CHAPTER 77 SOLUTION OF FIRST-ORDER DIFFERENTIAL EQUATIONS BY SEPARABLE VARIABLES

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1 CHAPTER 77 SOLUTION OF FIRST-ORDER DIFFERENTIAL EQUATIONS BY SEPARABLE VARIABLES EXERCISE 95 Page 84. Skeh a famil of urves represened b eah of he following differenial equaions: (a) = 6 (b) = () = + (a) If = 6, hen = 6d = 6 + There are an infinie number of graphs of = 6 + ; hree urves are shown below. (b) If =, hen = d = + A famil of hree pial urves is shown below. 9 04, John Bird

2 () If = +, hen = ( + )= + + A famil of hree pial urves is shown below.. Skeh he famil of urves given b he equaion = + deermine he equaion of one of hese urves whih passes hrough he poin (, ). If = +, hen = ( d ) + = + + If he urve passes hrough he poin (, ) hen = = Hene, = ( ) ( ) + + = = + A famil of hree urves is shown below, inluding = +, whih passes hrough he poin (, ) 9 04, John Bird

3 EXERCISE 96 Page 86. Solve: = os 4 Sine = os 4 = ( ) os 4 = sin Solve: = Sine = = = = d Hene, = d = ln 6 +. Solve: + =, given = when = If + =, hen = = ( ) d= + If = when =, hen = + from whih, = Hene, = 4. Solve: dθ + sin θ = 0, given = when θ = π Sine d dθ + sin θ = 0 dθ = sin θ = sinθd θ = ( os θ) + = osθ , John Bird

4 = when θ = π π hene, = os + = + = = Hene, = osθ + 5. Solve: e + =, given = when = 0 If + = hen e d Hene, ( ) e d = = ( e ) = e d e = + + If = when = 0, hen = ( ) + = Thus, = + e + or e 6. The gradien of a urve is given b + =. Find he equaion of he urve if i passes hrough he poin (, ). If d Hene, + =, hen d = = d= + 6 If i passes hrough,, = = Thus, Hene, = 6 + from whih, = + = 6 6 = 94 04, John Bird

5 7. The aeleraion a of a bod is equal o is rae of hange of veloi, d v. Find an equaion for v in d erms of, given ha when = 0, veloi v = u. d v d = a hene, v = a d = a + When = 0, veloi v = u, hene, u = 0 + from whih, = u Hene, veloi, v = a + u or v = u + a 8. An obje is hrown veriall upwards wih an iniial veloi u of 0 m/s. The moion of he obje follows he differenial equaion d s d = u g, where s is he heigh of he obje in meres a ime seonds g = 9.8 m/s. Deermine he heigh of he obje afer seonds if s = 0 when = 0 If d s d = u g, hen s = ( ) g u g d = u + Sine s = 0 when = 0, hen = 0 Hene, g s = u if u = 0 s = 9.8, hen s = u = The heigh when =, s = (0) 4.9( ) heigh = = 5.9 m 95 04, John Bird

6 EXERCISE 97 Page 87. Solve: = + Rearranging = + gives: d = d + Inegraing boh sides gives: = + Thus, b using he subsiuion u = ( + ), = ln( + ) +. Solve: = os If os = hen d os = se = an = +. Solve: ( + ) = 5, given = when = If ( ) + = 5 hen d + = 5 + = 5d + ln = 5+ = when =, hene, 5 ln + = + from whih, = 5 = + ln = , John Bird

7 4. The urren in an eleri irui is given b he equaion Ri + L d i d = 0, where L R are onsans. Show ha i = Ie R/L, given ha i = I when = 0 If Ri + L d i d from whih, = 0, hen d i L = Ri d di Ri = d L di R = d i L Thus, R ln i = + L i = I when = 0, hus ln I = Hene, ln i = R + ln I L ln i ln I = R L ln i = R I L Taking ani-logarihms gives: i I d i = R d i L e R = L e R L i = I 5. The veloi of a hemial reaion is given b d = k(a ), where is he amoun ransferred in ime, k is a onsan a is he onenraion a ime = 0 when = 0. Solve he equaion deermine in erms of. If d = k(a ), hen d a = k a = kd ln(a ) = k + = 0 when = 0, hene Thus, ln a = ln(a ) = k ln a 97 04, John Bird

8 ln a ln(a ) = k a e k an = a e k a ln = k a a a = e k = a ae k = a a = a( e k ) 6. (a) Charge Q oulombs a ime seonds is given b he differenial equaion R d Q d + Q C = 0, where C is he apaiane in farads R he resisane in ohms. Solve he equaion for Q given ha Q = Q 0 when = 0 (b) A irui possesses a resisane of 50 0 ohms a apaiane of farads, afer 0. seonds he harge falls o 8.0 C. Deermine he iniial harge he harge afer seond, eah orre o signifian figures. d Q Q (a) If R + = 0 hen d C dq Q = d RC dq = d Q RC ln Q = + k RC Q = Q 0 when = 0, hene, ln Q 0 = k Hene, ln Q = + ln Q0 RC ln Q ln Q 0 = RC ln Q = Q RC 0 Q Q 0 = e RC Q= Q0 e (b) R = 50 0 Ω, C = F, = 0. s Q = 8.0 C 98 04, John Bird

