Differential equations and linear algebra solutions manual pdf

DIFFERENTIAL EQUATIONS

AND

LINEAR ALGEBRA

MANUAL FOR INSTRUCTORS

Gilbert Strang

Massachusetts Institute of Technology

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2 Chapter 1. First Order Equations

1

Problem Set 1, page 3

1 Draw the graph ofy=etby hand, for 21 ftf 1. What is its slopedy/dtat t= 0? Add the straight line graph ofy=et. Where do those two graphs cross? Solution The derivative ofethas slope 1 att= 0. The graphs meet att= 1where their value ise. They don’t actually “cross” because the line is tangent to the curve: both have slopey 2 =eatt= 1.

2 Draw the graph ofy 1 =e 2 ton top ofy 2 = 2et. Which function is larger att= 0? Which function is larger att= 1?

Solution From the graphs we see that att= 0, the function 2 etis larger whereas at t= 1, e 2 tis larger. (etimeseis larger than 2 timese).

3 What is the slope ofy=e 2 tat t= 0? Find the slopedy/dtatt= 1.

Solution The slope ofe 2 tis 2 e 2 t. Att= 0this is 21. The slope att= 1is 2 e 21.

4 What “logarithm” do we use for the numbert(the exponent) whenet= 4?

Solution We use the natural logarithm to findtfrom the equationet= 4. We get that t= ln 4j 1. 386.

5 State the chain rule for the derivativedy/dtify(t) =f(u(t))(chain offandu). Solution The chain rule gives:

dy dt

df(u(t)) du(t)

du(t) dt

6 The second derivative ofetis againet. Soy=etsolvesd 2 y/dt 2 =y. A sec- ond order differential equation should have another solution, different fromy=Cet. What is that second solution? Solution The second solution isy=e 2 t. The second derivative is 2 ( 2 e 2 t) =e 2 t.

7 Show that the nonlinear exampledy/dt = y 2 is solved byy = C/(1 2 Ct) for every constantC. The choiceC= 1gavey= 1/(1 2 t), starting fromy(0) = 1. Solution Given thaty=C/(1 2 Ct), we have:

y 2 =C 2 /(1 2 Ct) 2 dy dt=C·( 2 1)·( 2 C)1/(1 2 Ct)

2 =C 2 /(1 2 Ct) 2

8 Why will the solution tody/dt =y 2 grow faster than the solution tody/dt =y (if we start them both fromy= 1att= 0)? The first solution blows up att= 1. The second solutionetgrows exponentially fast but it never blows up. Solution The solution of the equationdy/dt=y 2 fory(0) = 1isy= 1/(1 2 t), while the solution tody/dt=yfory(0) = 1isy=et. Notice that the first solution blows up att= 1while the second solutionetgrows exponentially fast but never blows up.

1. The Exponentialsetandeat 3

9 Find a solution tody/dt= 2 y 2 starting fromy(0) = 1. Integratedy/y 2 and 2 dt. (Or work withz= 1/y. Thendz/dt= (dz/dy) (dy/dt) = ( 21 /y 2 )( 2 y 2 ) = 1. Fromdz/dt= 1you will knowz(t)andy= 1/z.) Solution The first method has dy y 2

= 2 dt

y

y(0)

du u 2

t

0

dv (u, vare integration variables)

y

y(0)

= 2 t

21 y

= 2 t 21

y=

1 +t The approach usingz= 1/yleads todz/dt= 1andz(0) = 1/ 1. Thenz(t) = 1 +tandy= 1/z=1+ 1 t.

10 Which of these differential equations are linear (iny)?

(a)y 2 +siny=t (b)y 2 =t 2 (y 2 t) (c)y 2 +ety=t 10. Solution (a) Since this equation solves asinyterm, it is not linear iny. (b) and (c) Since these equations have no nonlinear terms iny, they are linear.

11 The product rule gives what derivative forete 2 t? This function is constant. Att= 0 this constant is 1. Thenete 2 t= 1for allt. Solution (ete 2 t) 2 =ete 2 t 2 ete 2 t= 0soete 2 t is a constant(1).

12 dy/dt=y+ 1is not solved byy=et+t. Substitute thatyto show it fails. We can’t just add the solutions toy 2 =yandy 2 = 1. What numbercmakesy=et+cinto a correct solution? Solution dy dt =y+ 1

d(et+c) dt =e

t+c+ 1

Wrong d(e

t+t) dt =e

t+t+ 1 Correct c= 21

Problem Set 1, page 15

1 Sett= 2in the infinite series fore 2. The sum must beetimese, close to 7. 39. How many terms in the series to reach a sum of 7? How many terms to pass 7. 3?

Solution The series fore 2 hast= 2:e 2 = 1 + 2 +

If we include five terms we get:e 2 j1 + 2 + 2 +

If we include seven terms we get:e 2 j1+2+

4 Chapter 1. First Order Equations

2 Starting fromy(0) = 1, find the solution tody/dt=yat timet= 1. Starting from thaty(1), solvedy/dt = 2 yto timet = 2. Draw a rough graph ofy(t)from t= 0tot= 2. What does this say aboute 21 timese? Solution y =etup tot= 1, so thaty(1) =e. Then fort > 1 the equation dy/dt= 2 yhasy=Ce 2 t. Att= 1, this becomese=Ce 21 so thatC =e 2. The solution ofdy/dt= 2 yup tot= 2isy=e 22 t. Att= 2we have returned to y(2) =y(0) = 1. Then(e 21 )(e) = 1. 3 Start withy(0) = $5000. If this grows bydy/dt=. 02 yuntilt= 5and then jumps to a=. 04 per year untilt= 10, what is the account balance att= 10? Solution tf5 :

dy dt

=. 02 y 5 ftf10 :

dy dt

=. 04 y givesy=Ce. 04 t y= 5000e. 02 t y(5) =Ce 22 = 5000e. 1 givesC= 5000e 2. 1 y(5) = 5000e. 1 y(t) = 5000(e. 04 t 20. 1 ) y(10) = 5000e. 3 4 Change Problem 3 to start with$5000growing atdy/dt=. 04 yfor the first five years. Then drop toa=. 02 per year until yeart= 10. What is the account balance att= 10? Solution dy dt

=. 04 y

dy dt

=. 02 yfor 5 ftf 10 y =C 1 e. 04 t y =C 2 e. 02 t y(0) =C 1 = 5000 y(5) =C 2 e. 1 = 5000e. 2 y(t) = 5000e. 04 tfortf 5 C 2 = 5000e. 1 y(5) = 5000e. 2 y(t) = 5000(e. 02 t+0. 1 ) y(10) = 5000e. 3 =same as in 1.3.

Problems 5–8 are about y=eat and its infinite series. 5 Replacetbyatin the exponential series to findeat: eat= 1 +at+

(at) 2 +···+

n!

(at)n+··· Take the derivative of every term (keep five terms). Factor outa to show that the derivative of eat equals aeat. At what timeTdoeseatreach 2? Solution The derivative of this series is obtained by differentiating the terms individ- ually: dy dt

=a+at+···+

(n 2 1)!

antn 21 +···

=a

1 +at+

(at) 2 +···+

(n 2 1)!

an 21 tn 21 +···

=aeat .IfeaT= 2 thenaT =ln 2 andT =ln 2 a

6 Start fromy 2 =ay. Take the derivative of that equation. Take thenthderivative. Construct the Taylor series that matches all these derivatives att= 0, starting from 1 +at+ 12 (at) 2. Confirm that this series fory(t)is exactly the exponential series for eat. Solution The derivative ofy 2 =ayisy 22 =ay 2 =a 2 y. The next derivative is y 222 =ay 22 which isa 3 y. Wheny(0) = 1, the derivatives att= 0area, a 2 , a 3 ,.. .so the Taylor series isy(t) = 1 +at+

a 2 t 2 +···=eat.

1. The Exponentialsetandeat 5

7 At what timestdo these events happen? (a) eat=e (b) eat=e 2 (c) ea(t+2)=eate 2 a.

Solution (a)eat=eat t= 1/a. (b)eat=e 2 at t= 2/a. (c)ea(t+2)=eate 2 a at allt. 8 If you multiply the series foreatin Problem 5 by itself you should get the series for e 2 at. Multiply the first 3 terms by the same 3 terms to see the first 3 terms ine 2 at.

Solution (1 +at+

a 2 t 2 )(1 +at+

a 2 t 2 ) = 1 + 2at+

a 2 t 2 +···

This agrees withe 2 at= 1 + 2at+

(2at) 2 +···

9 (recommended) Findy(t)ifdy/dt = ayandy(T) = 1(instead ofy(0) = 1).

Solution

dt dt

=ay givesy(t) =Ceat = 1att= T, this gives C=e 2 aTand y(t) =ea(t 2 T).

10 (a) Ifdy/dt= (ln2)y, explain whyy(1) = 2y(0).

(b) Ifdy/dt= 2 (ln2)y, how isy(1)related toy(0)? Solution

(a)

dy dt

= (ln 2)y³y(t) =y(0)et(ln 2)³y(1) =y(0)eln 2= 2y(0).

(b)

dy dt

= 2 (ln 2)y³y(t) =y(0)e 2 t(ln 2)³y(1) =y(0)e 2 ln 2=

y(0).

11 In a one-year investment ofy(0) = $100, suppose the interest rate jumps from 6%to10%after six months. Does the equivalent rate for a whole year equal8%, or more than8%, or less than8%? Solution We solve the equation in two steps, first fromt= 0tot= 6months, and then fromt= 6months tot= 12months. y(t) =y(0)eat y(t) =y(0)eat y(0) = $100e 0. 06 × 0. 5 = $100e. 03 y(1) = $103. 05 e 0. 1 × 0. 5 = $103. 05 e. 05 = $103. 05 = $108. 33 If the money was invested for one year at 8% the amount att= 1would be: y(1) = $100e 0. 08 × 1 = $108. 33. The equivalent rate for the whole year is indeed exactly 8%.

