Solutions Manual for A Course in Ordinary Differential Equations. by Randall J. Swift Stephen A. Wirkus

Transcript

1 Solutions Manual for A Course in Ordinary Differential Equations by Randall J. Swift Stephen A. Wirkus

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3 Preface This solutions manual is a guide for instructor s using A Course in Ordinary Differential Equations. Many problems have their solution presented in its entirety while some merely have an answer and few are skipped. This should provide sufficient guidance through the problems posed in the tet. As with the book, code for Matlab, Maple, or Mathematica is not given. It is our eperience that the synta given in the book is sufficient to learn the relevant commands used to obtain solutions to the various problems in the book. Please give Appendi A a chance, if you have not done so already. This solutions manual was put together by many people and we note a few of them here. We owe a big thanks to our former students David Monarres, for help in preparing portions of this book, and Walter Sosa and Moore Chung, for their help in preparing solutions. More recently Scott Wilde has helped tremendously in shaping this manual. Jenny Switkes and other colleagues and students have also given feedback on various drafts of this manual and all have been helpful. This book has evolved over the last few years and we have tried to make this solution manual stay in step. However, we realize that there are probably many typos throughout and we encourage you to contact us with your corrections. Hopefully future printings of this manual will have an eponentially decreasing number of such errors. We would appreciate any comments that you might have regarding the book and the manual. Randall J. Swift ( rjswift@csupomona.edu Stephen A. Wirkus ( swirkus@csupomona.edu URL for typos and errata: swirkus/acourseinodes

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5 Contents Traditional First-Order Differential Equations. Some Basic Terminology Separable Differential Equations Physical Problems with Separable Equations Eact Equations Linear Equations Chapter : Additional Problems Geometrical and Numerical Methods for First-Order Equations 5. Direction Fields the Geometry of Differential Equations Eistence and Uniqueness for First-Order Equations First-Order Autonomous Equations Geometrical Insight Population Modeling: An Application of Autonomous Equations Numerical Approimation with the Euler Method Numerical Approimation with the Runge-Kutta Method An Introduction to Autonomous Second-Order Equations Chapter : Additional Problems Elements of Higher-Order Linear Equations 9. Some Terminology Essential Topics from Linear Algebra Reduction of Order The Case of n Operator Notation Numerical Consideration for nth Order Equations Chapter : Additional Problems Techniques of Higher-Order Linear Equations 5 4. Homogeneous Equations with Constant Coefficients A Mass on a Spring Cauchy-Euler (Equidimensional Equation Nonhomogeneous Equations Method of Undetermined Coefficients via Tables Method of Undetermined Coefficients via the Annihilator Method Variation of Parameters Chapter 4: Additional Problems

6 5 Fundamentals of Systems of Differential Equations Systems of Two Equations Motivational Eamples Useful Terminology Linear Transformations and the Fundamental Subspaces EigenvaluesandEigenvectors Matri Eponentials Chapter 5: Additional Problems Techniques of Systems of Differential Equations A General Method, Part I: Solving Systems with Real, Distinct Eigenvalues A General Method, Part II: Solving Systems with Repeated Real or Comple Eigenvalues Solving Linear Homogeneous and Nonhomogeneous Systems of Equations Nonlinear Equations and Phase Plane Analysis Epidemiological Models Chapter 6: Additional Problems Laplace Transforms 7. Fundamentals of the Laplace Transform PropertiesoftheLaplaceTransform Step Functions, Translated Functions, and Periodic Functions The Inverse Laplace Transform Laplace Transform Solution of Linear Differential Equations Solving Linear Systems using Laplace Transforms The Convolution Chapter 7: Additional Problems Series Methods 8. Power Series Representations of Functions ThePowerSeriesMethod Ordinary and Singular Points The Method of Frobenius Bessel Functions Chapter 8: Additional Problems B Graphing Factored Polynomials C Selected Topics from Linear Algebra C. A Primer on Matri Algebra C. Gaussian Elimination, Matri Inverses, and Cramer s Rule C. Coordinates and Change of Basis

7 Chapter Traditional First-Order Differential Equations. Some Basic Terminology. With y(,wehavey ( 6. Substituting into the ODE gives (6 (, which is true for (,.. y, dy d d dy (. Substituting y and d d into the ODE, we get (,which is true for all. Thusy is a solution to dy d (y on (,.. y( 5 +4 (5 +4, dy d (5 +4 ( dy. Substituting y and 5 d 5 ( dy. y and (5 +4 d into the ODE, we get ( ( ( (5 +4 do not eist when which is true for { 4 5 }.Thusy 5 +4 is a solution to the ODE on ( 4 5,. 4. Since y( e, wecalculatey ( e. Substituting into the ODE gives (e + (e e +( e + e +e e + e + e e + which is true for all (,. 5. y(, dy d. Substituting y and dy into the ODE, we get d ( /

8 Section. which is true for all. Thusy is a solution to the ODE on (,. 6. y(, dy d ( Substituting y and dy d dy. y and ( d into the ODE, we get ( ( ( ( ( do not eist when. which is true for { }. Thusy is a solution to the ODE on (,. 7. Taking the derivative of y( gives us y ( and y ( +. Substitution gives ( + ( + which is true where both sides are defined, which occurs when. 8. y( sin +cos, dy d cos sin, y andd d sin cos. Substituting y and d y d into the ODE, we get ( sin cos+(sin +cos which is true for all. Thusy sin +cos is a solution to the ODE on (,. 9. y(, dy d,and d y d. Substituting y and d y into the ODE, we get d ( + ( which is true for all. Thusy is a solution to the ODE on (,.. y( + C sin, dy d +Ccos, and d y d C sin. Substituting y and d y d ODE, we get ( C sin +( + C sin into the which is true for all. Thus y + C sin is a solution to the ODE on (, for any constant C.. (a y e, y e,y e y y +y e e +e (b y e, y e, y 4e y y +y 4e 6e +e

9 Section.. (a y e, y e,y e y y + y e e + e (b y e, y e + e,y e +e y y + y e +e e e + e. (a y sin, y cos, y 9sin y +9y 9sin +9sin YES (b y sin, y cos, y sin y +9y sin +9sin NO (c y cos, y sin, y 9cos y +9y 9cos +9cos YES (d y e, y e, y 9e y +9y 9e +9e NO (e y, y,y 6 y +9y 6 +9 (unless NO 4. y +6y +9y (a y e, y e,y e e +6e +9e NO (b y e, y e y 9e 9e 8e +9e YES (c y e, y e + e y 9e 6e 9e 6e 8e +6e +9e YES (d y 4e,y e, y 6e 6e +7e +6e NO (e y e ( +, y e ( 5, y e (9 + e ( NO 5. y 7y +y (a y e, y e, y 4e 4e 4e +e NO (b y e, y e, y 9e 9e e +e YES (c y e 4, y 4e 4, y 6e 4 6e 4 8e 4 +e 4 YES (d y e 5, y 5e 5, y 5e 5 5e 5 5e 5 +e 5 NO (e y e +e 4,y e +8e 4, y 9e +e 4 e 4 +9e e 56e +e +4e 4 YES 6. y +4y +5y (a y e, y e,y 4e 4e 8e +5e NO (b y e sin, y e cos e sin, y 8e cos 8e cos +8e cos 8e sin +5e sin NO

