 at we is included for comparison Figure 4 Amplitude of the steady state response as a function of w for several values of 39y The curve associated with 39y 0 has a vertical asymptote at w 3 4 Brief Summary H to 03 F men The second order linear harmonic oscillator damped or undamped with sinusoidal forcing can be solved by using the method of undetermined coefficients In the undamped case beats occur when the forcing frequency is close to but not equal to the natural frequency of the oscillator This is because the homogeneous solution and the particular solution are both sinusoidal functions and their frequencies are close to each other Whenever two sinusoidal functions with close frequencies are added beats will occur In the undamped case resonance occurs when the forcing frequency is the same as the natural frequency of the oscillator In the damped case 39y gt 0 the homogeneous solution decays to zero as 25 increases so the steady state behavior is determined by the particular solution In the damped case the steady state behavior does not depend on the initial conditions The amplitude and phase of the steady state solution depend on all the parameters in the problem Words to Know harmonic oscillator damped undamped resonance beats transient steady state amplitude phase Math 308 Differential Equations Summary of the Method of Undetermined Coef cients The Method of Undetermined Coef cients is a method for nding a particular solution to the second order nonhomogeneous differential equation my 52 ky 905 when gt has a special form involving only polynomials exponentials sines and cosines In the following table Pnt is a polynomial of degree n Pnt ant an71tn 1 a1t a0 got mt rst guess ken Aert k coswt g k sinwt A sinwt B coswt Pnt Ant An1t 1 A1t A0 Pnte Antn An1t 1 A1t Ade Pntequot coswt g Pntequot sinwt Antn An71tn 1 A1t Ade coswt Bntn Bn1t 1 B12 B0e sinwt If any term in the rst guess is also a solution to the corresponding homogeneous equation multiply the whole guess by 25 If any term in this second guess is still a solution to the homogeneous equation multiply by 25 again ie multiply the rst guess by 252 To nd the coef cients substitute yp into the differential equation and collect the coef cients of the different functions of 25 Example Consider y 33 2y 252 We nd yht 154 k254i Since gt 252 a second degree polynomial we use the third line in the above table and we guess ypt Agt2 A1t A0 None of the three terms in this guess also solves the homogeneous equations so this guess will work Example Consider y 63 10 te gtcos We nd yht kle gtcos kge StsinOf Now gt te St cost so we use the fth line in the table above 71 1 a1 1 a0 O r 73 w 1 to make the rst guess ypt A1t A0e 3t cost B125 B0e 3t sint Alte gt cost A0573 cost Blte gt sint B0575 sint However the terms Aoe St cost and Boe gt sint both solve the homogeneous equation so we must multiply the rst guess by 25 Our guess is now 34105 t 1412 A0e 3 cost B12 Bo sint Altzeigt cost Aoteigt cost Bltzeigt sint Boteigt sint None of the terms in this guess solves the homogeneous equation so this guess will work Math 308 Differential Equations Fall 2002 Solved Problems Section 12 8 107 12 287 33 35 128 l7m going to do this one in gory detail7you donlt have to do all these steps explicitly every time you solve a problem like this First we note that yt E 2 is an equilibrium solution Now suppose yt 2 in fact letls assume that yt lt 2 Then d d 27y gt dt gt 7ln 27y tC1 gt 27y 7170102 it Note that C2 6 01 gt 0 Now since yt lt 2 we have 2 7 y gt 0 thus 2 7 y 2 7 y Thus 27y Cgequot gt 27yCgequot gt y27C2 quot Suppose on the other hand that yt gt 2 then 2 7 y lt 0 so 2 7 y 72 7 Then 2 7 y C26quot gt 72 7 y C26quot gt y 2 C26quot The two cases lt 2 and yt gt 2 have given us corresponding families of solutions y 2 7 C26quot and y E 2 02647 where in each case the constant C2 is positive We can combine the two cases into 9 2 0367 where C3 is an arbitrary constant possibly positive or negative In fact by setting C3 0 we obtain the equilibrium solution so this formula gives us all the solutions 1210 First note that yt E 0 is an equilibrium solution Now suppose y f 0 i t33Cl 7 t302 ty2t2y2 gt t2dt gt gag301 gt y So the solutions are t 7 3 y 7 is 02 where C2 is an arbitrary constant or yt 0 d d 1212 yt E 0 is an equilibrium solution If yt 0 d7 tf tylS gt Tyg t22 C1 gt y23 t23 C2 gt y2 