12 If you invest y(0) = $100 at 4% interest compounded continuously, then dy/dt=. 04 y. Why do you have more than$104at the end of the year? Solution The quantitative reason for why this is happening is obtained from solving the equation: dy dt

= 0. 04 y³y(t) =y(0)e. 04 t y(1) = 100e 0. 04 j$104. 08. .The intuitive reason is that the interest accumulates interest.

6 Chapter 1. First Order Equations

13 What linear differential equationdy/dt=a(t)yis satisfied byy(t) =ecost?

Solution The chain rule forf(u(t))hasy(t) =f(u) =euandu(t) = sint: dy dt

df(u(t)) dt

df dt

du dt

=eucost=ycost. Thena(t) = cos(t).

14 If the interest rate isa= 0. 1 per year iny 2 =ay, how many years does it take for your investment to be multiplied bye? How many years to be multiplied bye 2? Solution If the interest rate isa= 0. 1 ,theny(t) =y(0)e 0. 1 t= 10,the value is y(t) =y(0)e= 20,the value isy(t) =y(0)e 2.

15 Write the first four terms in the series fory=et

2 . Check thatdy/dt= 2ty. Solution y=et

2 = 1 +t 2 +

t 4 +

t 6 +··· dy dt

= 2t+ 2t 3 +t 5 +···= 2t

1 +t 2 +

t 4 +···

= 2tet

2 .

16 Find the derivative ofY(t) =

1 +nt

n . Ifnis large, thisdY /dtis close toY!

Solution The derivative ofY(t) =

1 +nt

n with respect totisn( 1 n)

1 +nt

n 21 = 1 +nt

n 21 . For largenthe extra factor1 +ntis nearly 1 , anddY /dtis nearY.

17 (Key to future sections). Suppose the exponent iny=eu(t)isu(t) =integral ofa(t). What equationdy/dt= ydoes this solve? Ifu(0) = 0what is the starting valuey(0)? Solution Differentiatingy=e

∫a(t)dt with respect totby the chain rule yieldsy 2 = a(t)e

∫ a(t)dt. Thereforedy/dt=a(t)y. Ifu(0) = 0we havey(0) =eu(0)= 1.

18 The Taylor series comes fromed/dxf(x), when you write outed/dx= 1 +d/dx+ 1 2 (d/dx)

2 +···as a sum of higher and higher derivatives. Applying the series tof(x)

atx= 0would give the valuef+f 2 + 12 f 22 +···atx= 0. The Taylor series says : This is equal tof(x)atx=. Solution f(1)=f(0) + 1f 2 (0) +

12 f 22 (0) +··· This is exactly

f(1) =

d dx

d dx

f(x)atx= 0.

19 (Computer or calculator, 2 is close enough) Find the timetwhenet = 10. The initialy(0)has increased by an order of magnitude—a factor of 10. The exact statement of the answer ist=. At what timetdoesetreach 100? Solution The exact time whenet= 10ist= ln 10. This istj 2. 30 or 2. 3026. Then the time wheneT= 100isT = ln 100 = ln 10 2 = 2 ln 10j 4. 605. Note that the time whenet= 101 ist= 2 ln 10and nott=ln 10 1.

20 The most important curve in probability is the bell-shaped graph of e 2 t

2 / 2 . With a calculator or computer find this function att = 22 , 21 , 0 , 1 , 2. Sketch the graph ofe 2 t

2 / 2 fromt=2>tot=>. It never goes below zero. Solution Att= 1andt= 21 , we havee 2 t

2 / 2 =e 21 / 2 = 1/

ej. 606

Att= 2andt= 22 , we havee 2 t

2 / 2 =e 22 j. 13.

1. Four Particular Solutions 7

21 Explain whyy 1 = e(a+b+c)tis the same asy 2 = eatebtect. They both start at y(0) = 1. They both solve what differential equation? Solution The exponent rule is used twice to finde(a+b+c)t=eat+bt+ct=eat+btect= eatebtect. This function must solvedydt = (a+b+c)y. The product rule confirms this.

22 Fory 2 =ywitha= 1, Euler’s first step choosesY 1 = (1 + ∆t)Y 0. Backward Euler choosesY 1 =Y 0 /(1 2 ∆t). Explain why1 + ∆tis smaller than the exacte∆t and 1 /(1 2 ∆t)is larger thane∆t. (Compare the series for 1 /(1 2 x)withex.) Solution 1 + ∆tis certainly smaller thane∆t= 1 + ∆t+ 12 (∆t) 2 + 16 (∆t) 3 +··· 1 12 ∆t= 1+∆t+(∆t)

2 +(∆t) 3 +···is larger thane∆t, because the coefficients drop

below 1 ine∆t.

Problem Set 1, page 27

1 All solutions tody/dt= 2 y+ 2approach the steady state wheredy/dtis zero and y=y>=. That constanty=y>is a particular solutionyp. Whichyn = Ce 2 tcombines with this steady stateypto start fromy(0) = 4? This question choseyp+ynto bey>+ transient (decaying to zero).

Solution y>= 2 =ypat the steady state whendydt = 0. Thenyn= 2e 2 tgives y=yn+yp= 2 + 2e 2 t= 4att= 0. 2 For the same equationdy/dt= 2 y+ 2, choose the null solutionynthat starts from y(0) = 4. Find the particular solutionypthat starts fromy(0) = 0. This splitting choosesynandypaseaty(0) +integral ofea(t 2 T)qin equation (4). Solution For the same equation as 11.4,yn= 4e 2 thas the correcty(0) = 4. Now ypmust be 222 e 2 tto start atyp(0) = 0. Of courseyn+ypis still2 + 2e 2 t. 3 The equationdy/dt= 22 y+8also has two natural splittingsyS+yT=yN+yP: 1. Steady (yS=y>) + Transient (yT³ 0 ). What are those parts ify(0) = 6? 2. (yN 2 = 22 yNfromyN(0) = 6) + (yP 2 = 22 yP+ 8starting fromyP(0) = 0). Solution 1. yS= 4(whendydt= 0: steady state) andyT= 2e 22 t. 2. yN= 6e 22 tandyP= 4 24 e 22 tstarts atyP(0) = 0. AgainyS+yT=yN+yP: two splittings ofy. 4 All null solutions tou 22 v= 0have the form(u, v) = (c, ). One particular solution tou 22 v= 3has the form(u, v) = (7, ). Every solution tou 22 v= 3has the form(7, ) +c(1, ). But also every solution has the form(3, ) +C(1, )forC=c+ 4. Solution All null solutions tou 22 v= 0have the form(u, v) =(c, 12 c). One particular solution tou 22 v= 3has the form(u, v) =(7,2). Every solution tou 22 v= 3has the form(7,2) +c(1, 12 ).

But also every solution has the form(3,0) +C(1, 12 ). HereC=c+ 4.

8 Chapter 1. First Order Equations

5 The equationdy/dt= 5withy(0) = 2is solved byy=. A natural split- tingyn(t) = andyp(t) = comes fromyn=eaty(0)andyp=

ea(t 2 T) 5 dT. This small example hasa= 0(soayis absent) andc= 0(the source isq= 5e 0 t). Whena=cwe have “resonance.” A factortwill appear in the solutiony. Solution dy/dt= 5withy(0) = 2is solved byy= 2+5t. A natural splittingyn(t) = 2 andyp(t) = 5tcomes fromyn(0) =y(0)andyp=

ea(t 2 s) 5 ds= 5t(sincea= 0). Starting with Problem 6, choose the very particular yp that starts from yp(0) = 0.

6 For these equations starting aty(0) = 1, findyn(t)andyp(t)andy(t) =yn+yp. (a)y 229 y= 90 (b)y 2 + 9y= 90 Solution (a) Since the forcing function isawe use equation 6: yn(t) =e 9 t

yp(t) = 909 (e 9 t 2 1) = 10(e 9 t 2 1)

y(t) =yn(t) +yp(t) =e 9 t+ 10(e 9 t 2 1) = 11e 9 t 210. .(b) We again use equation 6, noting thata= 29. The steady state will bey>= 10. yn(t) =e 29 t

yp(t) = 2909 (e 29 t 2 1) y(t) =yn(t) +yp(t) =e 29 t 2 10(e 29 t 2 1) = 10 29 e 29 t.

7 Find a linear differential equation that producesyn(t) =e 2 tandyp(t) = 5(e 8 t 2 1). Solution yn=e 2 tneedsa= 2. Thenyp= 5(e 8 t 2 1)starts fromyp(0) = 0, telling us thaty(0) =yn(0) = 1. Thisypis a response to the forcing term(e 8 t+ 1). So the equation fory=e 2 t+ 5e 8 t 25 must bedydt= 2y+ (e 8 t+ 1). Substitutey:

2 e 2 t+ 40e 8 t= 2e 2 t+ 10e 8 t 2 10 + (e 8 t+ 1). Comparing the two sides,C= 30andD= 10. Harder than expected. 8 Find a resonant equation(a=c)that producesyn(t) =e 2 tandyp(t) = 3te 2 t. Solution Clearlya=c= 2. The equation must bedy/dt= 2y+Be 2 t. Substituting y=e 2 t+ 3te 2 tgives 2 e 2 t+ 3e 2 t+ 6te 2 t= 2(e 2 t+ 3te 2 t) +Be 2 tand thenB= 3. 9 y 2 = 3y+e 3 thasyn=e 3 ty(0). Find the resonantypwithyp(0) = 0. Solution The resonantyphas the formCte 3 tstarting fromyp(0) = 0. Substitute in the equation: dy dt

= 3y+e 3 t isCe 3 t+ 3Cte 3 t= 3Cte 3 t+e 3 t and thenC= 1.