10 4 Section. (c y e cos, y e cos e sin, y 8e sin 8e sin 8e cos 8e sin +5e cos NO (d y cos, y sin, y 4cos 4cos 8sin +5cos NO 7. y e r, y re r,andy r e r.theodegives r e r +re r +e r e r (r +r + e r (r +(r + r, 8. y +4y +4y y e r, y re r + e r,y r e r +re r e r ( r +r +4r +4+4 r +(+4r +(4+4 r 4 ± or or We can disregard the value involving, which leaves only r. 9. (a y + y sin (i nd order (ii linear (iii N/A (b y +siny (i nd order (ii nonlinear (iii N/A (c y ( +(siny ( + y, y(, y (, y ( (i rd order (ii linear (iii IVP (d y + e y y 4, y( (i st order (ii nonlinear (iii IVP (e y + y y (i nd order (ii linear (iii N/A (f y + e y + y,y(, y(π (i nd order (ii nonlinear (iii BVP. (a y yy (i nd order (ii nonlinear (iii N/A (b y sin (i nd order (ii linear (iii N/A (c y +y,y(, y ( (i nd order (ii linear (iii BVP (d y,y(, y ( (i nd order (ii linear (iii IVP (e y 4y +4y,y(, y ( (i nd order (ii linear (iii IVP (f y + y +(lny (i nd order (ii linear (iii N/A

11 .. SEPARABLE DIFFERENTIAL EQUATIONS 5. Separable Differential Equations.. 4y d +( +dy 4 + d y dy 4 dy + d y Letting u +, du dy u y ln u + C ln y ln( ++C ln y e ln( ++C y y e C ( +, since e C > y C (, + where C ±ec tan dy+yd dy d y tan ln y ln sin + C y sin e C y C csc, c ±e C. (e +cosydy+ e (sin y +d,y(. Then cos y sin y + dy e e + d Substitute Then Therefore, u siny + u e + du cosydy du e d [ ln sin y + ln e + + C sin y +A (e + A Apply IC: (sin + (e A (sin+ + sin y + +sin e +

12 6 Section. 4. We have (y +d +( 4 +dy,y(. Then Therefore, Finally, ( 4 +ds (s +dr dy y d u du d arctan(y arctan(u+c arctan(y arctan( +C arctan( arctan( + C + arctan( C }{{} C π π/4 u + du 5. y d +( +dy have ln y 6. arctan(y arctan( + π π y tan( arctan( dy y y A( +e d. Substituting u +, u, du d, we + ( (u du u u du ln + + C (y + / d y dy d d y y + dy y y + dy Let u y +anddu ydy,so du ydy.thenwehave d du u ln + C (/ u (/ ln + C y +

13 Section We have ( y +y andy(. Then ( dy d y dy y d dy y y y d ln + C ln + C IC: ln + C C y ln + 8. We have y cot + y andy(. Then dy cot y + d dy tan d y ln y ln cos + C y e C cos y ±e C (cos +A(cos + Since y(, A + A. Thus y (cos + 9. y +y dy d y dy y d y dy d Use a u e u ln a.then y + C y + C y log + C

14 8 Section.. d dt + t d ( t dt d ( t dt t t + C t t +C ± t t + C, C C. Let u y du d dy d du d +cosu du d cosu du cos u d tan(u/ tan( y y +tan (/. Let u + y du d +dy d u u u u + du u + d ln(u + e u + + y y e +. We have ( +yy,y(. Let u y +. Then du d dy d + Substituting into the ODE gives ( u u This is separable: u u u +u + u u u +u du d

15 Section. 9 Note that Then ( u +u u + +u +u du +u d u ln +u + C Substitute back u +y: ( +y ln + +y + C Apply IC: + ( ln ++( +C 4 ln C The solution is given in implicit form as y ln + +y +4+ln 4. y 4 +y. Let u 4 +y du d 4+dy d u u +4 du d u +4 ( ln ( u + u ( ( ln 4 +y + 4 +y 5. We have (y +d + y( +4dy,y(. Then ( +4ydy (y +d y y + dy d +4 ( d dy y + +4 y ln y + ln +4 + C e y ln y+ ln +4 +C e y + ey e C Apply IC: y( +4 e y (y + ± ec ( +4 A ( +4 A e e ( + A ( +4 e y (y + e ( We have 8 cos yd+csc dy,y(π/ π/4. Then csc dy 8cos yd sec ydy 8 sin d ( tan y 8 4 sin + C where A ±e C tan y sin 4 + C, where C 8C

16 Section. Using y(π/ π/4, we solve for C: ( π tan 4 ( π sin π 6 + C C π So tan y sin 4 + π 7. We have dy d y +y, y(. Separating variables and integrating gives + y +y dy + d ( y (y + + ( dy y ( + + Now apply IC: The solution is then 4 ln y + + ln y ln + + ln + C 4 ln( ++ ln( ln( + + ln( + C C ln 4 4 ln 4 ln(y ++ ln(y ln + + ln + ln 4 4 ln 8. We have y e, y(. Then dy d e dy e d y e d Apply IC: ydy y( y( y( e t dt e t dt e t dt

17 Section. 9. We have y ye, y(. Then dy e d y ln y e + C C ln y e, when,y Thus C y is even since we have only an term. ln y y e / e ( / y ±e / e. (a We have y cos y, lim + y 9π/4. Then dy cosy + d First we use a trigonometric identity as follows: cos y + (cos y sin y+(cos y +sin y cos y Thus we obtain dy cos y sec ydy d d tany + C tan y + C C tany + ( 9π 4 lim + C tan + Thus tan y ( or y arctan

18 Section. (b We have y y +6 y.then y dy d (y 8 y y 8 dy d ln y 8 + C y 8 e C e y ±e C e +8 y [Ce +8 / where C ±e C Notice that no value for C bounds y since e C.. dn rn dt dn rdt N N dn rdt ln N rt + C Note N is number of population, i.e. >. e ln N e rt+c N e C e rt N C e rt Note C is >, dealing with possible population. Thus N C e rt is a solution to the ODE. Case. For values r we have eponential growth as t. Case. For values r< we have eponential decay as t.

19 Section.. lim N K t. We have ( + y d dy. Then dn dt ( r N N K dn rn ( N K dt (Let u N/K du dn dt K dt K du dt ruk( u du dt ru(u dt du ru(u ln u ln u t r ( r ln ( N K e tr ( N K N K e tr N K K e tr Let Then Substituting for u gives 4. We have Let Then ( + y dy d + y dy d u y y u u u + u +u u u ln + C y (ln + C dy d y +y y + y u y y u + u u + u u +u u u + u

20 4 Section. This equation is separable: Use partial fractions to rewrite Thus u(u + A u + du u + u d ln + C u + u. ( u(u + du u u + Plugging in u y/ then gives 5. We have B A(u ++Bu u(a + B+A u + } A + B B A du ln u ln u + ln u u + ln u u + ln + C u u + eln +C e C u C u C(u + u + u( CC u C C y Let u y y u + u. Substituting gives C C dy d + y dy d +y (u + u +u u +u u (u This is separable: Letting w u, dw du, du (u d u ln + C u ln + C u ln + C Plugging in u y/ gives y ln + C

