EXERCISES FOR SECTION 1.5

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1 1.5 Exisence and Uniqueness of Soluions v c v c Graph of approximae soluion obained using Euler s mehod wih = 0.1. Graph of approximae soluion obained using Euler s mehod wih = 0.1. EXERCISES FOR SECTION Since he consan funcion y 1 () = 3 for all is a soluion, hen he graph of any oher soluion y() wih y(0) <3 canno cross he line y = 3 by he Uniqueness Theorem. So y() <3 for all in he domain of y(). 2. Since y(0) = 1 is beween he equilibrium soluions y 2 () = 0andy 3 () = 2, we mus have 0 < y() <2 for all because he Uniqueness Theorem implies ha graphs of soluions canno cross (or even ouch in his case). 3. Because y 2 (0) <y(0) <y 1 (0), we know ha 2 = y 2 () <y() <y 1 () = + 2 for all. This resrics how large posiive or negaive y() can be for a given value of (ha is, beween 2 and + 2). As, y() beween 2 and + 2(y() as a leas linearly, bu no faser han quadraically). 4. Because y 1 (0) <y(0) <y 2 (0), he soluion y() mus saisfy y 1 () <y() <y 2 () for all by he Uniqueness Theorem. Hence 1 < y() <1 + 2 for all. 5. The Exisence Theorem implies ha a soluion wih his iniial condiion exiss, a leas for a small -inerval abou = 0. This differenial equaion has equilibrium soluions y 1 () = 0, y 2 () = 1, and y 3 () = 3 for all. Sincey(0) = 4, he Uniqueness Theorem implies ha y() >3 for all in he domain of y(). Also,dy/ > 0 for all y > 3, so he soluion y() is increasing for all in is domain. Finally, y() 3as. 6. Noe ha dy/ = 0ify = 0. Hence, y 1 () = 0 for all is an equilibrium soluion. By he Uniqueness Theorem, his is he only soluion ha is 0 a = 0. Therefore, y() = 0 for all. 7. The Exisence Theorem implies ha a soluion wih his iniial condiion exiss, a leas for a small -inerval abou = 0. Because 1 < y(0) <3andy 1 () = 1andy 2 () = 3 are equilibrium soluions

2 44 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS of he differenial equaion, we know ha he soluion exiss for all and ha 1 < y() <3 for all by he Uniqueness Theorem. Also, dy/ < 0for1< y < 3, so dy/ is always negaive for his soluion. Hence, y() 1as,andy() 3as. 8. The Exisence Theorem implies ha a soluion wih his iniial condiion exiss, a leas for a small - inerval abou = 0. Noe ha y(0) <0. Since y 1 () = 0 is an equilibrium soluion, he Uniqueness Theorem implies ha y() <0 for all. Also,dy/ < 0ify < 0, so y() is decreasing for all,and y() as increases. As, y() (a) To check ha y 1 () = 2 is a soluion, we compue dy 1 = 2 and y1 2 + y 1 + 2y = ( 2 ) 2 + ( 2 ) + 2( 2 ) = 2. To check ha y 2 () = is a soluion, we compue dy 2 = 2 and y2 2 + y 2 + 2y = ( 2 + 1) 2 + ( 2 + 1) + 2( 2 + 1) 2 = (b) The iniial values of he wo soluions are y 1 (0) = 0andy 2 (0) = 1. Thus if y() is a soluion and y 1 (0) = 0 < y(0) <1 = y 2 (0), hen we can apply he Uniqueness Theorem o obain y 1 () = 2 < y() < = y 2 () for all. Noe ha since he differenial equaion saisfies he hypohesis of he Exisence and Uniqueness Theorem over he enire y-plane, we can coninue o exend he soluion as long as i does no escape o ± in finie ime. Since i is bounded above and below by soluions ha exis for all ime, y() is defined for all ime also. (c) y y 2 () y 1 ()

3 1.5 Exisence and Uniqueness of Soluions (a) If y() = 0 for all, hendy/ = 0and2 y() =0 for all. Hence, he funcion ha is consanly zero saisfies he differenial equaion. (b) Firs, consider he case where y > 0. The differenial equaion reduces o dy/ = 2 y.ifwe separae variables and inegrae, we obain y = c, where c is any consan. The graph of his equaion is he half of he parabola y = ( c) 2 where c. Nex, consider he case where y < 0. The differenial equaion reduces o dy/ = 2 y. If we separae variables and inegrae, we obain y = d, where d is any consan. The graph of his equaion is he half of he parabola y = (d ) 2 where d. To obain all soluions, we observe ha any choice of consans c and d where c d leads o a soluion of he form (d ) 2, if d; y() = 0, if d c; ( c) 2, if c. (See he following figure for he case where d = 2andc = 1.) 4 y (c) The parial derivaive f/ y of f (, y) = y does no exis along he -axis. (d) If y 0 = 0, HPGSolver plos he equilibrium soluion ha is consanly zero. If y 0 = 0, i plos a soluion whose graph crosses he -axis. This is a soluion where c = d in he formula given above. 11. The key observaion is ha he differenial equaion is no defined when = 0. (a) Noe ha dy 1 / = 0andy 1 / 2 = 0, so y 1 () is a soluion.