Problems 10–13 are about y 22 ay= constant source q**.**

10 Solve these linear equations in the formy=yn+ypwithyn=y(0)eat.

(a)y 224 y= 28 (b)y 2 + 4y= 8 Which one has a steady state? Solution (a)y 224 y= 28 has a= 4 and yp= 2. But 2 is not a steady state at t=>because the solutionyn=y(0)e 4 tis exploding. (b)y 2 + 4y= 8has a= 24 and again yp= 2. This 2 is a steady state because a < 0 andyn³ 0.

1. Four Particular Solutions 9

11 Find a formula fory(t)withy(0) = 1and draw its graph. What isy>?

(a)y 2 + 2y= 6 (b)y 2 + 2y= 26 Solution (a)y 2 + 2y= 6has a= 22 and y>= 3and y=y(0)e 22 t+ 3. (b)y 2 + 2y= 26 has a= 22 andy>= 23 andy=y(0)e 22 t 23.

12 Write the equations in Problem 11 asY 2 = 22 YwithY=y 2 y>. What isY(0)?

Solution WithY =y 2 y>andY(0) = y(0) 2 y>, the equations in 1.4 are Y 2 = 22 Y. (The solutions areY(t) =Y(0)e 22 twhich isy(t) 2 y>= (y(0) 2 y>)e 22 t ory(t) =y(0)e 22 t+y>(1 2 e 22 t).

13 If a drip feedsq= 0. 3 grams per minute into your arm, and your body eliminates the drug at the rate 6 ygrams per minute, what is the steady state concentrationy>? Then in = out andy>is constant. Write a differential equation forY=y 2 y>. Solution The steady state hasyin=youtor 0 .3 = 6y> ory>= 0. 05. The equa- tion forY =y 2 y> isY 2 =aY = 26 Y. The solution isY(t) =Y(0)e 26 t or y(t) =y>+ (y(0) 2 y>)e 26 t. Problems 14–18 are about y 22 ay= step function H(t 2 T) :

14 Why isy>the same fory 2 +y=H(t 2 2)andy 2 +y=H(t 2 10)?

Solution Noticea= 21. The steady states are the same because the step functions H(t 2 2)andH(t 2 10)are the same after timet= 10.

15 Draw the ramp function that solvesy 2 =H(t 2 T)withy(0) = 2.

Solution The solution is a ramp withy(t) = y(0) = 2up to timeTand then y(t) = 2 +t 2 Tbeyond timeT.

16 Findyn(t)andyp(t)as in equation (10), with step function inputs starting atT= 4.

(a)y 225 y= 3H(t 2 4) (b)y 2 +y= 7H(t 2 4) ( What is y>? ) Solution (a)yp(t) = 35 (e5(t 2 4) 2 1)fortg 4 with no steady state.

(b)yp(t) = 271 (e 2 (t 2 4) 2 1)fortg 4 witha= 21 and y>= 7.

17 Suppose the step function turns on atT = 4and off atT = 6. Thenq(t) = H(t 2 4) 2 H(t 2 6). Starting fromy(0) = 0, solvey 2 + 2y =q(t). What is y>? Solution The solution has 3 parts. Firsty(t) =y(0) = 0up tot= 4. ThenH(t 2 4) turns on andy(t) = 212 (e 2 2(t 2 4) 2 1). This reachesy(6) = 212 (e 242 1)at time t= 6. Aftert= 6, the source is turned off and the solution decays to zero:y(t) = y(6)e 2 2(t 2 6). Method 2: We use the same steps as in equations (8) - (10), noting thaty(0) = 0. (e 2 ty) 2 =e 2 tH(t 2 4) 2 e 2 tH(t 2 6)

e 2 ty(t) 2 e 2 ty(0) =

t

4

e 2 xdx 2

t

6

e 2 xdx

e 2 ty(t) = 212 (e 2 · 42 e 2 t)H(t 2 4) + 12 (e 2 · 62 e

2 t )H(t 2 6)

y(t) = 212 (e 822 t 2 1)H(t 2 4) + 12 (e 1222 t 2 1)H(t 2 6) .

10 Chapter 1. First Order Equations

Fort³>, we have:

y>=

(e 822 ·> 2 1)H(t 2 4) +

(e 1222 ·> 2 1)H(t 2 6) = 0.

18 Supposey 2 =H(t 2 1) +H(t 2 2) +H(t 2 3), starting aty(0) = 0. Findy(t).

Solution We integrate both sides of the equation. t

0

y 2 (t)dt=

t

0

(H(t 2 1) +H(t 2 2) +H(t 2 3))dt

y(t) 2 y(0) =R(t 2 1) +R(t 2 2) +R(t 2 3) y(t) =R(t 2 1) +R(t 2 2) +R(t 2 3)

.R(t)is the unit ramp function= max(0, t).

Problems 19–25 are about delta functions and solutions to y 22 ay=q ·(t 2 T).

19 For allt > 0 find these integralsa(t), b(t), c(t)of point sources and graphb(t):

(a)

t

0

·(T 2 2)dT (b)

t

0

(·(T 2 2) 2 ·(T 2 3))dT (c)

t

0

·(T 2 2)·(T 2 3)dT

Solution Fort < 2 , the spike in·(t 2 2)does not appear in the integral from 0 tot:

(a)

t

0

·(T 2 2)dT=

0 ift < 2 1 iftg 2

The integral (b) equals 1 for 2 ft < 3. This is the differenceH(t 2 2) 2 H(t 2 3). The integral (c) is zero because·(T 2 2)·(T 2 3)is everywhere zero.

20 Why are these answers reasonable? (They are all correct.)

(a)

2>

et·(t)dt= 1 (b)

2>

(·(t)) 2 dt=> (c)

2>

eT·(t 2 T)dT=et

Solution (a) The differenceet·(t) 2 ·(t)is everywhere zero (notice it is zero att= 0). Soet·(t)and·(t)have the same integral (from2>to>that integral is 1 ). This reasoning can be made more precise. (b) This is the difference between the step functionsH(t 2 2)andH(t 2 3). So it equals 1 for 2 ftf 3 and otherwise zero. (c) As in part (a), the difference betweeneT·(t 2 T)andet·(t 2 T)is zero att=T (and also zero at every other> t). So

2>

eT·(t 2 T)dT=et

2>

·(t 2 T)dT=et.

21 The solution to y 2 = 2y+·(t 2 3) jumps up by 1 at t= 3. Before and aftert= 3, the delta function is zero andygrows likee 2 t. Draw the graph ofy(t)when (a) y(0) = 0and (b)y(0) = 1. Write formulas fory(t)before and aftert= 3.

1. Four Particular Solutions 11

Solution (a)y(0) = 0givesy(t) = 0untilt= 3. Theny(3) = 1from the jump. After the jump we are solvingy 2 = 2yandygrows exponentially fromy(3) = 1. So y(t) =e2(t 2 3). (b)y(0) = 1givesy(t) =e 2 tuntilt= 3. The jump producesy(3) =e 6 + 1. Then exponential growth givesy(t) =e2(t 2 3)(e 6 + 1) =e 2 t+e2(t 2 3). One part grows fromt= 0, one part grows fromt= 3as before.

22 Solve these differential equations starting aty(0) = 2:

(a)y 22 y=·(t 2 2) (b)y 2 +y=·(t 2 2). ( What is y>?) Solution (a)y 22 y=·(t 2 2)starts withy(t) =y(0)et= 2etup to the jump at t= 2. The jump brings another term intoy(t) = 2 et+et 22 fortg 2. Note the jump ofet 22 = 1att= 2. (b)y 2 +y=·(t 2 2)starts withy(t) =y(0)e 2 t= 2e 2 tup tot= 2. The jump of 1 att= 2starts another exponentiale 2 (t 2 2)(decaying becausea= 21 ). Then y(t) = 2 e 2 t+e 2 (t 2 2).

23 Solvedy/dt=H(t 2 1) +·(t 2 1)starting fromy(0) = 0: jump and ramp.

Solution Nothing happens andy(t) = 0untilt= 1. ThenH(t 2 1)starts a ramp iny(t)and there is a jump from·(t 2 1). Soy(t) = ramp+constant = max(0, t 2 1) + 1.

24 (My small favorite) What is the steady statey>fory 2 = 2 y+·(t 2 1) +H(t 2 3)?

Solution dy/dt= 0at the steady stateyss. Then 2 y+·(t 2 1) +H(t 2 3)is 2 y>+ 0 + 1andy>= 1.

25 Whichqandy(0)iny 223 y=q(t)produce the step solutiony(t) =H(t 2 1)?

Solution We simply substitute the particular solutiony(t) =H(t 2 1)into the original differential equation withy(0) = 0): ·(t 2 1) 23 H(t 2 1) =q(t) Notice how·(t 2 1)inq(t)produces the jumpH(t 2 1)iny, and then 23 H(t 2 1)in q(t)cancels the 23 yand keepsdy/dt= 0aftert= 1. Problems 26–31 are about exponential sources q(t) =Qect and resonance.

26 Solve these equationsy 22 ay=Qectas in (19), starting from y(0) = 2 : (a)y 22 y= 8e 3 t (b)y 2 +y= 8e 23 t ( What is y>?) Solution (a)a= 1 , c= 3 and y(0) = 2 (b)a= 21 , c= 23 and y(0) = 2

y(t) =y(0)eat+ 8

ect 2 eat c 2 a

y(t) =y(0)eat+ 8

e 23 t 2 e 2 t c 2 a

y(t) = 2et+ 8

e 3 t 2 et 321

y(t) = 2e 2 t+ 8

e 23 t 2 e 2 t 232 ( 2 1) y(t) = 2et+ 4(e 3 t 2 et) y(t) = 2e 2 t 2 4(e 23 t 2 e 2 t) y(t) = 4e 3 t 22 et y(t) = 24 e 23 t+ 2e 2 t ygoes to>as t³> y goes to 0 ast³> .