21 Section y y + y y + y + y y + y + y ( y ( y + + (homogeneous Let y v, y v + v. Then dv d + v +v + v dv d +v +v dv d arcsinh v ln + C Substituting back in, we have y sinh(ln + C y sinh(ln + C 7. We have ( +y d dy. Then +y dy d +y dy d dy ( y d + Let y v, y v + v. Then (homogeneous dv d + v +v dv d +v dv v + d ln v + + C v + Ce y Ce y Ce 8. We have (y y d + dy dy y y y ( y ( d

22 6 Section. Let u y, y u + u. Then u + u u u u u u ln u +ln u ln + C u u eln +C C y/ (y/ C y y C y y C y Cy C y C +C 9. We have y y( y. Then dy d y ( y dy ( y d ( ( y Let u y, dy d du + u. Then d du d + u u ( u du d u u du u ln + C y ln + C y ± ln + C d

23 Section. 7. We have ( + y y y. Then. dy d y + y dv d + v v + v dv d + v v +v dv d v +v v dv v v v d +v dv d v v +v +v v v dv dv v( v dv + v ( v dv d ln v ln v ln + C v v C y ( y C y C y C y Cy C y C(y ( y v, y v + v,v y y y tan(y/ y y ( y +tan (Homogeneous v + v v +tanv dv tan v d ln ln sin v + C e C sin v e C sin y y sin e C y sin e C. ( + y d (4 +y dy dy d y + C

24 8 Section.. We have y + y yy.then Let y v + v. Then y y (y y y y y 4 y dy d y (y (y dy yd (y dy +( y d dv d + v v v y v dv v d v v v dv v v(v d v dv d v v + v v dv v v d v v v dv d ln(v ln v ln + C ln +ln v ln + C v v(v C (y(y ( y C ( y C y(y C 4. ( y+(y y y (fory y + C. If y, the equation is true trivally.

25 Section ( +4yy +y y +y +4y +(y/ +4(y/ ( y v, y v + v, v y v + v +v +4v d ( + 4vdv +v 4v (Partial fractions yields: ln [ ln( + v( v 5 6 ln [ ln( + (y/( (y/ We have Let u y, y u + u. Then ( y d +( + y dy dy d y y + y ( +u + u +u du u + u u +u du d u +u u + y du u d +u +u +u du d d arctan u + ln +u ( y arctan + ( ( y ln +

26 Section. 7. yd ( + y dy dy d (y/ +(y/ (y v, y v + v v + v v +v v v + v v + +v v + v dv d ( v + ln ln + C v v + v e C (y/+ (y/ e C 8. ( ( y (y/ y + y +(y/ ( y v, y v + v, v y v + v v ( + v v v v ( + v ( + v v v dv d ( (v 4 ln v ( ((y/ 4 ln (y/ + v ln + C +(y/ ln + C 4. (a M(, y t Homog. deg. n M(t, ty t n M(, y ( n M(, y M(, y n M(, y (b Repeat above echanging M for N. (c M(, yd + N(, ydy dy M(, y M(, y d N(, y N(, y Define g (t M(,t, g (t N(,t, g(t g (t g (t dy d M(, y N(, y g ( y g ( y g( y

27 Section. 4. In (.F (, y ( + yd dy F (t, ty (t + tytd t t dy t (( + yd dyt F (, y Homogeneous of degree. The others are similar. 4. Given that M(, y d + N(, y dy is homogeneous, then dy M(, y M(, (y/ and the substitution d N(, y N(, (y/ r cos t d r sin tdt+costdr y r sin t dy r cos tdt+sintdr dy M(, tan t M d N(, tan t N dy r cos tdt+sintdr d r sin tdt+costdr M r cos tdt+sintdr N r sin tdt+costdr (r cos tdt+sintdrn M( r sin tdt+costdr N sin tdr+ M cos tdr Nrcos tdt+ Mrsin tdt ( dr N cos t + M sin t r dt N sin t + M cos t Since the function M and N are functions of t, the right hand side is a function of r times a function of t. Hence, it is seperable. 44. (a From 4. with r ( + y and t tan y [ ( + y ep ln ( dr r cos t +(cost +sintsint r dt cos t sin t +(cost +sintcost ( cos t +sintcos t +sin t r cos t ( +sintcos t r cos t ( dr +sintcos t r cos dt t ( ln r ln cos +tant t [ ( r ep ln cos +tant + C t, we finally have cos (tan ( y ( + tan(tan y +C (b

28 Section. 45. (a dy d y y + (y/ (y/+ which is homogeneous. As before, y v and v + v v v + v v + (v +dv d v + v + C ln Ke (v /4 (v/ K ep( y 4 y (b i. ( y d +(y + +5dy M(, y y M(t, ty t ty tm(, y. Hence, M is not homogeneous. Similarly for N, and so the equation is not homogeneous. ii. X, y Y, d dx, dy dy and dy d dy dx Y X Y +X iii. X K ep( Y 4X Y + X (y + K ep( 4( + y + ( Lines are parallel, so there is no point of intersection. 47. Lines are parallel, so there is no point of intersection. 48. dy +y 5 which gives point of intersection to be (4,. Thus, we let X 4and d +4y y Y +andgetto dy dx X +Y X +4Y +(Y/X +4(Y/X Y vx Y v X + v v X v +4v ln X 5ln v 4X ln +4 5ln y +4 4( +4

29 Section. 49. Point of intersection is (,. Thus we let X +andy Y andget dy dx Y X + Y dx X + Y dy Y (/y+ X uy dx dy u Y + u u Y + u u + du u + dy Y ln u + ln Y + C Y K u + K (X/Y + ( y + ±K y + + ( y + 5. y and as in 49., + y ( dy Y dx X + Y dx ( X dy Y + X uy dx dy u Y + u u + u X u + u + u X (u + du u + dx X tan (u ln X + C ( tan ln + C y +

30 4 Section.. Physical Problems with Separable Equations. r kv, v e 6m/s, v,m 75kg, s 8m, s e 5m m dv dt mg kv 75v 75(9.8 kv 75 kv 75 dv dt 75 kv.8 ln kv 7. kv+7. t + C k kv 7. [ kt + C ep kv From the given information lim v(t 6 t k6 7. k.47 k v C.45v 7. Thus, we have.45v +7. e (.6t and after some simplification ( +e.6t v(t 6 e.6t. r kv, r.48n when v k.48.4 dv dt (.4(9.8.48v dv dt 9.8.v dv 9.8.v dt If air resistance is neglected, then s(t 4.9t +t, v(t 9.8t + t.4, s(.4.4ft 4. v(t gt + v, v( g + v v g s(ft/s 96ft/s 5. (a v,v(t gt t, v(4 8ft/s 4 (b Ave velocity tdt64ft/s 4 (c v(t t, s(t6t + s,s(4 s(4 6 + s s 56 cliff was 56 ft tall 6. a(t,v(t t + v t, s(t 6t + s, s(5 6(5 + s s 4 cliff was 4 ft tall

31 Section s 5, v,s(t 6t 5 t average velocity km 5/4 5 8km 4.48km/s 5s 8. T (t (T T s e kt + T s,t s, T 5,T( 5 T ( 7e k 5 k ( 7 ln 7. Then to find eating time (t eat, T (eat 7e k(eat eat ln(/7 4.8hrs 5pm ln(7/7 9. Let t bepm. T ( 79, T( 68, T s 6. T (t 9e kt +6,T( 9e k +668 Thus the person died about :pm 9e k 8 k ln.4.88 T (t 9e.88t +6 (Solve for t when T t.46. T (t 8e kt +,T( 8e k +6 k.69 T (t 8e.69t ln 6 +5 t 4min ln. N(t amount of N in tank after t seconds. N( 6. dn dt rate in - rate out. N (. dn N dt N(t e( t/+c, N( 6 e C C ln4.9 N(t e ( t/+.9. We want to solve N(t ( fort. t (ln( seconds min. 4. s(tamount ( of salt at time t. s(. ds s(tkg dt (5L/m s(t L s(t e ( t/+c C ln s(t e ( t/+ln s( kg 5. (a ds s(t (4lb/gal(gal/m (s(tlb/5gal(gal/min8 dt 5 s(t 6+9e ( t/5 concentration s(t/5. (b ds dt s 8 s(t e( t/5 conc s(t/5 t. 6. s(t amount of salt in pond, s,v(t volume of pond, v, dv dt 5 v(t 5t. ( ( ds 5kg dt s(tkg (5m /d m ( 5tm (m /d 6.5 s(t 5t Yes, after days, there is no water in the pond which will result in no salt in the pond.