t23 Cg3 gt y id t23 C2 Thus the solutions are yt0 or ytt23C23 or yt7t23C23 323 tdt gt yT 1226 yt E 0 is an equlibrium solution If yt 07 dy 2 2 2 dy 1 2 1 7t 2 t 2 77 Zdt 777 2 2t 0 7 dt yy gtyy2 gt w y 1 y t222t01 We require y017 which gives us y07ci11 gt C171 Thus 7 7 1 7 72 y t222t717t24t72 1228 We see that zt E 0 is an equilibrium solution Now if zt 07 ii in j if itdt lnlzl 422 01 m Lg01 02642 m I 036722 I Thus the general solution is zt 0367122 where C3 is an arbitrary constant We want the solution for which 10 177r7 so we have 10 0360 C3 So the solution to the initial value problem is equot22 zt d 1233 First7 recall from Calculus that 21 1 arctany Then 9 d9 2 d9 2 2 a y 1t gt y21tdt gt arctany t 2C1 gt ytant 2C1 We want the solution where y0 17 so we must have y0 tan C1 1 gt C1 7r4 Thus the solution to the initial value problem is yt tan t227r4 1235 Let St be the amount of salt measured in pounds in the bucket at time t7 where t is the time in minutes There are two processes here that are causing St to change First7 we are dumping salt into the bucket at the rate 14 lbsmin this will cause the amount of salt to increase at the rate 14 lbsmin The amount of salt is also changing because of the water that is leaving through the spigot We are told that the water leaves at the rate 12 galmin This water contains salt7 so this causes the amount of salt to decrease7 but at what rate The concentration of the salt in the water is St5 lbsgal we are told that the bucket holds ve gallons The rate at which the salt is leaving is given by the formula concentration gtlt flow rate7 which in this case gives us St5 X 12 St10 Thus7 the bucket is gaining salt at the rate of 14 lbsmin7 but is also losing it at the rate of St10 lbsmin Therefore7 the net rate of change of the salt is 14 7 5107 and the corresponding mathematical version of that statement is d5 1 S dt 7 4 10 We are told that initially the bucket is full of pure water7 which means 50 0 We now have an initial value problem to solve The differential equation is separable Here we go d5 10dS 17 gt 17dt gt mdt gt 7101nl2i575lt01 gt 2575 Z 10 67 0010 Cge Elo gt 25 7 S Cge Elo gt S 2 5 7 Cge Eloi So the general solution to the differential equation is St 7 2 5 7 agar10 We want the solution where 50 07 so 50 7 2 5 7 C3 7 0 7 C3 7 25 The solution to the initial value problem is St 7 2 5 7 256V To answer parts acl7 we just plug in numbers a 51 E 02379 b 510 1580 C 560 2494 d 51000 7 2500 To answer e7 we can take a very7 very long time77 to mean t 7gt 00 Then e tlim St 7 tlim 2 5 7 256W 7 25 Notes on the Periodically Forced Harmonic Oscillator Warren Weckesser Math 308 Differential Equations 1 The Periodically Forced Harmonic Oscillator By periodically forced harmonic oscillator we mean the linear second order nonhomogeneous dif ferential equation my by kg Fcoswt 1 where m gt O b 2 O and k gt 0 We can solve this problem completely the goal of these notes is to study the behavior of the solutions and to point out some special cases The parameter b is the damping coe icient also known as the coe icient of friction We consider the cases I 0 undamped and b gt 0 damped separately 2 Undamped b 0 When I O we have the equation my kg Fcoswt 2 k w0 7 m This is the natural frequency77 of the undamped unforced harmonic oscillator To solve 2 we must consider two cases to 7 we and w 4 For convenience de ne 21 to 31 Loo By using the method of undetermined coefficients we nd the solution of 2 to be yt c1 cosw0t 02 sinw0t coswt 3 mw2 7 Lug where as usual cl and 62 are arbitrary constants Note that the amplitude of yp becomes larger as w approaches 4 This suggests that something other than a purely sinusoidal function may result when w we Beats and Resonan ce 1 i Figure 1 Solutions to 2 for several values ofw The initial conditions are y0 0 and y 0 0 The solid curves are the actual solutions while the dashed lines show the envelope or modulation of the amplitude Beats Let us consider the initial conditions y0 0 and y 0 0 We must have 01 7Fmw2 7 w and 02 0 which gives W in By using some trigonometric identities we may rewrite this as yt m 3F 2 sin 0 3W sin 0 J 4 w07w mcoswt 7 coswot Now if w x we we can think of this expression as the product of sin and L sin 2 mwg 7 LUZ 2 Since w x we lwo 7 w is small the rst expression has a much higher frequency than the second We see that the solution