12 Chapter 1. First Order Equations

27 Whenc= 2. 01 is very close toa= 2, solvey 222 y=ectstarting fromy(0) = 1. By hand or by computer, draw the graph ofy(t): near resonance. Solution We substitute the valuesa= 2, c= 2. 01 andy(0) = 1into equation (18):

y(t) =y(0)eat+

ect 2 eat c 2 a

y(t) = 2eat+

e 2 t 2 e 2. 01 t 2. 0122 y(t) = 2e 2 t+ 100(e 2 t 2 e 2. 01 t)

y(t) = 101e 2 t 2100 e 2. 01 t The graph of this function shows the “near resonance” whencja.

28 Whenc= 2is exactly equal toa= 2, solvey 222 y=e 2 tstarting fromy(0) = 1. This is resonance as in equation (20). By hand or computer, draw the graph ofy(t). Solution We substitutea= 2, c= 2(resonance) andy(0) = 1into equation (19): y(t) =y(0)eat+teat=e 2 t+te 2 t.

29 Solvey 2 + 4y= 8e 24 t+ 20starting fromy(0) = 0. What isy>?

Solution We havea= 24 , c= 24 andy(0) = 0. Equation (19) with resonance leads to 8 te 24 t. The constant source 20 leads to 20(e 24 t 2 1). By linearity y(t) = 8te 24 t+ 20(e 24 t 2 1). The steady state isy>= 220.

30 The solution toy 22 ay=ectdidn’t come from the main formula (4), but it could. Integratee 2 asecsin (4) to reach the very particular solution(ect 2 eat)/(c 2 a). Solution y(t) =eaty(0) +eat

t

0

e 2 aTq(T)dT

=eaty(0) +eat

t

0

e 2 aTecTdT

=eaty(0) +eat

t

0

e(c 2 a)TdT

=eaty(0) +eat

e(c 2 a)t 2 e 0 c 2 a

=eaty(0) +

ect 2 eat c 2 a

=yn+yvp .

1. Four Particular Solutions 13

31 The easiest possible equation y 2 = 1 has resonance! The solutiony=tshows the factort. What number is the growth rateaand also the exponentcin the source? Solution The growth rate iny 2 = 1or dy/dt=e 0 tisa= 0. The source isect withc= 0. Resonance a=c. The resonant solutiony(t) =teat isy=t, certainly correct for the equationdy/dt= 1.

32 Suppose you know two solutionsy 1 andy 2 to the equationy 22 a(t)y=q(t).

(a) Find a null solution toy 22 a(t)y= 0. (b) Find all null solutionsyn. Find all particular solutionsyp. Solution (a)y=y 12 y 2 will be a null solution by linearity. (b)y=C(y 12 y 2 )will give all null solutions. Theny=C(y 12 y 2 )+y 1 will give all particular solutions. (Alsoy=c(y 12 y 2 ) +y 2 will also give all particular solutions.)

33 Turn back to the first page of this Section 1. Without looking, can you write down a solution toy 22 ay=q(t)for all four source functionsq,H(t),·(t),ect? Solution Equations (5), (7), (14), (19).

34 Three of those sources in Problem 33 are actually the same, if you choose the right values forqandcandy(0). What are those values? Solution The sourcesq= 1andq=H(t)andq=e 0 tare all the same fortg 0.

35 What differential equationsy 2 =ay+q(t)would be solved byy 1 (t)andy 2 (t)? Jumps, ramps, corners—maybe harder than expected (math.mit.edu/dela/Pset1).

a

0 1 2

y 1 (t) y 2 (t) et− 1 e 2 −t− 1

0 1 2

Solution (a)

dy 1 dt

= 1 2 ·(t 2 1) 2 ·(t 2 2)witha= 0.

(b)

dy 2 dt

=y 2 + 1up tot= 1. Add in 22 e ·(t 2 1)to drop the slope frometo 2 eat t= 1. Aftert= 1we needdy 2 /dt= 2 y 221 to keepy 2 =e 22 t 21.

14 Chapter 1. First Order Equations

Problem Set 1, page 37

Problems 1-6 are about the sinusoidal identity (9). It is stated again in Problem 1.

1 These steps lead again to the sinusoidal identity. This approach doesn’t start with the usual formula cos(Ët 2 Ç) = cosËt cosÇ+ sinËt sinÇfrom trigonometry. The identity says :

If A+iB=R eiÇ then A cos Ët+B sin Ët=R cos (Ët 2 Ç). Here are the four steps to find that real part ofRei(Ët 2 Ç). Explain Step 3 whereR e 2 iÇ equalsA 2 iB: Rcos (Ët 2 Ç)=Re

R ei(Ët 2 Ç)

=Re

eiËt(R e 2 iÇ)

= ( what is R e 2 iÇ?) =Re[(cosËt+isinËt) (A 2 iB)] =A cos Ët+B sin Ët. Solution The key point is that ifA+iB=ReiÇthenA 2 iB=Re 2 iÇ(the complex conjugate). 2 To expresssin 5t+ cos 5tasRcos (Ët 2 Ç), what areRandÇ? Solution The sinusoidal identity hasA= 1, B= 1, and Ë= 5. Therefore: R 2 =A 2 +B 2 = 2³R=

2 and tanÇ=

.Answer

2 cos

5 t 2

3 To express6 cos 2t+ 8 sin 2tasRcos (2t 2 Ç), what areRand tanÇandÇ? Solution Use the Sinusoidal Identity withA= 6, B= 8and Ë= 2. R 2 =A 2 +B 2 = 6 2 + 8 2 = 100and R= 10

tanÇ=BA= 86 = 43 andÇ is in the positive quadrant 0 toà 2

not à to 32 Ã

6 cos(2t) + 8 sin(2t) = 10 cos

2 t 2 arctan

4 IntegratecosËtto find(sinËt)/Ëin this complex way. (i) dyreal/dt= cosËtis the real part ofdycomplex/dt=eiËt.

(ii) Take the real part of the complex solution.

Solution (i) The complex equationy 2 =eiËtleads toy=

eiËt iË

(ii) Take the real part of that solution (since the real part of the right side iscosËt).

Re

eiËt iË

=Re

cosËt iË

sinËt Ë

sinËt Ë

5 The sinusoidal identity forA= 0andB= 21 says that 2 sinËt=Rcos(Ët 2 Ç). FindRandÇ. Solution R 2 =A 2 +B 2 = 0 2 + 1 2 = 1³R= 1

tanÇ= 10 =>³Ç=à 2 or 32 Ã: Here it is 32 Ã, sinceA+iB= 2 i Therefore we have SOLUTION: 2 sinËt= cos(Ët 232 Ã) CHECK: t= 0 gives0 = 0, Ët=à 2 gives 2 1 = 21.

1. Real and Complex Sinusoids 15

6 Why is the sinusoidal identity useless for the sourceq(t) = cost+ sin 2t? Solution The sinusoidal identity needs the sameËin all terms. But the first term has Ë= 1while the second term hasË= 2. 7 Write2+3iasreiÇ, so that2+3 1 i= 1 re 2 iÇ. Then writey=eiËt/(2+3i)in polar form. Then find the real and imaginary parts ofy. And also find those real and imaginary parts directly from(2 23 i)eiËt/(2 23 i)(2 + 3i). Solution r=: 22 + 3 2 =: 13 andÇ= arctan(3/2)

2 + 3i=

13 eiarctan(3/2) y=eiËt/(2 + 3i) =

13 eiarctan(3/2)+iËt Writing this in cartesian (rectangular) form gives real part=

13 cos(arctan(3/2) +Ët) = 2 cos(Ët) 2 3 sin(Ët)

imag part=

13 sin(arctan(3/2) +Ët) = 3 cos(Ët) + 2 sin(Ët) .We can also find the real and imaginary parts from: (2 23 i)eiËt (2 23 i)(2 + 3i)

223 i 13

eiËt=

223 i 13

(cos(Ët) +isin(Ët)).

8 Write these functionsAcosËt+BsinËtin the formRcos(Ët 2 Ç): Right triangle with sidesA,B,Rand angleÇ. (1)cos 3t 2 sin 3t (2)

3 cosÃt 2 sinÃt (3)3 cos(t 2 Ç) + 4 sin(t 2 Ç) Solution (1)cos 3t 2 sin 3t=

2 cos(3t 274 Ã) =

2 cos(3t+Ã 4 ).

Checkt= 0 : 1 =

2 cos( 274 Ã) =

2 cos(Ã 4 ).

(2)

3 cosÃt 2 sinÃt= 2 cos(Ãt+Ã 6 ).

Check:(

1. 2 + ( 2 1) 2 = 2 2 Att= 0 :

3 = 2 cos 30ç. (3)3 cos(t 2 Ç) + 4 sin(t 2 Ç) = 5 cos(t 2 Ç 2 tan 2143 ).

Problems 9-15 solve real equations using the real formula (3) for M and N**.**

9 Solvedy/dt= 2y+ 3 cost+ 4 sintafter recognizingaandË. Null solutionsCe 2 t. Solution dydt= 2y+ 3 cost+ 4 sint= 2y+ 5 cos(t 2 Ç)with tanÇ= 43. Method 1: Look fory=Mcost+Nsint. Method 2: SolvedYdt = 2Y+ 5ei(t 2 Ç)and theny=real part ofY.

Y=i 252 ei(t 2 Ç)= 55 ( 2 i 2 2)ei(t 2 Ç) and y= 2 2 cos(t 2 Ç) + sin(t 2 Ç).

10 Find a particular solution tody/dt= 2 y 2 cos 2t.

Solution Substitutey=Mcost+Nsintinto the equation to findMandN 2 Msint+Ncost= 2 Mcost 2 Nsint 2 cos 2t Match coefficients ofcostandsintseparately to findMandN.

N= 2 M 21 and 2 M= 2 N giveM=N= 2

Note: This is called the “method of undetermined coefficients” in Section 2.

16 Chapter 1. First Order Equations

11 What equationy 22 ay=AcosËt+BsinËtis solved byy= 3 cos2t+ 4 sin 2t?

Solution ClearlyË= 2. Substituteyinto the equation: 2 6 sin 2t+ 8 cos2t 23 acos 2t 24 asin 2t=Acos 2t+Bsin 2t. Match separately the coefficients ofcos 2tandsin 2t: A= 8 23 aand B= 2624 a

.