32 6 Section. ( ( 7. C(t amount of CO, C (.5(.m, dc dt m Cm min (.4 m (m /min.8.c This gives us that the concentration will be half of what it started (i.e. there will be.5m of CO in.45 minutes. 8. da dt ka, da A kdt A(t ekt+c A e kt. A( A /A e k e k k ln(/. A(t.A A e.t t ln(.. days 9. A(.56 e k k.5798 t.95 years. p amount of decayed tin, Moriginal amount of tin. dp k(m p(p wherem p is the amount of tin left dt dp kdt p(m p ( p M ln kt + C M p M p(t M +e Mkt+C where C would be determined by the original amount of decayed tin in the organ.. Let (, y be the point of tangency and c be the -intercept of the tangent line. Then c y a c a y. dy d y c y a y y y a dy y a d y ± a Because of symmetry, either graph will give the desired result.. Let (, y be the point of tangency and c be the -intercept of the tangent line. Then c + y b. dy d y c y b y b ln y y This is an implicitly defined function.

33 .4. EXACT EQUATIONS 7. Let A(t be the amount of snow on the ground. Then A(t kt since it is falling at a constant rate and there was none when it started. Let P (t be the distance plowed. Then dp dt m A(t m kt. Thus dp m kt dt P m k ln t + C, P ( C P (t m k ln t +. P ( m k ln + m k.44. P (t.44 lnt +. Set P (t togett 4which says it started snowing at about 8am..4 Eact Equations. We have y +(+ y dy d. UsingM y and N + y,wehave M y 6y, N 6y so the equation is eact. We need to solve f f M and y N: f f d y d y + g(y This f y + g(y mustalsosatisfy f y N. + y y ( y + g(y y + g (y g (y g(y y Thus the solution is given by f C: y + y C. We have (y ++( +4y dy,y(. Use M y +andn +4y. Then M y N so the equation is eact. We set f f M and y N and solve. f f y dy ( +4y dy y +y + h( This f must also satisfy f M: y + ( y +y + h( y + h ( h ( h( The solution is given by Initial conditions yield C,so y +y + C y +y +

34 8 Section.4. We have (y sec +sectan + (tan +y dy withy(. Setting M (ysec + sec tan andn tan +y leads to M y sec N so the ODE is eact. Then, F (y sec +sectan d F y tan +sec + g(y tan +y f y tan + g (y y g (y g(y y The solution is y tan +sec + y C. With initial conditions, ( + + C C, so y tan +sec + y 4. We have (y sin cos + y sin +(sin y cos dy with y(. Let M y sin cos + y sin and N sin y cos. Then M y sin cos +ysin N sin cos y( sin so the equation is eact. We set f Then f f M and y N and solve. f y dy (sin y cos dy y sin y cos + h( y sin cos + y sin (f y sin cos + y sin + h ( h ( h( C The solution is given by y sin y cos C Initial conditions give us sin cos, so C 9 and y sin y cos 9 5. We have y +( y dy. Setting M y and N y, Then M y N f y f(, y y + g(y f y + g (y y g (y y g(y y

35 Section.4 9 Thus the solution satisfies y y C 6. We have ( 9y +(4y 6 ydy. UsingM ( 9y and N (4y 6 y gives M y 8 y N Then f 9 y f(, y y + g(y f y 6 y + g (y 4y 6 y g(y y 4 Thus the solution satisfies y + y 4 C With initial conditions y(, ( ( + 4 C gives C, so y + y 4 7. We have e y (y + e 4 dy withy(. Using M e y and N y e y we have M y e y N We need to solve f Md e y d e y + g(y Substituting into f y N, y (e y + g(y e y + g (y y e y g (y y g(y y The solution is given by e y + y C. Now apply initial conditions: e ( C, so C 9 e. The solution is then e y y e 9 8. We have ( + y sin +( y cos dy. Then so the equation is not eact. y [ + y sin y sin [ y cos 4y sin

36 Section.4 ( 9. We have ( ln y+ y y dy. Then y [ + ln y y [ y y y so the equation is eact. f + ln y f(, y + ln y + φ(y f y y + φ (y y y φ(y y + C Then f(, y + ln y y + C. (+ y ( d + y dy M y, N F the equation is eact. Thus + y F y + g(y. F y + g (y y g (y y, g(y y /+C F (, y y + y + C (. We have sin y + d + ( +cosy dy. Then cos y [ y sin y + cot[ycsc[y [ ( +cosy cos y cos y cos y We find cos + cos(y cos y cos cos y cos y So so it is eact. Then N cos y (cos cos y sin cot y csc y y f sin y f(, y + + φ(y sin y f y cot y csc y + φ (y cos y +cosy cos y cos y sin y + cos y siny cot y csc y cot y csc y

37 Section.4 So and φ (y cot y csc y φ(y csc(y f(, y csc y + + csc(y+c + C. Solve a a y a d + (( a a y a dy. (i As a separable equation, (ii As an eact equation, a a y a d (a a y a dy a a (a y a a d y a dy a (a d dy y a ln + C (a ln(y y (a C a y [aa y a a( a a y a y [( aa y a a( a a y a f aa y a f(, y a y ( a + φ(y f y ( aa y a + φ (y ( a a y a φ (y C f(, y a y ( a + C. Determine A R such that the equation is eact. Then solve the resulting equation. (a ( +y d +(A +4y dy. Then so A anda.then y [ +y [A +4y A f +y f(, y + y + φ(y f y + φ (y +4y φ(y y + C So f(, y + y +y + C

38 Section.4 (b ( Ay + y ( d + dy. Wehave [ Ay y + y [ A + + so A. Then Therefore f y y f(, y y + y + φ(y f y + + φ (y φ(y C f(, y y y + C 4. Assume M y N F. We want to find F (, y such that M, F N. As in the proof y of Thm..4., proceed replacing the argument of F M(, y withonefor F N(, y and continuing in the same manner. y 5. Determine the most general function N(, y sothat is eact. For the equation to be eact, ( + y d + N(, y dy so y [ + y y N N(, y y d y + φ(y 6. Determine the most general function M(, y such that M(, y d +(ye + y e dy is eact. We have So M y [ye + y e ye +y e M(, y ye +y e dy y e + y e + φ( 7. Show that (A + By d +(C + Dy dy is eact if B C. To prove this, assume that B C. Then [A + By y B [C + Dy C