given in 4 is a high frequency oscillation with an amplitude that is modulated by a low frequency oscillation In Figure 1 we consider an example where F 1 m 1 and we 3 In the rst three graphs the solid lines are yt given by 4 and the dashed lines show the envelope or modulation of the amplitude of the solution Note that the vertical scale is different in each graph the amplitude increases as w approaches we Remember that the solutions in Figure 1 are for the special initial conditions y0 0 and y 0 O Beats occur with other initial conditions but the function that modulates the amplitude of the high frequency oscillation will not be a simple sine function Figure 2 shows a solution to 2 with we 35 but with y0 1 and y 0 72 Beats with Nonzero Initial Conditions i i i i Figure 2 Solution to 2 for w 35 with the initial conditions y0 1 and y 0 72 22 w woz Resonance When w we in 2 the method of undetermined coefficients tells us that ypt Acosw0t i Bsinw0t can not be the particular solution because each term also solves the homogeneous equation In this case we multiply the guess by t to obtain ypt At cosw0t i Bt sinw0t This means that the particular solution is an oscillation whose amplitude grows linearly with t This phenomenon is known as resonance When we substitute yp into 2 and simplify we obtain 72mw0A sinw0t i 2mw0B cosw0t Fcosw0t The terms containing a factor of t automatically cancel For this equation to hold for all t we must have A 0 and B F2mw0 Thus the general solution to 2 when w we is yt 01 cosw0t i 62 sinw0t i tsinw0t 2mw0 Special Case For the special initial conditions y0 0 and y 0 O we obtain 01 02 0 and so yt Ft sinw0t2w0 This solution is the fourth graph shown in Figure 1 with m 1 and we 3 3 Damped b gt 0 To solve 1 with the method of undetermined coefficients we rst guess ypt Acoswt i B sinwt We then check that neither term in yp solves the homogeneous equation When I gt 0 there are three possible forms for the homogeneous solution underdamped critically damped and overdamped but in all cases the homogeneous solutions decay to zero as t increases so neither term in yp can be a solution to the homogeneous equation So our rst guess for yp will work Substituting this guess into 1 and choosing the coefficients to make yp a solutions yields w2 7 wgFm wa 9P W COW t W 81W 5 Recall that an expression of the form A coswt i B sinwt may be written as R coswt 7 gt where R xA2 i B2 and tan gt BA R is the amplitude and j is the phase angle With this we have F t y m2w2 way bzwz coswt 7 gt 6 3 where 7 b j arctan The general solution to 1 is the usual W 24W 241205 As mentioned earlier when I gt O the homogeneous solution will decay to zero as t increases For this reason the homogeneous solution is sometimes called the transient solution and yp is called the steady state response Example Consider the initial value problem 24 2210y 0080737 240 127 20 4 We have m 1 b 2 k 10 F 1 and w 2 We also nd wo wkm V10 The homogeneous solution is yht cleft cos3t cge t sin3t and using 5 we nd the particular solution to be 3 1 i ypt cos2t sinat The solution to the initial value problem is yt ige t cos3t ge t sin3t cos2t T13 sin2t Figure 3 shows yh yp and y y yp 31 Amplitude of the steady state Let us consider The amplitude of the steady state is F mzwz wggt2 bzwz39 We see that the amplitude depends on all the parameters in the differential equation In particular the amplitude of the steady state solution depends on the frequency w of the forcing Let s take F1m1w03sowehave 1 wz g2 bzwz39 The amplitude as a function of w for several values of b is shown in Figure 4 Note that as 1 gets smaller the peak in the response becomes larger and the location of the peak approaches we The graph of the amplitude when I O for which the response blows up77 at we is included for comparison t h 0 05 o 1 2 3 4 5 6 7 8 9 10 05 n a o5 o 1 2 3 4 5 6 7 8 9 10 05 s u 39 0 0 o 1 2 3 4 5 6 7 8 9 10 Figure 3 Graphs of mu 317 and the full solution to 1 with parameter values m 17 b 27 k 107 F 17 and With initial conditions 340 712 and y 0 4 Amplitude ofthe Steady State 1 1 1 1 Figure 4 Amplitude of the steady state response as a function of w for several values of b The curve associated with b 0 has a vertical asymptote at w 3 4 Brief Summary H to 03 F m0 The second order linear harmonic oscillator damped or undamped with sinusoidal forcing can be solved