12 The particular solution toy 2 =y+ costin Section 4 isyp=et

e 2 scoss ds. Look this up or integrate by parts, froms= 0tot. Compare thisypto formula (3). Solution That integral goes from 0 tot, and it leads toyp= 12 (sint 2 cost+et) If we use formula (3) witha= 1, Ë= 1, A= 1, B= 0we get

M= 2

aA+ËB Ë 2 +a 2

ËA 2 aB Ë 2 +a 2

This solutiony=Mcost+Nsint=

2 cost+ sint 2

is a different particular solution (not starting fromy(0) = 0). The difference is a null solution 12 et.

13 Find a solutiony=McosËt+NsinËttoy 224 y= cos 3t+ sin 3t.

Solution Formula (3) witha= 4, Ë= 3, A=B= 1gives

M= 2

14 Find the solution toy 22 ay=AcosËt+BsinËt starting from y(0) = 0.

Solution One particular solutionMcosËt+NsinËtcomes from formula (3). But this starts fromyp(0) =M. So subtract off the null solutionyn=M eatto get the very particular solutionyvp=yp 2 ynthat starts fromyvp(0) = 0.

15 Ifa= 0show thatMandNin equation (3) still solvey 2 =AcosËt+BsinËt.

Solution Formula (3) still applies witha= 0and it gives

M= 2

y= 2

cosËt+

sinËt.

This is the correct integral ofAcosËt+BsinËtin the differential equation.

1. Real and Complex Sinusoids 17

Problems 16-20 solve the complex equation y 22 ay=Rei(Ët 2 Ç).

16 Write down complex solutionsyp=Y eiËtto these three equations:

(a)y 223 y= 5e 2 it (b)y 2 =Rei(Ët 2 Ç) (c)y 2 = 2y 2 eit Solution (a)y 223 y= 5e 2 ithasiË Y eiËt 23 Y eiËt= 5e 2 it. SoË= 2andY= 2 i 523.

(b)y 2 =Rei(Ët 2 Ç) hasiËY eiËt=Rei(Ët 2 Ç). SoY=iËRe 2 iÇand the solution is y=Y eiËt=iËRei(Ët 2 Ç).

(c)y 2 = 2y 2 eit hasË= 1andiY eit= 2Y eit 2 e 2 it. ThenY=i 2212 = 221 i=2+ 5 iandy=Y eit.

17 Find complex solutionszp=ZeiËtto these complex equations: (a)z 2 + 4z=e 8 it (b)z 2 + 4iz=e 8 it (c)z 2 + 4iz=e 8 t Solution (a)z 2 + 4z=e 8 it has z=Ze 8 itwith 8 iZ+ 4Z= 1and Z=4+8 1 i= 428 i 16+64=

1 20 (1 22 i). (b)z 2 + 4iz=e 8 itis like part (a) but 4 changes to 4 i. ThenZ= 4 i+8 1 i= 121 i= 212 i.

(c)z 2 + 4iz=e 8 t has z=Ze 8 t. Then 8 Ze 8 t+ 4iZe 8 tgivesZ=8+4 1 i= 8822 +4 4 i 2.

18 Start with the real equationy 22 ay=Rcos (Ët 2 Ç). Change to the complex equation z 22 az=Rei(Ët 2 Ç). Solve forz(t). Then take its real partyp=Rez.

Solution Putz=Zei(Ët 2 Ç)in the complex equation to findZ:

iËZ 2 aZ=RgivesZ=

2 a+iË

R( 2 a 2 iË) a 2 +Ë 2

The real part ofz=Z(cos(Ët 2 Ç) +isin(Ët 2 Ç)) is R a 2 +Ë 2 ( 2 acos(Ët 2 Ç) +Ësin(Ët 2 Ç)).

19 What is the initial valueyp(0)of the particular solutionyp from Problem 18? If the desired initial value isy(0), how much of the null solutionyn = Ceat would you add toyp?

Solution That solution to 18 starts fromyp(0) =a 2 R+Ë 2 ( 2 acos( 2 Ç) +Ësin( 2 Ç))at t= 0. So subtract that number timeseatto get the very particular solution that starts fromyvp(0) = 0.

20 Find the real solution toy 222 y= cosËtstarting fromy(0) = 0, in three steps : Solve the complex equationz 222 z = eiËt, take yp = Rez, and add the null solutionyn=Ce 2 twith the rightC. Solution Step 1 222 Z=eiËtis solved byz=ZeiËtwithiËZ 22 Z= 1and Z= 2 2+ 1 iË= 2 4+ 22 ËiË 2.

Step 2. The real part ofZeiËtisyp=4+ 1 Ë 2 ( 2 2 cosËt+ËsinËt).

Step 3. yp(0) = 4+ 2 Ë 22 soyvp=yp+4+ 2 Ë 2 e 2 tincludes the rightyn=Ce 2 tfor yvp(0) = 0.

18 Chapter 1. First Order Equations

Problems 21-27 solve real equations by making them complex a note on ³**.**

Example 4 wasy 22 y= cost 2 sint, with growth ratea= 1and frequencyË= 1. The magnitude ofiË 2 ais

2 and the polar angle hastan³= 2 Ë/a= 21. Notice : Both ³= 3Ã/ 4 and ³= 2 Ã/ 4 have that tangent! How to choose the correct angle³? The complex numberiË 2 a=i 21 is in the second quadrant. Its angle is³= 3Ã/ 4. We had to look at the actual number and not just the tangent of its angle.

21 Findrand³to write eachiË 2 aasrei³. Then write 1 /rei³asGe 2 i³. (a)

3 i+ 1 (b)

3 i 21 (c)i 2

Solution (a)

3 i+ 1is in the first quadrant (positive quarter 0 f»fÃ/ 2 ) of the complex plane. The angle with tangent

3 / 1 is 60 ç=Ã/ 3. The magnitude of

3 i+ isr= 2. Then

3 i+ 1 = 2 eiÃ/ 3. (b)

3 i 21 is in the second quadrantÃ/ 2 f»fÃ. The tangent is 2

3 , the angle is »= 2Ã/ 3 , the number is 2 e 2 Ãi/ 3. (c)i 2

3 is also in the second quadrant (left from zero and up). Now thetangent is 21 /

3 , the angle is»= 150ç= 5Ã/ 6. The magnitude is still 2 , the number is 2 e 5 Ãi/ 6.

22 UseGand³from Problem 21 to solve (a)-(b)-(c). Then take the real partof each equation and the real part of each solution. (a)y 2 +y=ei

: 3 t (b)y 22 y=ei

: 3 t (c)y 22 : 3 y=eit

Solution (a)y 2 +y=ei

: 3 tis solved byy=Y ei

: 3 twheni: 3 Y+Y= 1. Then Y=: 31 i+1= 12 e 2 iÃ/ 3 from Problem 21(a). The real partyreal= 12 cos(

3 t 2 Ã/3) ofY ei

: 3 tsolves the real equationy 2 real+yreal= cos(

3 t).

(b)y 22 y=ei

: 3 tis solved byy=Y ei

: 3 twheni: 3 Y 2 Y= 1. ThenY= 1 2 e

22 Ãi/ 3

from Problem 21(b). the real partyreal= 12 cos(

3 t 22 Ã/ 3 ) solves the real equation yreal 22 yreal= cos(

3 t).

(c)y 22

3 y =eitis solved byy =Y eitwheniY 2

3 Y = 1. ThenY = 1 2 e

25 Ãi/ 6 from Problem 21(c). The real partyreal= 1 2 cos(t 25 Ã/6)ofY e

itsolves

yreal 2

3 yreal= cost.

23 Solvey 22 y= cosËt+ sinËtin three steps : real to complex, solve complex, take real part. This is an important example. Solution Note: I intended to choose Ë= 1. Theny 22 y= cost+ sinthas the simple solutiony= 2 sint. I will apply the 3 steps to this case and then to the harder problem for anyË. (1) FindRandÇin the sinusoidal identity to write cosËt+ sinËtas the real part of Rei(Ët 2 Ç). This is easy for anyË. tanÇ=

soÇ=

cosËt+ sinËt=

2 cos

Ët 2

(2) Solvey 22 y = eiËt byy = Ge 2 i³eiËt. Multiply by Re 2 iÇto solve z 22 z=Rei(Ët 2 Ç).

1. Real and Complex Sinusoids 19

Ë= 1 y 22 y=eithasy=Y eitwithiY 2 Y= 1. ThenY=i 211 =: 12 e 3 Ãi/ 4 = Ge 2 i³.

z=

2 ei(t 2 Ã/4)

: 2 e

3 Ãi/ 4

=eiteÃi/ 2 =ieit. The real part ofzisy= 2 sint.

Any Ë y 22 y=eiËtleads toiËY 2 Y= 1andY=

iË 21

e 2 i³

with tan³=Ë. Thenz(t) =

1 1+Ë 2 e

2 i³

2 ei(Ët 2 Ã/4)

(3) Take the real party(t) =Rez(t). Check thaty 22 y= cosËt+ sinËt.

y(t) =Rez(t) =

: 2 1+Ë 2 cos(Ët 2 ³ 2

à 4 ). Now we needtan³=Ë,cos³= : 1 1+Ë 2 , sin³=:1+ËË 2. Finallyy=

: 2 1+Ë 2 [cos(Ët 2

à 4 ) cos³+ sin(Ët 2

à 4 ) sin³].

24 Solvey 22

3 y= cost+ sintby the same three steps witha=

3 andË= 1.

Solution (1)cost+ sint=

2 cos(t 2 Ã 4 ).

(2)y=Y eit withiY 2

3 Y= 1and Y=i 21 : 3 = 12 e 25 Ãi/ 6 from 1.5(c).

Thenz(t) = (

2 ei(t 2 Ã/4))( 12 e 25 Ãi/ 6 ).