4 46 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS (b) Separaing variables, we have dy y = 2. Solving for y we obain y() = ce 1/,wherec is any consan. Thus, for any real number c, define he funcion y c () by 0 for 0; y c () = ce 1/ for > 0. For each c, y c () saisfies he differenial equaion for all = 0. 4 y There are infiniely many soluions of he form y c () ha agree wih y 1 () for < 0. (c) Noe ha f (, y) = y/ 2 is no defined a = 0. Therefore, we canno apply he Uniqueness Theorem for he iniial condiion y(0) = 0. The soluion y c () given in par (b) acually represens wo soluions, one for < 0 and one for > (a) Noe ha and dy 1 dy 2 = d = d ( ) 1 = 1 1 ( 1) 2 = (y 1()) 2 ( ) 1 = 1 2 ( 2) 2 = (y 2()) 2, so boh y 1 () and y 2 () are soluions. (b) Noe ha y 1 (0) = 1andy 2 (0) = 1/2. If y() is anoher soluion whose iniial condiion saisfies 1 < y(0) < 1/2, hen y 1 () <y() <y 2 () for all by he Uniqueness Theorem. Also, since dy/ < 0, y() is decreasing for all in is domain. Therefore, y() 0as, and he graph of y() has a verical asympoe beween = 1an = 2.

5 48 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS 15. (a) The equaion is separable. We separae, inegrae (y + 2) 2 dy =, and solve for y o obain he general soluion y() = (3 + c) 1/3 2, where c is any consan. To obain he desired soluion, we use he iniial condiion y(0) = 1 and solve 1 = (3 0 + c) 1/3 2 for c o obain c = 27. So he soluion o he given iniial-value problem is y() = (3 + 27) 1/3 2. (b) This funcion is defined for all. However, y( 9) = 2, and he differenial equaion is no defined a y = 2. Sricly speaking, he soluion exiss only for > 9. (c) As, y().as 9 +, y() (a) The equaion is separable. Separaing variables we obain (y 2) dy =. Solving for y wih help from he quadraic formula yields he general soluion y() = 2 ± 2 + c. To find c, wele = 1andy = 0, and we obain c = 3. The desired soluion is herefore y() = (b) Since is always posiive and y() <2 for all, he soluion y() is defined for all real numbers. (c) As ±, Therefore, lim y() =. ± 17. This exercise shows ha soluions of auonomous equaions canno have local maximums or minimums. Hence hey mus be eiher consan or monoonically increasing or monoonically decreasing. A useful corollary is ha a funcion y() ha oscillaes canno be he soluion of an auonomous differenial equaion. (a) Noe dy 1 / = 0a = 0 because y 1 () has a local maximum. Because y 1 () is a soluion, we know ha dy 1 / = f (y 1 ()) for all in he domain of y 1 (). In paricular, so f (y 0 ) = 0. 0 = dy 1 = f (y 1 ( 0 )) = f (y 0 ), =0

6 1.5 Exisence and Uniqueness of Soluions 49 (b) This differenial equaion is auonomous, so he slope marks along any given horizonal line are parallel. Hence, he slope marks along he line y = y 0 mus all have zero slope. (c) For all, dy 2 = d(y 0) = 0 because he derivaive of a consan funcion is zero, and for all f (y 2 ()) = f (y 0 ) = 0. So y 2 () is a soluion. (d) By he Uniqueness Theorem, we know ha wo soluions ha are in he same place a he same ime are he same soluion. We have y 1 ( 0 ) = y 0 = y 2 ( 0 ). Moreover, y 1 () is assumed o be a soluion, and we showed ha y 2 () is a soluion in pars (a) and (b) of his exercise. So y 1 () = y 2 () for all. Inoherwords,y 1 () = y 0 for all. (e) Follow he same four seps as before. We sill have dy 1 / = 0a = 0 because y 1 has a local minimum a = (a) Solving for r,wege Consequenly, r = ( ) 3v 1/3. 4π s() = 4π ( ) 3v 2/3 4π = cv() 2/3, where c is a consan. Since we are assuming ha he rae of growh of v() is proporional o is surface area s(),wehave dv = kv 2/3, where k is a consan. (b) The parial derivaive wih respec o v of dv/ does no exis a v = 0. Hence he Uniqueness Theorem ells us nohing abou he uniqueness of soluions ha involve v = 0. In fac, if we use he echniques described in he secion relaed o he uniqueness of soluions for dy/ = 3y 2/3, we can find infiniely many soluions wih his iniial condiion. (c) Since i does no make sense o alk abou rain drops wih negaive volume, we always have v 0. Once v > 0, he evoluion of he drop is compleely deermined by he differenial equaion. Wha is he physical significance of a drop wih v = 0? I is emping o inerpre he fac ha soluions can have v = 0 for an arbirary amoun of ime before beginning o grow as a saemen ha he rain drops can sponaneously begin o grow a any ime. Since he model gives no informaion abou when a soluion wih v = 0 sars o grow, i is no very useful for he undersanding he iniial formaion of rain drops. The safes asserion is o say is he model breaks down if v = 0.