by using the method of undetermined coefficients In the undamped case beats occur when the forcing frequency is close to but not equal to the natural frequency of the oscillator In the undamped case resonance occurs when the forcing frequency is the same as the natural frequency of the oscillator In the damped case b gt 0 the homogeneous solution decays to zero as 25 increases so the steady state behavior is determined by the particular solution In the damped case the steady state behavior does not depend on the initial conditions The amplitude and phase of the steady state solution depend on all the parameters in the problem Words to Know harmonic oscillator damped undamped resonance beats transient steady state amplitude phase Math 308 Differential Equations Fall 2002 Solving The Coffee Cup Problem The differential equation for Ht the temperature at time t is dH 7 k 70 7 H l a Note that H t 70 1 lie Ht is the function with constant value 70 is a solution to the differential equation Now assume that Ht 70 We treat as if it really is a fraction and rewrite the differential equation as dH H770 dH 7m 1an 7 70 7 7kt 01 7 7m and then integrate both sides This gives us Note the absolute value on the left recall from Calculus that f d ln cl Now exponentiate both sides to obtain H 7 70 e k lrc1 601671 0267 where C2 601 Note that C2 gt 0 Now if H 7 70 gt 0 then H 7 70 H 7 70 and we have Ht 7 70 7 025163 2 If H 7 70 lt 0 then H 7 70 7H 7 70 and so Ht 7 70 7 026163 3 Thus the solutions to the differential equation are given by l 2 and For this example we see that we can combine these three formulas into one expression that includes all the solutions Ht 7 70 036 where C3 is an arbitrary constant If we are given an initial condition H 0 H0 we can nd C3 in terms of H0 by evaluating the solution at t 0 and setting it equal to H0 H0 7 70 Cgeo 7 70 03 7 H0 so C3 H0 7 70 and the solution to the initial value problem is Ht 7 70 H0 7 70 The following plot shows the solutions for H0 90 H0 70 and H0 50 with k Oil for the time interval 0 S t S 25 100 Ht vs t Getting Started with Differential Equations in Maple September 2003 7 In this Maple session we see some of the basic tools for working with differential equations in Maple We will only deal with first order equations here 7 First we need to load the quotDEtoolsquot library gt with DEtools 7 Let39s de ne a differential equation As an example we consider y y4y This is a first order ordinary differential equation It is Problem 11 of Section 11 in the text The following command defines a variable called quoteqquot that holds the differential equation gt eq diffytt yt4Yt 2 eq I ytyt 4 yt 7 A few points 1 The derivative of y is specified with the quotdiffquot command 2 We can not drop the quottquot from the dependent variable y Maple treats quotyquot and quotytquot differently and our equation is for quotyt Two useful commands are quotDEplotquot and quotdsolvequot Direction Fields with DEplot Let39s create a direction field for this equation One Maple command that does this is called quotDEplotquot In the first example given here the first argument is the differential equation the second is the unknown function in the differential equation the third is the range of the independent variable to show in the plot and the fourth is the range of the dependent variable gt DEploteqyt tl lyl 5 Here is the same example with some additional arguments included For example the quotdirgridquot option i lets you decide how many alrows to include and quotcolorblackquot makes the arrows black 7 gt DEplot eqy t tl lyl 5 title Direction Field colorblack dirgrid 25 25 Direction Field 5gt 2 37 7 M 07 7 77 7 i i i b 7 Notice that a quick glance at the diection eld let s us predict how the solutions will behave Before proceeding print a copy of the above direction eld and sketch the solution for which y0l Be sure to consider negative t as well as positive Also sketch the equilibrium solutions How to print the direction eld I In the menu bar select Options gt Plot Display gt Window When you select this option subsequent plots will appear in a separate window rather than in Maple session 2 Move the cursor back to the previous DEplot command and hit enter This will execute the command again and now it will appear in a separate window 3 Click on the printer icon or use the menu bar File gt Print Note