(3) The real part ofz(t)isy(t) =: 12 cos(t 21312 Ã).

25 ( Challenge ) Solvey 22 ay = AcosËt+BsinËt in two ways. First, find RandÇon the right side andGand³on the left. Show that the final real solution RGcos (Ët 2 Ç 2 ³)agrees withMcosËt+NsinËtin equation (3).

Solution The first way hasR=

A 2 +B 2 andtanÇ=B/Afrom the sinusoidal identity. On the left side 1 /(iË 2 a) =Ge 2 i³from equation (8) withG= 1/

Ë 2 +a 2 andtan³= 2 Ë/a. Combining, the real solution isy=RGcos(Ët 2 Ç 2 ³). This agrees withy=McosËt+NsinËt(equation (3) givesMandN).

26 We don’t have resonance fory 22 ay=ReiËtwhenaandË= 0are real. Why not? (Resonance appears whenyn=Ceatandyp=Y ectshare the exponenta=c.) Solution Resonance requires the exponentsaandiËto be equal. For realathis only happens ifa=Ë= 0.

27 If you took the imaginary party=Imzof the complex solution toz 22 az=Rei(Ët 2 Ç), what equation wouldy(t)solve? Answer first withÇ= 0.

Solution Assumingais real, the imaginary part ofz 22 az=Rei(Ët 2 Ç)is the equation y 22 ay=Rsin(Ët 2 Ç). WithÇ= 0this isy 22 ay=RsinËt.

Problems 28-31 solve first order circuit equations : not RLC but RL and RC.

VcosËt L R

currentI(t)

• • • (-

VcosËt R C

q(t) = integral ofI(t)

20 Chapter 1. First Order Equations

28 SolveL dI/dt+RI(t) =V cosËtfor the currentI(t) =In+Ipin the RL loop.

Solution Divide the equation byLto producedI/dt 2 aI=XcosËtwitha= 2 R/L andX=V /L. In this standard form, equation (3) gives the real solution:

I=McosËt+NsinËtwithM= 2

aX Ë 2 +a 2

and N=

Ë 2 +a 2

29 WithL= 0andË = 0, that equation is Ohm’s LawV =IRfor direct current. The complex impedance Z=R+iËLreplacesRwhenL= 0andI(t) =IeiËt.

L dI/dt+RI(t) = (iËL+R)IeiËt=V eiËt gives Z I=V.

What is the magnitude|Z|=|R+iËL|? What is the phase angle inZ=|Z|ei»? Is the current|I|larger or smaller because ofL? Solution |Z|=

R 2 +Ë 2 L 2 and tan»=ËLR. Since|Z|increases withL, the current|I|must decrease.

30 SolveR

dq dt

q(t) =V cosËtfor the chargeq(t) =qn+qpin the RC loop.

Solution Dividing byRproducesdqdt 2 aq=XcosËt witha= 2 RC 1 andX=VR. As in Problem 28, equation (3) givesMandNfromË and thisa.

31 Why is the complex impedance nowZ = R+iËC 1? Find its magnitude|Z|.

Note that mathematics prefers i=

21 , we are not conceding yet to j=

Solution The physicalRCequation for the currentI = dqdt isRI+C 1

Idt= VcosËt=Re(V eiËt). The solutionIhas the same frequency factorXeiËt, and the integral has the factor eiËt/iË. Substitute into the equation and match coefficients ofeiËt: RX+iËC 1 X=VisZX=Vwith impedanceZ=R+iËC 1.

Problem Set 1, page 50

1 Solve the equationdy/dt=y+ 1up to timet, starting fromy(0) = 4. Solution We use the formulay(t) =y(0)eat+sa(eat 2 1)witha= 1ands= 1and y(0) = 4: y(t) = 4et+et 2 1 = 5et 21

2 You have$1000to invest at ratea= 1 = 100 %. Compare after one year the result of depositingy(0) = 1000immediately with no source (s= 0), or choosingy(0) = 0and s= 1000/year to deposit continually during the year. In both casesdy/dt=y+q. Solution We substitute the values for the different scenarios into the solution formula: y(t) = 1000et = 1000eat one year y(t) = 1000et 2 1000 = 1000(e 2 1)at one year

You get more for depositing immediately rather than during the year.

1. Models of Growth and Decay 21

3 Ifdy/dt=y 21 , when does your original deposity(0) = 12 drop to zero?

Solution Again we use the equationy(t) =y(0)eat+as(eat 2 1)witha= 1ands= 21. We sety(t) = 0and find the timet: y(t) =y(0)et 2 et+ 1 =et(y(0) 2 1) + 1 = 0

et=

12 y(0)

= 2andt= ln 2.

Notice! Ify(0)> 1 , the balance never drops to zero. Interest exceeds spending.

4 Solve

dy dt

=y+t 2 fromy(0) = 1with increasing source termt 2.

Solution Solution formula (12) witha= 1andy(0) = 1gives

y(t) =et+

t

0

et 2 ss 2 ds=et 2 t(t+ 2) + 2et 2 2 = 3et 2 t(t+ 2) 22

Check:

dy dt

= 3et+ 2t 22 equalsy+t 2.

5 Solve

dy dt

=y+et(resonancea=c!) fromy(0)=1with exponential sourceet.

Solution The solution formula witha= 1and sourceet(resonance!) gives :

y(t) =et+

t

0

et 2 sesds=et+

t

0

etds=et(1 +t)

Check by the product rule :

dy dt

=et(1 +t) +et=y+et.

6 Solve

dy dt

=y 2 t 2 from an initial deposity(0) = 1. The spendingq(t) = 2 t 2 is growing. When (if ever) doesy(t)drop to zero? Solution

y(t) =et 2

t

0

et 2 ss 2 ds=et+t(t+ 2) 22 et+ 2 = 2 et+t(t+ 2). This definitely

drops to zero (I regret there is no nice formula for that timet).

Check:

dy dt

= 2 et+ 2t+ 2 =y 2 t 2.

7 Solve

dy dt

=y 2 etfrom an initial deposity(0) = 1. This spending term 2 etgrows at the sameetrate as the initial deposit (resonance). When (if ever) doesydrop to zero?

Solution y(t) =et 2

t

0

et 2 sesds=et 2

t

0

etds=et(1 2 t)( this is zero at t= 1)

Check by the product rule :dydt=et(1 2 t) 2 et=y 2 et.

22 Chapter 1. First Order Equations

8 Solve

dy dt

=y 2 e 2 tfromy(0) = 1. At what timeTisy(T) = 0?

Solution y(t) =et 2

t

0

et 2 se 2 sds=et 2

t

0

et+sds=et+et(1 2 et) = 2et 2 e 2 t

This solution is zero when 2 et=e 2 tand2 =etandt= ln 2. Check thaty= 2et 2 e 2 tsolves the equation : dydt= 2et 22 e 2 t=y 2 e 2 t. 9 Which solution (yorY) is eventually larger ify(0) = 0andY(0) = 0? dy dt

=y+ 2t or

dY dt

= 2Y+t.

Solution dy dt

=y+ 2t

dY dt

= 2Y+t

y(t) =

t

0

et 2 s· 2 sds Y(t) =

t

0

e 2 t 22 s·sds

y(t) = 2( 2 t+et 2 1) Y(t) =

e 2 t 21 2 In the long runY(t)is larger thany(t), since the exponent 2 tis larger thant.

10 Compare the linear equationy 2 =yto the separable equationy 2 =y 2 starting from y(0) = 1. Which solutiony(t)must grow faster? It grows so fast that it blows up to y(T) =>at what timeT? Solution dy dt

=y

dy dt

=y 2 dy y

=dt

dy y 2

=dt

y(t)

y(0)

du u

t

0

dt

y(t)

y(0)

du u 2

t

0

dt

ln(y(t)) 2 ln(y(0)) =t 2

y(t)

y(0)

=t

y(t) y(0)

=et y(t) =

1 y(0) 2 t

12 t y(t) =y(0)et=et The second solution grows much faster, and reaches a vertical asymptote atT= 1.

11 Y 2 = 2Y has a larger growth factor (becausea = 2)thany 2 = y+q(t). What sourceq(t)would be needed to keepy(t) =Y(t)for all time? Solution dYdt = 2Y+ 1with for exampleY(0) =y(0) = 0

Y(t) =

t

0

e 2 t 22 sds=

e 2 t 21 2

1. Models of Growth and Decay 23

Put this solution intodydt=y+q(t):

e 2 t=

e 2 t 21 2

+q(t) e 2 t+ 1 2

=q(t)

12 Starting fromy(0) =Y(0) = 1, doesy(t)orY(t)eventually become larger?

dy dt

= 2y+et

dY dt

=Y+e 2 t.

Solution dy dt

= 2y+et

y(t) =e 2 t+

t

0

e 2 t 22 sesds=e 2 t+e 2 t 2 et= 2e 2 t 2 et

Solving the second equation: dY dt

=Y+e 2 t

Y(t) =et+

t

0

et 2 se 2 sds=et+e 2 t 2 et=e 2 t is always smaller thany(t).

Questions 13-18 are about the growth factor G(s, t) from time s to time t**.**

13 What is the factorG(s, s)in zero time? FindG(s,>)ifa= 21 and ifa= 1.

Solution The solution doesn’t change in zero time soG(s, s) = 1. (Note that the integral ofa(t)fromt=stot=sis zero. ThenG(s, s) =e 0 = 1. We are talking about change in the null solution, withy 2 =a(t)y. A source term with a delta function does produce instant change.) Ifa= 21 , the solution drops to zero att=>. SoG(s,>) = 0. Ifa= 1, the solution grows infinitely large ast³>. SoG(s,>) =>.

14 Explain the important statement after equation (13): The growth factor G(s, t) is the solution to y 2 =a(t)y+·(t 2 s). The source·(t 2 s)deposits$1at times. Solution When the source term·(t 2 s)deposits $1 at times, that deposit will grow or decay toy(t) =G(s, t)at timet > s. This is consistent with the main solution formula (13).