39 .5. LINEAR EQUATIONS Then [A + By [C + Dy y which by Theorem.8. implies that the equation is eact. Now assume that the equation is eact. Then by Theorem.8. we know that B [A + By [C + Dy C y or that B C. 8. Show that the homogeneous equation (A + By + Cy d +(D + Ey + Fy dy is eact if and only if B D and E C. To prove this, assume that B D and E C. Then y [A + By + Cy B +Cy D + Ey [D + Ey + Fy Therefore, by Theorem.8., the equation is eact. Now assume that the equation is eact. Then by Theorem.8., B +Cy y [A + By + Cy [D + Ey + Fy D + Ey which implies that B D and C E..5 Linear Equations. y + y : Linear equation with a(, b(. Let A( d ln. Then e A( and e A(. The solution is given by ( y( [ d + C + C if < ( + C if Then y( + C for all.. y + y. Linear with a( +, b(.then A( + d ln + ln( + because + is always positive. Substituting u +, du d leads to e A( e ln(+ + ; e A( e ln(+ +

40 4 Section.5 Then b(e A( d ( ( + d + d arctan The solution is therefore. ( +y 4 +y leads to Using P where + y (+ ( arctan +(+ +C y y and Q + gives d ln + + e p( e p( e ln + + e ln d sgn( + sgn( + The solution is y + ( sgn( ++C or { if +< if + y ( + ( + C 4. We have y + y tan sec with y(π. Then so y sec tan + C. N( sec d d [y sec sec 5. We have dy So y sin + C cos sinπ + C cos π C C y sin cos ( + y d. Then Let u du/ d. Then dy d y Q( N( e (/( d N( e (/ (/u du e ln (/

41 Section.5 5 So [ d y d Let u. Then du/ d and d u du y u / du +C so y +C and y ( + C 6. y y 4e and y( 4. Let P (, Q( 4e.Then d e p( e ; e p( d e 4e e d 4 d 4 The solution is y e (4 + C With initial conditions, 4 e (4( + C C 4,thus y 4e ( + 7. y + y, P (, Q(, μ( e / e / d e / + C. ( Thus the general solution is y e / e / + C 8. P (, Q( e,μ( e ln e d e e. Thus the general solution is y e ln (e e + C (e e + C 9. P (,Q(, μ( e e d e + 4 e + C. Thus the general solution is ( e y e + 4 e + C. We have dy d + y cos ( π with y 4 π and >. This time Q( which implies that d [y cos( y sin(+c d So with y( π 4 π,wehavethat π 4 ( π π sin + C +C so C and y( sin( + so N(,

42 6 Section.5. P ( Q( μ( e Q(μ( d e + C ( y e e + C. P ( Q( μ( e Q(μ( d ( +e y + C. P ( Q( cos μ( e Q(μ( d e (cos +sin+c ( e y e (cos +sin+c 4. P ( Q( e μ( e Q(μ( d e + C ( e y e + C

43 Section y +y withy(. Then we have A( d e A( e, e A( e 4 b(e A( e d y(te A(t b(te At 6. We have ( + y dy yd gives + y y d dy or d dy y y Thus a(y y, b(y y, A(y y dy ln y. Then Then y[y+cy and e A(y e ln y y, e A(y y ( b(ye A(y dy y dy dy y y y + Cy 7. We have (e y y or(e y dy d. Soey d dy we have N(y e y. so the solution is d dy [ey e y and e y e y + C e y + Ce y 8. We have (sin y + cot yy. Rewriting,wehave (sin y + cot y d dy d cot y siny dy With Q( cot y we have N(y e ln(sin csc y, so d sin y [ csc y dy sin y d sinycos y [ csc y dy sin y csc y cosydy 9. (a y,y, y + P (y + csc y siny + C (y sin y + C sin y and d dy + ey.withq(y

44 8 Section.5 (b y y is a solution. y ky, y ky. Then ky + P (ky k(y + P (y k( (c y y + y, y y + y Then y + y + P ((y + y (y + P (y +(y + P (y +(Sincey and y are solutions. (a y y + y, y y + y Then y + y + P ((y + y (y + P (y +(y + P (y +r( (Since y and y are solutions r( (b y y + y, y y + y Then y + y + P ((y + y (y + P (y +(y + P (y r(+q( (Since y and y are solutions (c Solution to ( y +y e is y e (e + C andy +y cos is e y e ( cos +sin+c. 5 ( Thus y e e + e ( cos +sin+c is a solution to the original equation. 5. (a y P (y dy P ( d y ln y P ( d y c ±e P ( d (b y A(y c, y A (y c + A(y c. Then Q( y + P (y A (y c + A(y c + P (A(y c A (y c + A((y c + P (y c A (y c (c A (e P ( d Q( A ( Q(e P ( d A( Q(e P ( d d (d y p y c A(, ( y y c + y p y c + y c A( y c ( + A( e P ( d Q(e P ( d d. Let y be the amount of pollutant at time t. Then ( y y ( 5 + t 4 y 5 + t Then y + y 4 is linear with solution 5 + t ( y(t 4(t (t C. C 4(5 6. y(t 4(t + 5 4(56,y( 87 (t Concentration g/gal

45 Section.5 9. (a dn dt rn + ra dn dt rn ra N(t a + Cert. N( A C N(t Ae rt A (b r ln. So,withA,N(t e rt t. N(4 5,, 645 N(6 6, 58, 4, 5 d 7. dt + a(t f(t, P(t a(t, μ(t e a(t dt ( (t e a(t dt e a(t dt f(t dt + C. ( ( lim (t lim e a(t dt e a(t dt f(t dt + C sincea(t >, e a(t dt t t ( 9. (a See section.5.. μ( ep ( (y d which gives an eact y DE and F (, y y + y + C ( M (b N y N ( (. So μ( which gives an eact De and F (, y ln y + C. M and N change roles, but the argument is eactly the same as (. M. y y, N ( M N y N M y +, only in so μ( N N + (y which a function only in so μ( e y ( M y N which a function + M y +y, N y ( N M M y y which a function only in y so μ(y y M y y +, M y, N N y +Which does not yield anything of a function in or y ( M N y N which a function only in so μ( ( M y N which a function only in so μ( e ( M y N which a function only in so μ( cos y ( M N y N 4 ( which a function only in + M y, N e N M y y +, N y N M y cos y, N so μ( ( + 4. n,v y v v which is linear and gives solution v ++Ce y ++Ce

46 4 Section.5 4. y y 4 cos + y tan, so Let v y.then So dy d +(y tan(y y4 cos dv dy y 4 [ d d 7dv d y4 dy d dv ( d +( tan(y y 4 cos dv +(tan(v cos d N e tan d e ln(/(cos sec d d [v sec cos d d [v sec + cos v sec 4 sin + C y sec 4 sin + C 44. n 4,v y v 9v which is linear and gives v Ce 9 + y Ce 9 + y Ce We have y y + y.then y y y y y y y This is a Bernoulli equation with r, so r (. Thus we need to solve u + ( u (. Take a(, b(, anda( d ln. Then e A( e ln, e A( e ln ( b(e A( d d ( ( d d < ( d d ( u + C + C y + C y ( + C / 46. y is a trivial solution, so if y y y ln y 4 + C y ±e 4 +C 4 dy y (4 d