that whether you use the icon or the pulldown menu the quotactivequot window is printed If you have just created a plot it will be the active window Otherwise click on the window first before printing it 4 You may want to restore the default plot output to Inline by using the menu bar to select Options gt PlotDisplay gt Inline 7 Let39s add a solution curve to this plot To do this we must specify an initial condition Let s try y0l To specify initial conditions in DEplot you put them in a list contained in square brackets just after the t range The added option quotlinecolorbluequot tells the command to draw the solution curve in black The default line color is yellow gt DEploteqyt tl 1 y0 1 yl 5linecolorblue V n w L571 N LN Vr V I l V u l A V V L 14VVl C Note that I used double square brackets around the initial condition This is because you can enter i several initial conditions each must be in square brackets Here is an example gt DEplot quYt Itl lI YO l I y0 0 I Y0 1 I Y0 3 I Y0 4 y0 5 yl 5 linecolorblue 4 7 r 7 7 7 7 r7 7 7 r7 7 r 7 77 7 7 7 17 Observe how the arrows drawn in the direction field are tangent to the solution curvesthat is the whole point of a direction eld Important Note The solution curves plotted by the DEplot command have been computed numerically using an algorithm similar to Euler39s Method They are generally good approximations but they are not exact and they may occasionally be very bad approximations The Dsolve Command The command quotdsolvequot can solve some differential equations analytically Let s start out with a simple example If you haven t done so before solve this by hand before continuing It is both linear and separable First we ll define the equation and save it in a Maple variable called quoteq2quot l l l l l i a W Igytyt Now we39ll use quotdsolvequot to find the general solution The first argument of quotdsolvequot is the equation and the second is the function to solve for gt dsolve eq2 y t yt 7C1 eH This should look familiar Notice that Maple has called the arbitrary constant quot7C1quot This is a common Maple convention when it creates a constant it typically puts an underscore in the beginning of the name We can also use quotdsolvequot to solve an initial value problem For example suppose we have the initial condition y02 We specify this in the command by combining the differential equation with the initial condition in quotcurly bracketsquot gt dsolve eq2y0 2Yt yt2 eH Look at that example carefully The first argument of dsolve is eq2y02 In Maple curly brackets are commonly used to create lists of things In this case we make a list of the equation and the initial condition and pass this to quotdsolvequot as its first argument The second argument yt stays the same How do we plot this solution Unfortunately the solution given above is not in a format that is easily plotted by the quotplo quot command The plot command plots expressions but the above solution is actually an equation it has an equals sign sitting the middle The expression that we want namely 2expt is the righthandside of the equation Fortunately there is a command in Maple that lets us take the righthandside from an equation The function is called quotrhsquot L So here is one way to plot the solution above First save the solution in a variable say quotsol2quot because it solves eq2 gt 3012 dsolveeq2y0 2Yt i if 3012 yt 2 e Now take out the righthandside and save it in a new variable called rhsSol2 gt rhsSolZ rhs 012 rhsSolZ 2 em Now plot it 7 gt p10trhsSolZtl 2 You can combine several of those steps gt 3012 rhs dsolveeq2y0 2Yt i sol2 2 eH 7 gt plot3012tl 2 1 0 5 0 055 4 U1 M E Let39s try another example Recall that quoteqquot is still de ned gt eq a EYtYf4 Yt This equation is separable and can be solved by hand The y integral can be done using partial fractions i Let39s give this to quotdsolvequot gt 3011 rhs dsolveeqyt l sol 4 74 l 4 e 7C This is probably simiar to what you would get if you solved it by hand Note however that Maple has given no indication that the equilibrium solution yt0 is also a possibility Let39s try the initial condition y0l 7 gt solla rhsdsolveeqy0lyt solla4 4t 3e 7 gt plotsollatl 1 1 0 8 0 6 0 4 0 2 0 052 054 0 6 0 8 1 t That looks like what we expected from the direction eld Let39s try the initial condition y00 to make sure that Maple will give us the correct solution if we 7 