15 Now explain this meaning ofG(s, t)whent is less than s. We go backwards in time. For t < s , G(s, t) is the value at time t that will grow to equal 1 at time s. Whent= 0,G(s,0)is the “present value” of a promise to pay$1at times. If the inter- est rate isa = 0 = 10 % per year, what is the present value G(s,0) of a million dollar inheritance promised ins= 10years? Solution In factG(t, s) = 1/G(s, t). In the simplest casey 2 =yof exponential growth,G(s, t)is the growth factoret 2 sfromstot. ThenG(t, s)ises 2 t= 1/et 2 s. That numberG(t, s)would be the “present value” at the earlier timetof a promise to pay $1 at the later times. You wouldn’t need to deposit the full $1 because your deposit will grow by the factorG(s, t). All you need to have at the earlier time is 1 /G(s, t), which then grows to 1.

24 Chapter 1. First Order Equations

16 (a) What is the growth factorG(s, t)for the equationy 2 = (sint)y+Qsint?

(b) What is the null solutionyn=G(0, t)toy 2 = (sint)ywheny(0) = 1?

(c) What is the particular solutionyp=

t

0

G(s, t)Qsins ds?

Solution (a) Growth factor:G(s, t) = exp

û

ý

t

s

sinT dT

þ

ø= exp(coss 2 cost).

(b) Null solution:yn=G(0, t)y(0) =e 12 cost.

(c) Particular solution:yp=

t

0

ecoss 2 costQsins ds

=Qe 2 cost[ 2 ecoss]t 0 =Q

e 12 cost 21

.Checkyp(0) =Q(e 02 1) = 0.

17 (a) What is the growth factorG(s, t)for the equationy 2 =y/(t+ 1) + 10?

(b) What is the null solutionyn=G(0, t)toy 2 =y/(t+ 1)withy(0) = 1?

(c) What is the particular solutionyp= 10

t

0

G(s, t)ds?

Solution (a)G(s, t) = exp

þ

ø

t

s

dT T+ 1

ù

û= exp [ln(t+ 1) 2 ln(s+ 1)] =t+ 1 s+ 1

Null solutionyn=G(0, t)y(0) = exp[ln(t+ 1)] =t+ 1sinceln(0 + 1) = 0.

Particular solutionyp= 10

t

0

exp [ln(t+ 1) 2 ln(s+ 1)]ds= 10(t+ 1)

t

0

ds s+ 1

10(t+ 1) ln(t+ 1).

18 Why isG(t, s) = 1/G(s, t)? Why isG(s, t) =G(s, S)G(S, t)?

Solution MultiplyingG(s, t)G(t, s)gives the growth factorG(s, s)from going up to timetand back to times. This factor isG(s, s) = 1. SoG(t, s) = 1/G(s, t). MultiplyingG(s, S)G(S, t)gives the growth factorG(s, t)from going up fromstoS and continuing fromStot. In the exampley 2 =y, this iseS 2 set 2 S=et 2 s=G(s, t).

Problems 19–22 are about the “units” or “dimensions” in differential equations.

19 (recommended) Ifdy/dt=ay+qeiËt, withtin seconds andyin meters, what are the units foraandqandË?

Solution ais in “inverse seconds”—for examplea=. 01 per second. qis in meters. Ëis in “inverse seconds” or 1/seconds—for exampleË= 2Ãradians per second.

1. The Logistic Equation 25

20 The logistic equationdy/dt=ay 2 by 2 often measures the timetin years (andy counts people). What are the units ofaandb? Solution ais in “inverse years”—for examplea= 1percent per year. bis in “inverse people-years” as inb= 1percent per person per year.

21 Newton’s Law ism d 2 y/dt 2 +ky=F. If the massmis in grams,yis in meters, andtis in seconds, what are the units of the stiffnesskand the forceF? Solution kyhas the same units asm d 2 y/dt 2 sokis in grams per (second) 2. Fis in gram-meters per (second) 2 —the units of force.

22 Why is our favorite exampley 2 =y+ 1very unsatisfactory dimensionally? Solve it anyway starting fromy(0) = 21 and fromy(0) = 0. The three terms iny 2 =y+ 1seem to have different units. The ratea= 1is hidden (with its units of 1/time). Also hidden are the units of the source term 1. Solution y(t) =y(0)et+ 11 (et 2 1).This iset 21 ify(0) = 0. The solution stays at steady state ify(0) = 21.

23 The difference equationYn+1=cYn+QnproducesY 1 =cY 0 +Q 0. Show that the next step producesY 2 =c 2 Y 0 +cQ 0 +Q 1. AfterNsteps, the solution formula forYN is like the solution formula fory 2 =ay+q(t). Exponentials ofachange to powers of c, the null solutioneaty(0)becomescNY 0. The particular solution

YN=cN 21 Q 0 +···+QN 21 is like y(t) =

t

0

ea(t 2 s)q(s)ds.

Solution Y 2 =cY 1 +Q 1 =c(cY 0 +Q 0 ) +Q 1 =c 2 Y 0 +cQ 0 +Q 1. The particular solutioncQ 0 +Q 1 agrees with the general formula whenN= 2. The null solutionc 2 Y 0 is Step 2 inY 0 , cY 0 , c 2 Y 0 , c 3 Y 0 ,.. .likeeaty(0).

24 Suppose a fungus doubles in size every day, and it weighs a pound after 10 days. If another fungus was twice as large at the start, would it weigha pound in 5 days? Solution This is an ancient puzzle and the answer is 9 days. Starting twice as large cuts off 1 day.

Problem Set 1, page 61

1 Ify(0) =a/ 2 b, the halfway point on theS-curve is att= 0. Show thatd=band

y(t) =

a d e 2 at+b

a b

e 2 at+ 1

. Sketch the classicS-curve — graph of1(e 2 at+ 1)

fromy2>= 0toy>=

a b

. Mark the inflection point.

Solution d= a y(0)

2 band y(0) =

a 2 b

lead tod=

a a 2 b

2 b= 2b 2 b=b

Thereforey(t) =

a d e 2 at+b

a b e 2 at+b

a b

e 2 at+ 1

26 Chapter 1. First Order Equations

2 If the carrying capacity of the Earth isK=a/b= 14billion people, what will be the population at the inflection point? What isdy/dtat that point? The actual population was 7. 14 billion on January 1 , 2014. Solution The inflection point comes wherey=a/ 2 b= 7million. The slopedy/dtis

dy dt

=ay 2 by 2 =a

a 2 b

2 b

a

2 b

a 2 4 b

.This is b

a

2 b

= 49b.

3 Equation (18) must give the same formula for the solutiony(t)as equation (16). If the right side of (18) is calledR, we can solve that equation fory:

y=R

b a

y

b a

y=R ³ y=

1 +Rab

Simplify that answer by algebra to recover equation (16) fory(t). Solution This problem asks us to complete the partial fractions method which inte- grateddy/(y 2 bay 2 ) =adt. The result in equation (18) can be solved fory(t). The right side of (18) is calledR:

R=eat

y(0) 12 bay(0)

=eata

y(0) a 2 by(0)

=eat

a d

Then the algebra in the problem statement gives

y=

1 +Rba

eatad 1 +eatbd

= multiply by

de 2 at de 2 at

a de 2 at+b

4 Change the logistic equation toy 2 =y+y 2. Now the nonlinear term is positive, and cooperation of y with ypromotes growth. Usez= 1/yto find and solve a linear equation forz, starting fromz(0) =y(0) = 1. Show thaty(T) =>whene 2 T= 1/ 2. Cooperation looks bad, the population will explode att=T.

Solution Puty= 1/zand the chain ruledydt = 2 z 21 dzdtinto the cooperation equation y 2 =y+y 2 :

2

z 2

dz dt

z

z 2

gives

dz dt

= 2 z 21.

The solution starting fromz(0) = 1isz(t) = 2e 2 t 21. This is zero when 2 e 2 T= 1 oreT= 2orT = ln 2.

At that timez(T) = 0meansy(T) = 1/z(T)is infinite: blow-up at timeT= ln 2.

5 The US population grew from 313 , 873 , 685 in 2012 to 316 , 128 , 839 in 2014. If it were following a logisticS-curve, what equations would give youa, b, din the formula (4)? Is the logistic equation reasonable and how to account for immigration? Solution We need a third data point to find all three numbersa, b, d. See Problem (23). There seems to be no simple formula for those numbers. Certainly the logistic equation is too simple for serious science. Immigration would give a negative value for hin the harvesting equationy 2 =ay 2 by 22 h.

1. The Logistic Equation 27

6 The Bernoulli equation y 2 = ay 2 byn has competition termbyn. Introduce z =y 12 nwhich matches the logistic case whenn = 2. Follow equation (4) to show thatz 2 = (n 2 1)( 2 az+b). Writez(t)as in(5)-(6). Then you havey(t). Solution We make the suggested transformation:

z=y 12 n z 2 = (1 2 n)y 2 ny 2 dz dt = (1 2 n)y

2 n(ay 2 byn) = (1 2 n)(ay 12 n 2 b) dz dt = (1 2 n)(az 2 b)

z(t) =e(1 2 n)atz(0) 2

b a

(e(1 2 n)at 2 1) =

de(1 2 n)at+b a d=az(0) 2 b=

a y(0)

2 b

y(t) =

a de(1 2 n)at+b

Problems 7–13 develop better pictures of the logistic and harvesting equations.

7 y 2 =y 2 y 2 is solved byy(t) = 1/(de 2 t+ 1). This is anS-curve wheny(0) = 1/ 2 andd= 1. But show thaty(t)is very different ify(0)> 1 or ify(0)< 0.

Ify(0) = 2thend= 122 1 = 212. Show thaty(t)³ 1 from above. Ify(0) = 21 thend= 2112 1 = 22. At what timeTisy(T) =2>?