47 Section We have ( +(y + y y. Then dy d + y y + dy d + y + y Let v y,so and that which implies that dv dy y d d dv d ( y dy d ( dv d ( y + y y + ( dv d y dv ( + d v + Q( +, N( e (/(+ d +,so [ d v d + + v + ln We have y y 4y. Then v (ln + ( + (ln + ( + y y ( +ln + dy d y 4 y ( 4 dy d + y y / (Bernoulli equation Let v y (/ y /,so dv d dy y / d.then ( dv d [y/ + 4 y y / ( dv d + v (N( [ d y d y ln + C y ln + C

48 4 Section We have y +y + 5 y e. Then dy d +y + 5 y e dy d + y ( 5 e y (Bernoulli equation dv d v y dv d ( y + y ( 5 e y y dy d dv d ( y dv d 4 y (5 e [ Now use N(,so d 4 d v [e. Then 4 v 4 e ( + C v 4 e ( + C 4 5. We have y dy (y + d. Then y ( Let v y + y,so dv d y dy Using N( e ( + C dy d dy d (y y + dy d y + y ( dy d + y y (Bernoulli equation dv d ( y we have d d + d.then ( [ v,so y y dv d + v + C y + C 5. Use the Bernoulli method to solve the logistic equation ( dn dt rn N K We have Let v N,so dv dt ( v dn dt rn r K N dn dt +( rn r K N N dn dt and dv dt ( N dn dt.then dv dt ( N +( rn r K N dv dt + rv r K

49 .6. CHAPTER : ADDITIONAL PROBLEMS 4 Using N e rt gives d dt [vert r K ert,so ve rt K ert + C v k + Ce rt N K + ce rt N(t.6 Chapter : Additional Problems. False - It is linear, but no initial condition is given. True - y makes it non-linear. False - y( f( is an eplicit solution K + Ce rt 4. The point is to get linear DE into an eact DE by multiplying by an integrating factor. So even if this were true, there would be no reason to do so. 5. False - Solutions can be defined on a restricted portion of the real line 6. True 7. True - dy f(y d 8. y C4 e 4 4 (y 4 ( + Ce 4 4 (Ce 4 y 9. y Ce + y + Ce Ce y. y C y ±(C /, y ± C, y ±(C / y. Valid when C (. y ( + C, (y ( + C ( + C y. We have yy + y. Then y dy y d ydy y d ydy y d ln y + C y e C e / y Ce /, where C ±C y ± Ce / +

50 44 Chapter Review. We have y y with y(. Rewrite the equation as y y /. Separate variables and integrate: y / dy d y / + C It is easiest to apply the initial conditions now: ( / (+C leads to C 6, so 4. We have y y y. Then ( / y y / 6 y / y ( dy d (y +y dy y d (by partial fractions +y + / dy d y ln y + ln y + C ln y y +C y e C e y ± e C e y ± e C y +C where C ±e C 5. We have e ( + d dt. Then d dt e e e d e dt Letting v e, du e, du u dt ln u t + C u e C e t e ±e C e t e ±e C e t ln ±e C e t ln Ce t where C ±e C

51 Chapter Review We have y + y y with y(.5. Then Apply initial conditions: y y y ln y ln y ln + C y y dy d y y ( y y A A( A y dy d Then we have + y + y 7. We have y y e y/. Rewriting we have y y ey/.letu y, dy d du d + u. Then du d + u u e u du d e u e u du d e u ln + C e y e / ln + C e y ln + C ( e / ln + C y ln e / 8. y y (+y ln(+ y which is Homogeneous. Let v y v + v v (+vln(+v dv ( + vln(+v d [u ln(+v du u d u e C ln( + y ec ln( + y ±ec

52 46 Chapter Review 9. We have y y cos ( ln ( y. Then ( ( y y y cos ln y y ( y (ln cos dv d + v v cos(ln(v dv d v(cos(ln(v v dv d v(cos(ln(v v(cos(ln(v dv d ( cot ln(v ln + C ( ( y cot ln ln + C (homogeneous. We have (y + y d dy.weusem(, y y + y and N(, y, so M(t, ty ty + t y t(y + y+tm(, y N(t, ty t t( tn(, y We need to solve dy d y y + dv d + v v + v dv d v dv v d v ln + C Then y ln +C y 4(ln y + C. We have dy d y + y. Then dy ( y ( y d + dv d + v v + v dv v d (homogeneous arcsin(v ln + C ( y arcsin ln + C

53 Chapter Review 47. dy d 4y + y 4(y/ +(y/ Homogeneous (y v v + v 4v +v +v v +v dv d ln ( v ( v ln + C [ ( (y/ ( (y/ ±e C. We have (y/+(y +ln dy. We find ( y y [y +ln so the equation is eact. Then, f 4 f(, y y ln + φ(y f y ln + φ (y y +ln φ(y 4 y4 + C So f(, y y ln + 4 y4 + C ( + y ( 5y dy 4. We have y + y. Then d [ + y y y 6 y [ 5y y 6 y

54 48 Chapter Review so the equation is eact. Solving, we get f y + f(, y y + + φ(y f y y φ (y 5 y φ(y 5 y + C f(, y y y + C + φ (y y 5 y 5. We have ( + y d ydy. Then [ + y y [ y y so the equation is eact. Solving, y y φ (y leadstoφ(y C, so f + y f(, y + yd f(, y + ( y / + φ(y f y y + φ (y y f(, y + ( y / + C 6. y + y y which is Bernoulli n, v y, dv dy y d d.thenwehavedv d + v which is linear with solution v y + C y + C 7. M y y, N y N ( M y N which gives μ( and F (, y y y + C y + (y + which is only a function of

55 Chapter Review dy d y y (Homog (y v v + v v v dv v d v + C y + C 9. Not linear, eact, homogeneous, Bernoulli, seperable. Not linear, eact, homogeneous, Bernoulli, seperable. y + y Linear y C [ e. Linear y e (cos +sin+c [ 5. Linear y e 8 e ( + C 4. Bernoulli n 4,v y,v v (linear v y + C 5. Bernoulli n,v [ y,v +v (linear e v y e ( + C 6. Bernoulli n,v ( y,v v 4 (linear v y e / 4e / + C 7. T (t (T T s e kt + T s 95e kt +5, k.55 soup will be edible in about 4 minutes. 8. T (t 5e kt +7,k.5 coffee is drinkable in about 5 minutes. 9. Let y(t amount of salt at time t, y( 6 dy 6+t dt 5 + t 4 t 5 + t y 9 ln t +5 t + C y( 6 C 46.8 The tank will fill in 5 minutes y(5.4lbs of salt

56 5 Chapter Review 4. y(t lbs of sugar at time t, y( 5 dy 5 + pt p dt t p 5 pt t dy p 5 pt dt t y (5 pln t p ln t pt + C C 5 5 ln We want y(95.5 p.74lbs/min 4. y(t amount of nitric acid, y( ( dy +.t d. 8 t 56 t t y(t 88 ln t +6t + 88 ln The solution reaches % when y(t.t which is when t 8 minutes. Since the tank empties in minutes, this will never happen. 4. If (, y is the point of tangency, then ( dy, is the point of intersection and d y y. Then dy y d y is such a curve.