9 0. Hene, = Q0e ( ) = Q from whih, iniial harge, Q 0 = = 9.0 C When = s, harge, Q = 0 e Q = 9.0e 50 0 = 5.8 C 7. A differenial equaion relaing he differene in ension T, pulle ona angle θ oeffiien of friion µ is d T = µt. When θ = 0, T = 50 N, µ = 0.0 as slipping sars. Deermine he ension dθ a he poin of slipping when θ = radians. Deermine also he value of θ when T is 00 N. Sine d T dθ = µt hen dt T = µ dθ dt = T µ dθ ln T = μθ + When θ = 0, T = 50 N, µ = 0.0, hene ln 50 = (0.0)(0) + from whih, = ln 50 Hene, ln T = μθ + ln 50 ln T ln 50 = μθ T ln = µθ 50 from whih, T µθ = e T = 50 e µθ 50 When θ = radians, ension, T = 50 e ( ) = 50e = 7. N 0.0 () When T = 00 N, 00 = 50 e(0.0) θ = e (0.0) θ = e(0.0) θ 50 Hene, ln = ln [e (0.0) θ ] = 0.0 θ from whih, ona angle, θ = ln 0.0 =. rad 99 04, John Bird

10 8. The rae of ooling of a bod is given b d θ = kθ, where k is a onsan. If θ = 60 C when = d minues θ = 50 C when = 5 minues, deermine he ime aken for θ o fall o 40 C, orre o he neares seond. If d θ = kθ hen d dθ k d θ = dθ θ = kd ln θ = k + When θ = 60 C, =, ln 60 = k + () When θ = 50 C, = 5, ln 50 = 5k + () () () gives: ln 60 ln 50 = k from whih, k = Subsiuing in (): 60 ln = ln 60 = ( ) + from whih, = ln 60 + ( ) = 4.59 Hene, ln θ = k + = When θ = 40 C, ln 40 = ime, = 4.59 ln 40 = 8.67min = 8 min 40 s , John Bird

11 EXERCISE 98 Page 80. Solve: = os Sine = os hen d os d = = os d ln = sin +. Solve: ( ) = ( + ), given = when = If ( ) ( ) = +, hen ( ) ( = + ) d d = + + = when =, hene, 4 = + + from whih, = 0 Thus, = +. Solve: = e, given = 0 when = 0 = e = (e )(e ), b he laws of indies Separaing he variables gives: e = e d e d = e d Inegraing boh sides gives: Thus he general soluion is: e d = e e = e + d When = 0, = 0, hus: e 0 = e , John Bird

12 from whih, = = Hene he pariular soluion is: e = e + 4. Solve: ( ) + ( + ) = 0, given = when = If ( ) + ( + ) = 0 hen ( + ) = ( ) = ( ) Thus, + ( ) d = d + = d ln + = ln + = when =, hene, ln + = ln + from whih, = Thus, ln + = ln or ln + ln = ln + ln = ( ) ln = 5. Show ha he soluion of he equaion + + = is of he form + + = onsan Sine + d = + d hen d = d = + + ln ( ) ln ( ) + + = ln ( ) ln ( ) + ln = , John Bird

13 or ln + + = + + = e = a onsan 6. Solve = ( ) for, given = 0 when = Sine = ( ) hen ( ) = = ( ) For = le u = ( ) from whih, d u = d = d u du u u Hene, = = d u = ln u = ln ( ) if = hen ln = ln( ) + = 0 when =, hene, ln = ln + from whih, = 0 Hene he pariular soluion is: ln = ln( ) ln ln = ( ) = ( ) = ( ) = ( ) 0 04, John Bird

14 7. Deermine he equaion of he urve whih saisfies he equaion =, whih passes hrough he poin (, ). Sine = hen = d= d = ln + If he urve passes hrough (, ) hen = = hene, Thus, or = ln + = ln+ from whih, = = ln + 8. The p.d., V, beween he plaes of a apaior C harged b a sead volage E hrough a resisor R is given b he equaion d V d + V = E (a) Solve he equaion for V given ha a = 0, V = 0 (b) Calulae V, orre o signifian figures, when E = 5 vols, C = farads, R = 00 0 ohms =.0 seonds. (a) Sine dv V E d + = hen dv EV = d dv d = E V E V = + k from whih, ln ( ) A = 0, V = 0, hene, ln E = k E V = ln E Thus, ln ( ) = ln E ln ( E V) 04 04, John Bird

15 E ln = E V E EV = e E e = E V V = E E = E Ee e V = E e vols.0 6 (b) Volage, e 0 0 V = E = 5 e 00 0 = ( e ) =. V 9. Deermine he value of p, given ha = p, ha = 0 when = when = 6 Sine = p hen p p d = d= d d = ( ) p d = 0 when =, hene, = 0 when = 6, hene, () () gives: p = + p = + + p 0 = () p 0 = () 0 = p p 0 = + 9 p = from whih, p = , John Bird