Solution First,y(0) = 2is above the steady-state valuey>=a/b= 1/ 1. Then d= 212 andy(t) = 1/(1 212 e 2 t)is larger than 1 and approachesy(>) = 1/ 1 from above ase 2 tgoes to zero. Second,y(0) = 21 is below theS 2 curve growing fromy(2>) = 0toy(>) = 1. The valued= 22 givesy(t) = 1/( 22 e 2 t+ 1). Whene 2 tequals 12 this isy(t) = 1/ 0 and the solution blows up. That blowup time ist= ln 2. 8 (recommended) Show those 3 solutions toy 2 =y 2 y 2 in one graph! They start from y(0) = 1/ 2 and 2 and 21. The S-curve climbs from 12 to 1. Above that, y(t)descends from 2 to 1. Below theS-curve,y(t)drops from 21 to2>. Can you see 3 regions in the picture? Dropin curves above y= 1 and S -curves sandwiched between 0 and 1 and dropoff curves below y= 0. Solution The three curves are drawn in Figure 3 on page 157. The uppercurves and middle curves approachy>=a/b. The lowest curves reachy=2>in finite time: blow-up. 9 Graphf(y) =y 2 y 2 to see the unstable steady stateY = 0and the stableY = 1. Then graphf(y) =y 2 y 222 / 9 with harvestingh= 2/ 9. What are the steady statesY 1 andY 2? The 3 regions in Problem 8 now haveZ-curves abovey= 2/ 3 , S-curves sandwiched between 1 / 3 and 2 / 3 , dropoff curves belowy= 1/ 3. Solution The steady states are the points whereY 2 Y 2 = 0(logistic) andY 2 Y 22 2 9 = 0(harvesting). That second equation factors into(Y 2

1 3 )(Y 2

2 3 )to show the steady states 13 and 23.

28 Chapter 1. First Order Equations

10 What equation produces anS-curve climbing toy>=Kfromy2>=L?

Solution We can choosey 2 =ay 2 by 22 hwith steady statesKandL. Then aK 2 bK 22 h= 0andaL 2 bL 22 h= 0. If we divide byh, these two linear equations give a h

and

b h

Check :

a h

b h

= 0 and

a h

b h

11 y 2 =y 2 y 2214 = 2 (y 212 ) 2 shows critical harvesting with a double steady state

aty=Y = 12. The layer ofS-curves shrinks to that single line. Sketch a dropin curve that starts abovey(0) = 12 and a dropoff curve that starts belowy(0) = 12.

Solution The solution toy 2 = 2 (y 212 ) 2 comes from integrating 2 dy/(y 212 ) 2 =dt to get 1 /(y 212 ) =t+C. Theny(t) = 12 +t+ 1 C. Ify(0)> 12 thenC > 0 and this curve approachesy(>) = 12 ; it is a hyperbola coming down toward that horizontal line. Ify(0)< 12 thenCis negative and the above solutiony= 12 +t+ 1 Cblows up (or blows down! sinceyis negative) at the positive timet= 2 C. This is a dropoff curve below the horizontal liney= 12. (Ify(0) = 12 the equation isdy/dt= 0and the solution stays at that steady state.)

12 Solve the equationy 2 = 2 (y 212 ) 2 by substitutingv=y 212 and solvingv 2 = 2 v 2.

Solution This approach uses the solutions we know todv/dt= 2 v 2. Those solutions arev(t) =t+ 1 C. Thenv=y 212 gives the samey= 12 +t+ 1 Cas in Problem 11.

13 With overharvesting, every curvey(t)drops to2>. There are no steady states. SolveY 2 Y 22 h= 0(quadratic formula) to find only complex roots if 4 h > 1. The solutions forh= 54 arey(t) = 122 tan(t+C). Sketch that dropoff ifC= 0. Animal populations don’t normally collapse like this from overharvesting. Solution Overharvesting isy 2 =y 2 y 22 hwithhlarger than 14 (Problems 11 and 12 hadh= 14 and critical harvesting). The fixed points come fromY 2 Y 22 h= 0. The quadratic formula givesY = 12 (1±

124 h). These roots are complex forh > 14 : No fixed points. Forh= 54 the equation isy 2 =y 2 y 2254 = 2 (y 212 ) 221. Thenv=y 212 hasv 2 = 2 v 221. Integratingdv/(1 +v 2 ) = 2 dt gives tan 21 v= 2 t 2 Cor v= 2 tan(t+C). y=v+ 12 = 122 tan(t+C). The graph of 2 tantstarts at zero and drops to2>att=Ã/ 2.

14 With two partial fractions , this is my preferred way to findA=

r 2 s

s 2 r

(y 2 r)(y 2 s)

(y 2 r)(r 2 s)

(y 2 s)(s 2 r)

Check that equation: The common denominatoron the right is(y 2 r)(y 2 s)(r 2 s). The numerator should cancel ther 2 swhen you combine the two fractions.

1. The Logistic Equation 29

Separate

y 221

and

y 22 y

into two fractions

y 2 r

y 2 s

Note Whenyapproachesr, the left side of PF2 has a blowup factor 1 /(y 2 r). The other factor 1 /(y 2 s)correctly approachesA = 1/(r 2 s). So the right side of PF2 needs the same blowup aty=r. The first termA/(y 2 r)fits the bill. Solution 1

y 221

(y 2 1)(y+ 1)

y 21

y+ 1

y 21

y+ 1

The constants areA=

r 2 s

y 22 y

(y 2 1)y

y 21

y

y 21

y

r 2 s

15 The threshold equation is the logistic equation backward in time :

dy dt

=ay 2 by 2 is the same as

dy dt

= 2 ay+by 2.

NowY = 0is the stable steady state. Y = a/bis the unstable state (why ?). Ify(0)is below the thresholda/btheny(t) ³ 0 and the species will die out. Graphy(t)withy(0)< a/b(reverseS-curve). Then graphy(t)withy(0)> a/b. Solution The steady states ofdy/dt= 2 ay+by 2 come from 2 aY+bY 2 = 0so againY= 0orY=a/b. The stability is controlled by the sign of df/dy at y=Y:

f= 2 ay+by 2 tells howygrows

df dy

= 2 a+ 2by tells how∆ygrows

Y= 0has

df dy

= 2 a (STABLE) Y=

a b

has

df dy

= 2 a+2b

a

b

=a(UNSTABLE)

TheS-curves go downward fromY=a/btoward the lineY= 0(never touch).

16 (Cubic nonlinearity) The equationy 2 =y(1 2 y)(2 2 y)has three steady states : Y = 0, 1 , 2. By computing the derivativedf /dyaty = 0, 1 , 2 ,decide whether each of these states is stable or unstable. Draw the stability line for this equation, to showy(t)leaving the unstableY’s. Sketch a graph that showsy(t)starting fromy(0) = 12 and 32 and 52.

Solution y 2 =f(y) =y(1 2 y)(2 2 y) = 2y 23 y 2 +y 3 has slopedfdy= 2 26 y+3y 2.

Y= 0 has dydf= 2 (unstable) S 2 curves go up fromY= 0 toward Y= 1 Y= 1has dfdy= 21 (stable) S 2 curves fromY= 2go down toward Y= 1 Y= 2 has dydf= 2(unstable)

30 Chapter 1. First Order Equations

17 (a) Find the steady states of the Gompertz equation dy/dt=y(1 2 lny).

Solution (a)Y(1 2 lnY) = 0at steady statesY= 0andY=e. (b) Show thatz= lnysatisfies the linear equationdz/dt= 1 2 z. Solution (b)z= lnyhasdzdt= 1 ydydt=y(1 2 lny)/y= 1 2 lny= 1 2 z.

(c) The solutionz(t) = 1 +e 2 t(z(0) 2 1)gives what formula fory(t)fromy(0)? Solution (c)z 2 = 1/zgives thatz(t). Then sety(t) = 1/z(t):

y(t) =

1 +e 2 t(z(0) 2 1)

1 +e 2 t

y(0)

18 Decide stability or instability for the steady states of

(a)dy/dt= 2(1 2 y)(1 2 ey) (b)dy/dt= (1 2 y 2 )(4 2 y 2 ) Solution (a)f(y) = 2(1 2 y)(1 2 ey) = 0atY= 1andY= 0 df dy= 22 e

y(1 2 y) 2 2(1 2 eY)

AtY= 1 dfdy= 2 2(1 2 e)> 0 (UNSTABLE) AtY= 0 dydf= 22 (STABLE)

(b)f(y) = (1 2 y 2 )(4 2 y 2 ) = 0atY= 1, 21 , 2 , 22 dfdy= 210 y+ 4y 3

Y= 1givesdydf= 26 (STABLE) Y= 21 givesdydf= 6(UNSTABLE)

Y= 2givesdydf= 12(UNSTABLE) Y= 22 givesdydf= 212 (STABLE)

19 Stefan’s Law of Radiation isdy/dt=K(M 42 y 4 ). It is unusual to see fourth powers. Find all real steady states and their stability. Starting fromy(0) =M/ 2 , sketch a graph ofy(t). Solution f(Y) =K(M 42 Y 4 )equals 0 atY=MandY= 2 M(alsoY=±iM). df dy= 24 KY

3 = 24 KM 3 (Y=Mis STABLE) df dy= 4KM

3 (Y= 2 Mis UNSTABLE)

The graph starting aty(0) =M/ 2 must go upwards to approachy(>) =M.

20 dy/dt = ay 2 y 3 has how many steady statesY fora < 0 and thena > 0? Graph those valuesY(a)to see a pitchfork bifurcation —new steady states suddenly appear asapasses zero. The graph ofY(a)looks like a pitchfork. Solution f(Y) =aY 2 Y 3 =Y(a 2 Y 2 )has 3 steady statesY= 0,

a, 2

a. df dy=a 23 y

2 equalsaatY= 0, df dy= 22 aatY=

aandY= 2

a.

ThenY= 0is UNSTABLE andY=