57 Chapter Geometrical and Numerical Methods for First-Order Equations. Direction Fields the Geometry of Differential Equations. Looking at the point (,, y 9, which matches graph b.. Looking at the point (,, y 8, which matches graph c.. Looking at the point (,, y 9, which matches graph a. 4. Looking at the point (,, y 8, which matches graph d. 5. Graph B 6. Graph C 7. Graph D 8. Graph A 9. y y 4 #9 # y cosy - - 5

58 5 Section.. y siny y e y y

59 Section y cos y sin y e

60 54 Section. 7. y y y + y y ( +(y

61 Section. 55. y e. y y y y

62 56 Section.. y y(y y( + 5. y (y + y Eistence and Uniqueness for First-Order Equations. (a y y, y(, f y, f y eists and is unique on some interval. y. The theorem guarantees that a solution

63 Section. 57 (b y y, y(, f y, f y y.thetheoremdoesnot guarantee that a solution eists or is unique on some interval. (c y y +, y(, f y +, f. Thetheoremdoesnot guarantee that y a solution eists or is unique on some interval.. (a f(, y (y + / f, y, which is discontinuous when y, so a (y + / solution eists and is unique everywhere ecept possibly along y. (b f(, y ( y /5, f y, which is discontinuous when y, so solutions 5( y 4/5 eist and are unique everywhere ecept possibly along y. Alternatively, f(, y ( y, f 5 y, so solutions eist and are unique everywhere. 5 (c f(, y y, f y y, which are both discontinuous when y, so solutions eist and are unique everywhere ecept possibly along y. (d f(, y ( + y f, y, which are both discontinuous when y, so ( + y solutions eist and are unique everywhere ecept possibly along y.. (a f y 4 y( / which is not continuous at y. Hence a solution eists and is unique everywhere ecept possibly along y. Actual solution: y 6. This solution crosses y when. 7 (b f y (/ which, along with f(, y, are continuous everywhere. Hence solution eists and is unique. Actual solution: y. (c f y y( 5/ which is not continuous at y. Hence a solution eists and is unique everywhere ecept possibly along y. ( (/5 5 Actual solution: y (6/5 This solution crosses y when. 6 (d f y y( / which is not continuous at y. Hence a solution eists and is unique everywhere ecept possibly along y. Actual solution: y 6 which crosses y when f(, y y is continuous for y and f y is continuous for y>. Hence, by the y theorem, a solution eists and is unique for y(. Actual solution is given by: y ( f(, y 5(y /5 and y(. y y is discontinuous when y, so solutions (y /5 eist and are unique everywhere ecept possibly along y. Solving for the initial value, y 5(y /5 (y /5 dy 5 d 5 (y /5 5 + C

64 58 Section Applying initial conditions, 5 ( /5 5(+C C (y /5 y ±( 5/ Solutions passing through (, are NOT unique. y ± ( 5/ f y (/ ( y which is continuous everywhere (as is f, therefore, unique solution will eist everywhere. f y (y ( / which is continuous everywhere ecept y, so solutions eist and are unique everywhere ecept possibly along y. Actual solution: ± (/ f y y which is continuous everywhere ecept y, so solutions eist and are unique everywhere ecept possibly along y. ( / Actual solution: + 9. f(, y and f are not continuous at y, so solutions eist and are unique everywhere y ecept possibly along y. Actual solution: y + cos which IS unique.. f y f y lim h f f DNE at y ;ify>, then y andify<, then y. Hence, f(, y + h f(, y DNE. Therefore, the solutions will eist and be unique ecept h possibly at y. If y, y e C e which will never be. If y<, y e C e which will never be. Thus, the only solution to this DE is y which IS unique.

65 .. FIRST-ORDER AUTONOMOUS EQUATIONS GEOMETRICAL INSIGHT 59. First-Order Autonomous Equations Geometrical Insight. y y +,y + End behavior: +y (i y y y - y Root y Multiplicity (ii By the phase line diagram, y (iii y>, y as y<, y as (iv y-plane is an unstable equilibrium point

66 6 Section.. y y +4y +4,y +4y +4,(y + Root y Multiplicity End behavior: y (i y y y y (ii By the phase line diagram, y is a half-stable equilibrium point. (iii y>, y as y<, y as (iv y-plane

67 Section. 6. 6( ( + (i y y y - - y (ii is a stable equilibrium is an unstable equilibrium (iii If >, then (t as t. If < <, then (t ast. If <, then (t ast. (iv y-plane

68 6 Section. 4. ( +( Equilibrium Multiplicity Highest power and coefficient: (+(+ + (i y y y y (ii, unstable; stable (iii For (,, as t For (,, ast For (,, as t (iv y-plane

69 Section ( ( 9 ( ( +( Highest power coefficient: ( ( + 5 (i y y y y (ii, unstable, stable (iii For (,, as t For (,, ast For (,, as t (iv y-plane Equilibrium Multiplicity

70 64 Section. 6. y siny, π <y<π sin y, π <y<π End behavior: trigonometric sine wave (i y y y Roots y π y y π Multiplicity y (ii By the phase line diagram, y π is a stable equilibrium point. y is an unstable equilibrium point. y π is a stable equilibrium point. (iii For y>π, y π as For <y<π, y π as For π <y<, y π as For y< π, y π as (iv y-plane

71 Section y e y / e, e y / e. When y y 4 y 4 (i y y y y (ii By the phase line diagram, y is stable y is unstable (iii For y>, y For <y<, y For y<, y (iv y-plane

72 66 Section. 8. y y, y Root y Multiplicity End behavior: y (i y y y y (ii By the phase line diagram, y is a half-stable equilibrium point. (iii For y>, y as. For y<, y as. (iv y-plane

73 Section ( Highest power and coefficient: ( (i y y y y - Equilibrium Multiplicity (ii is a half-stable point; isastablepoint. (iii For (,, ast. For (,, ast (iv y-plane

74 68 Section.. ( ( ( + Equilibrium Multiplicity Highest power and coefficient: ( ( (+ + 6 (i y y y y - (ii isstable; is half-stable; is unstable. (iii If >, then as t. If < <, then ast. If < <, then ast. If <, then ast. (iv y-plane

75 Section. 69. y cosy +, π <y<π. cosy +,cosy. End behavior: cosine wave (i y y y Roots y π y π Multiplicity y (ii By the phase line diagram, y π is a half-stable equilibrium point. y π is a half-stable equilibrium point. (iii For y>π, y as. For π <y<π, y π as. For y< π, y π as. (iv y-plane

76 7 Section.. ( + (5 + 7 ( ( 4 Equilibrium 5 Multiplicity 7 4 Highest power and coefficient: (+(+ 7 ( ( 4 6 (i y y y y (ii 5, are unstable points;, are stable points; is a half-stable point. (iii If < 5, then as t. If 5 < <, then ast. If < <, then ast. If < <, then ast. If < <, then ast. If >, then ast. (iv y-plane

77 Section. 7. v g k m v. Equilibrium satisfies g (k/mv,sov gm k v g (i v v k m v (ii v gm is a stable equilibrium point. k (iii For v [,, v gm k as t (iv Figures will vary. With g,m.5, k as in eample in. then v g k m v. Equilibrium satisfies g (k/mv,sov gm/k yields v ± gm/k. Note that this is a free-fall problem where v> in the downward direction. We thus ignore v<. (i Graphs will vary gm (ii is a stable equilibrium point k (iii For v [,, v + gm/k as t (iv Graphs will vary 5. ( ( +, ( ( + End behavior: ( ( 5 Roots Multiplicity

