MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

Transcription

1 MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by y 1 y = e, > 0 µ = e 1 d = e ln = 1 Muliplying boh sides of he ODE by µ, we obain ( ) d 1 d y = e, which has he general soluion y() = e + C, where C is a consan As, we see ha y() if C > 0, y() 0 if C = 0, and y() if C < 0 Boyce and DiPrima, p 40, Problem 18 Soluion: Firs we pu he given linear firs-order ODE in sandard form: An inegraing facor is given by y + y = sin, > 0 µ = e d = e ln = Muliplying boh sides of he ODE by µ, we obain d ( y ) = sin d 1

2 Inegraing boh sides of he preceding equaion, we ge y() = sin d = d( cos ) = cos + cos d = cos + sin + C, where C is a consan From he iniial condiion y(π/) = 1, we deduce ha C = (π/) 1 Hence he soluion o he given iniial-value problem is: y = 1 ( ) π 1 + sin cos 4 3 Boyce and DiPrima, p 40, Problem 0 Soluion: In sandard form he given equaion reads: ( y ) y = 1, > 0 An inegraing facor is given by ( ( µ = exp ) Muliplying boh sides of he ODE by µ, we obain ) d = e +ln = e e ln = e d ( e y ) = e d Inegraing boh sides of he preceding equaion wih respec o yields or e y = e e + C, y = Ce, where C is a consan o be deermined by he iniial condiion From he iniial condiion ha y(ln ) = 1, we obain C = Hence he soluion o he given iniialvalue problem is: y = e

3 4 Boyce and DiPrima, p 40, Problem 8 Soluion: I is easy o solve he iniial-value problem y + 3 y = 1 1, y(0) = y o o ge he soluion y() = (y o 1 8 )e /3 Indeed, µ = e /3 is an inegraing facor Muliplying boh sides of he given equaion by µ, we obain (ye /3 ) = e /3 1 e/3 Inegraing boh sides of he preceding equaion, we ge ye /3 = 3 e/3 1 e /3 d = 3 e/3 1 ( 3 e/3 9 ) 4 e/3 + C, where we have used inegraion by pars Using he iniial condiion o pu C in erms of y o leads o he soluion given above If here is a = τ a which he soluion ouches he -axis bu does no cross i, hen a = τ we have y(τ) = 0 and y (τ) = 0 From he given differenial equaion, we observe ha τ = I follows ha from which we obain y() = (y o 1 8 )e 4/3 = 0, y o = 1 9e4/3 8 = Boyce and DiPrima, p 41, Problem 30 Soluion: Muliplying boh sides of he ODE y y = sin by he inegraing facor µ = e, we obain (e y) = e + 3e sin Inegraing boh sides of he preceding equaion, we ge e y = e + 3 e sin d = e + 3I, (1) 3

4 where I = e sin d Using inegraion by pars wice, we have I = e d( cos ) = e cos e cos d ( ) = e cos e sin + e sin d = e (sin + cos ) I + C 1 Hence I = 1 (e (sin + cos ) + C 1 ) Subsiuing his expression of I ino (1), we obain he general soluion of he given differenial equaion: y = 1 3 ( e (sin + cos ) ) + Ce, where C is a consan Imposing he iniial condiion y(0) = y o leads o he soluion of he given iniial-value problem: y() = 1 3 ( (sin + cos ) + y o + 5 ) e For a bounded soluion, we mus have y o = 5/ 6 Boyce and DiPrima, p 48, Problem Soluion: From y = x, we ge y(1 + x 3 ) I follows ha he general soluion is: ydy = x 1 + x 3 dx y = 1 3 ln 1 + x3 + C The given differenial equaion requires ha y 0 and x 1 7 Boyce and DiPrima, p 48, Problem 16 (a), (c) Soluion: From y = x(x + 1), we ge 4y 3 which has he general soluion 4y 3 dy = (x 3 + x)dx, y 4 = x4 4 + x + C 4

5 The iniial condiion y(0) = 1/ dicaes ha C = 1/4 Hence we have y 4 = x4 + x + 1 = (x + 1) 4 4 I follows ha he soluion o he given iniial-value problem is: x + 1 y = The domain of he soluion is clearly (, ) 8 Boyce and DiPrima, p 49, Problem Soluion: The given equaion y = 3x /(3y 4) is separable Separaing he variables and inegraing boh sides of he equaion, we ge (3y 4) dy = 3x dx + C or y 3 4y = x 3 + C, where C is a consan o be deermined from he iniial condiion From he iniial condiion y(1) = 0, we obain C = 1 Hence he required soluion is given implicily by he equaion y 3 4y = x 3 1 A glance a he given differenial equaion reveals ha y ± as y ±/ 3 When y = / 3, x = [1 16/(3 3)] 1/3 176; when y = / 3, x = [1 + 16/(3 3)] 1/ Hence he approximae inerval on which he soluion is defined is ( 176, 1598), which conains he poin x = 1 9 Boyce and DiPrima, p 49, Problem 4 Soluion: I is easy o solve he iniial-value problem The soluion is or in explici form y = ex, y(0) = y 3y + y = x e x + 1 y = x ex The soluion y assumes is maximum value when he funcion f(x) = 13/4 + x e x is a is absolue maximum Noe ha f (x) = 0 implies = e x or x = ln, and f (ln ) = < 0 Hence f has only one local maximum, which is locaed a x = ln Since f(ln ) = 13/4 + (ln 1) > f(0) = 9/4, f aains is maximum value a x = ln 5

6 10 Boyce and DiPrima, p 51, Problem 38 Soluion: (a) The given differenial equaion can be pu in he form dy dx = f(y/x), where f(y/x) = 3(y/x) 1 (y/x) Hence he given differenial equaion is homogeneous (b) The subsiuion y = vx reduces he given differenial equaion o he form which is equivalen o v x + v = 3 v 1 v, x dv dx = 1 ( v 1 ) v = 1 v 1 v I is easy o see ha v = 1 and v = 1 are special soluions of he preceding differenial equaion To seek oher soluions (ie, v ±1), we pu he separable equaion in he form v v 1 dv = dx x The general soluion o he preceding equaion is: ln v 1 = ln x + C 1 or ln y x = C 1, which can be pu in he form y x = C, where C = ec 1 is a posiive consan x 3 Noe ha if we pu C = 0, hen we ge y = ±x, which are none oher han he special soluions v = ±1 Thus he formula y x = C x 3, where he consan C 0, includes all soluions of he given differenial equaion x 3 6