speci cally ask for it gt sollb rhs dsolve eqy0 0Yt sollb 0 OK that worked Let39s try one more initial condition y0 12 gt sollc rhs dsolveeqy0 l2Yt 301104 4t l 9e 7 gt plotsollctl 1 1000 500 0 500 7i 1000 1500 7 20007 What happened Let39s force the plot command to use a smaller vertical range 7 gt plot sollc t4 4y3 5 37 7 It appears that the solution has a vertical asymptote If you look at the solution found above you ll see that this is the case In fact we can nd that value where it occurs by nding where the denominator is zero In the following we use the quotdenomquot command to get the denominator and then nd the root using the quotsolvequot command gt den denomsollc den l 9 elt740 ll 9 K74 n gt r solveden0t gt eva1fr 5493061442 Can Maple solve any first order differential equation Unfortunately no Here is a simple example gt eq3 diff yt t sinyt cos t E eq3 gya sinyt cost E gt dsolve eq3 y t The function returns nothing Maple can not solve this equation Math 308 Differential Equations Solutions to 432 and 4310 432 The equation is dzy dtz First find yh The homogeneous equation is 9y 0 so when we guess y 65 we find 52 9 0 Thus 5 i3z39 and we have 9y Ssin2t y k1 cos 3t k2 sin 3t Now we find the particular solution yp The first guess is yp Acos 2t Bsin 2t and neither term in yp solves the homogeneous equation so this will work We find y 72A sin 2t 2B cos 2t y 74A cos 2t 7 4B sin 2t Now put these into the original equation 74A cos 2t 7 4B sin 2t 9A cos 2t 9B sin 2t 5 sin 2t 5A cos 2t SE sin 2t 5 sin 2t and this implies A 0 and B 1 Thus yp sin 2t The general solution is yt y yp k1 cos 3t k2 sin3t sin 2t 4310 The initial value problem is dzy 7 4y 300s 2t y0 0 y 0 0 dzt2 First find yh following the usual procedure we find y k1 cos 2t k2 sin 2t Now we find the particular solution yp Our first guess is yp Acos 2t B sin 2t This will not work because each term in this guess also solves the homogeneous problem So we multiply by t to get the next guess yp At cos 2t Bt sin 2t Neither term solves the homogeneous problem so this will work We find y B 7 2At sin 2t A 2Bt cos 2t y 4B cos 2t 7 4At cos 2t 7 4A sin 2t 7 4Btsin 2t Put these into the original equation 4B cos 2t 7 4At cos 2t 7 4A sin 2t 7 4Btsin 2t 4At cos 2t 4Btsin 2t 300s 2t 4B cos 2t 7 4Asin 2t 300s 2t and we have A 0 and B 34 Thus yp itsin 2t The general solution is then 3 yt yh yp k1 cos 2t k2 sin 2t Ztsin 2t Now choose k1 and kg to satisfy the initial conditions y0 0 gt k100 0 gt k1 0 Now yt 2kg cos 2t 32tcos 2t 34 sin 2t and y 0 0 gt 2kg 0 0 0 gt k2 0 The solution to the initial value problem is 3 yt 1t sin 2t Math 308 Differential Equations Undetermined Coe icients More Examples Example Find the correct form of the particular solution to y 4y tsin2t but do not solve for the coef cients Solution First we find yht k1 cos2t k2 sin2t Since gt tsin2t our first guess for yp is A1t A0 cos2t B1t B0 sin2t Since the terms A0 cos2t and B0 sin2t solve the homogeneous equation we must multiply our initial guess by t to obtain ypt tA1t A0 cos2t B1t B0 sin2t Altz cos2t Aotcos2t Bltz sin2t Botsin2t None of the terms in yp solve the homogeneous equation so this will work Example Find the general solution to y 4y 4y teem Solution The characteristic polynomial is A2 4A 4 A 22 so the only root is A1 72 The solution to the homogeneous equation is then yht k16T2tk2t6T2t Now gt te Zt so our first guess is A1t A0e 2t but both Alte Zt and Aoe Zt solve the homogeneous equation Therefore we multiply by tto obtain tA1tA0e 2t Altze Zt A0te 2t However the second term Aote Zt is still a solution to the homogeneous equation so multiply by t again We obtain ypht t2 A1t A0e 2t A1t3672t A0t2672t7 and this guess will work To find the coefficients we will need yt 72A1t3 Aorze 2 t 3A1t2 2A0te 2t 2A1t3 7 2A0t2 2A0t T2t and gm 7272A1t3 3A1 i 2A0t2 Mme t 76A1t2 6A1 i 4A0t 2A0e 2 4A1t3 712A1 4A0t2 6A1 i 8A0t 2A0e 2 To solve for the coefficients put these into the differential equation The exponential factor in each term can be canceled and we are left with 4A1 i 8A1 4A1t3 712A1 4A0 12A1 i 8A0 4A0t2 6A1 i 8A0 8A0t 2A0 t The coefficients of t3 and t2 are automatically zero To make the remaining terms match for all t we must have 6A1 1 and 2A0 0 Thus A0 0 and A1 16 and ypt 16t36T2t The general solution is 1 yt yht ypt klc Zt kgtCTZt EtSETZt

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