78 7 Section (i (ii By the phase line diagram, is a stable equilibrium point; is a half-stable equilibrium point; is a half-stable equilibrium point. (iii For >, ast. For <<, ast. For <<, ast. For <, ast. (iv t-plane

79 Section ( ( + 4, ( ( +4 End behavior: ( ( 5 (i Roots ±i Multiplicity ignore (ii By the phase line diagram, is a stable equilibrium point. (iii For >, ast. For <, ast. (iv t-plane

80 74 Section. 7. y ( y (y +4 Root Multiplicity Highest power and coefficient: ( y (y y 7 (i (ii By the phase line diagram, y is a stable equilibrium point. (iii If y <, then y as. If y >, then y as. (iv t-plane

81 Section y y (4 y(9 y Roots 4 Multiplicity Highest power and coefficient: y ( y( y y 5 (i (ii y, 4arestable;y is half-stable; y is unstable. (iii If y <, y as. If <y <, y as. If <<, y as. If <<4, y 4as. If >4, y 4as. (iv t-plane

82 76 Section ( (, 5 ( ( End behavior: 5 ( ( 9 (i. Roots two normal Multiplicity 5 ignore (ii By the phase line diagram, is a half-stable equilibrium point; is an unstable equilibrium point. (iii For >, as t. For <<, ast. For <, as t. (iv t-plane

83 Section. 77. ( ( + (, ( ( + ( End behavior: (( ( 9 (i Roots two non-real ± Multiplicity ignore (ii By the phase line diagram, is an unstable equilibrium point; is a half-stable equilibrium point; is a stable equilibrium point; is an unstable equilibrium point. (iii If >, as t. If <<, ast. If <<, ast. If <<, ast. If <, as t. (iv t-plane

84 78 Section.. ( (, ( ( End behavior: ( ( 7 (i Roots ± Multiplicity (ii By the phase line diagram, is a stable equilibrium point; equilibrium point; is a stable equilibrium point. (iii If >, ast. If <<, ast. If <<, ast. If <, ast. (iv t-plane is an unstable

85 Section. 79. ( +5( 4 ( +5 End behavior: 8 (i 5 5 Roots 4 5 Multiplicity (ii By the phase line diagram, is a stable equilibrium point; equilibrium point; is a stable equilibrium point. (iii If >, ast. If <<, ast. If <<, ast. If <, ast. (iv t-plane is an unstable

86 8 Section.. y ±, f (y y, f ( Unstablef ( Stable 4. y ±, f (y y, f ( Unstablef ( Stable 5. y, ±π, f (y cosy, f ( Unstablef (±π Stable 6. y, f (y y,f ( Unstable 7. y,f (y y,f ( Inconclusive. However, from the phase line diagram, y is stable. 8. If we Taylor epand the function about y and keep the lowest order non-zero term, we see that we have y f ( (y y as the approimate solution near the equilibrium point. Phase line analysis then shows the equilibrium point is stable. 9. (a y ry y y y.4 y. sqrt r sqrt r y y sqrt r sqrt r y Figure.: Phase line for y ry y : r<, r,r> Figure.: Pitchfork bifurcation for y ry y

87 Section. 8 (b y ry + y y y.4 y. sqrt r sqrt r y y sqrt r sqrt r y Figure.: Phaseline for y ry + y : r<, r, r> Figure.4: Pitchfork bifurcation for y ry + y

88 8 Section.4.4 Population Modeling: An Application of Autonomous Equations. (a is half-stable, is stable. For logistic equation, is unstable and k is stable. Yes, the are different. (b For small, logistic model growth is larger.. -Stable, - Unstable, 6-Stable. -Stable, - Unstable, -Stable 4. (a - Half-stable, - Unstable, 4-Stable (b For small, Allee effect model has larger growth rate. 5. (a Eponential growth - Unlimited growth rate. No limitations placed on organisms. (b Logistic model - Growth rate dependent on factors such as population amount or food availability. (c Allee effect - Growth rate dependent on factors such as population amount or food availability as well as a sufficient population to sustain itself. 6. (a - Unstable, a -Stable, 5 - Unstable (b For > 5, bacteria grows uninhibited. (c The parameter a could represent the strength of the immune system or ability of the body to fight off the given bacteria. A healthy person would likely have an a-value that is closer to than to 5 because the lower value of a represents a lower value of the bacteria (that is stable. 7. (a For a>, - Unstable 5 a -Stable 5+ a - Unstable -Stable (b For a, - Unstable 5 - Half-stable -Stable (c For a<, - Unstable -Stable (d Saddle-node (e Bacteria grows unchecked to a level of. No, the bacteria population eventually reaches and levels off at, which above the fatal level. (f The parameter a could again represent the strength of the immune system or ability of the body to fight off the given bacteria. 8. ( ( 6( 9. ( ( ( ( 8 or ( ( ( ( 8. (a ( ( 4

89 .5. NUMERICAL APPROXIMATION WITH THE EULER METHOD 8 (b - Half-stable, - Half-stable, 4 - Unstable. (a r( a( r, a > (b i. <a<, - Unstable, a -Stable, - Unstable ii. a, - Unstable, - Half-stable iii. a>, - Unstable, -Stable, a - Unstable (c When a<, the bacteria goes to the stable level a if it starts with a level less than and grows without bound otherwise. When a, the bacteria goes to the half-stable level a if it starts with a level less than and grows without bound otherwise. When a>, the bacteria goes to the stable level of if it starts with a level less than a and grows without bound otherwise..5 Numerical Approimation with the Euler Method. dy/d, y( ; eplicit solution: y 4 (4 + Euler Eplicit i y i y( i dy/d 4 y, y( ; eplicit solution: y e (5 /5 Euler Eplicit i y i y( i dy d y cos, h., y( ; eplicit solution y +sin. +(.(. +(.(. +(.(. 4 +(.(4.4 y y +(.( ( cos(.9 y.9+(.( (.9 cos(..894 y (.( (.894 cos(..756 y (.( (.756 cos(..699

90 84 Section.5 Eplicit: y( y(. y(. y(. y(.4 +sin +sin( sin( sin( sin( Euler Eplicit i y i y( i y sin, y(π,h.. y π π +. π +. π +. 4 π +.4 y ( sin(π y +(. ( sin(π +. y +( ( sin(π +. y (. ( ( sin(π +. y (. ( Eplicit: y ( 4cos /4 y(π y(π +. ( 4cos(π +. / y(π +. ( 4cos(π +. / y(π +. ( 4cos(π +. / y(π +.4 ( 4cos(π +.4 /4.996

91 Section.5 85 Euler Eplicit i y i y( i π π π π π dy/d ye, y( ; eplicit solution: y ep( e Euler Eplicit i y i y( i dy/d y, y( ; eplicit solution: y Euler Eplicit i y i y( i dy/d y +cos, y( ; eplicit solution: y (sin cos + e Euler Eplicit i y i y( i dy/d y +sin, y( ; eplicit solution: y (cos +sin 5e

92 86 Section.5 Euler Eplicit i y i y( i dy/d e y, y( ; eplicit solution: y ln( + e Euler Eplicit i y i y( i dy/d + y, y( ; eplicit solution: y +e Euler Eplicit i y i y( i dy/d ( +(y +, Euler Eplicit y( ; eplicit solution: y tan( + i y i y( i