520 CHAPTER TEN /SOLUTIONS Solutions for Section 10.4 dP = 0:02P implies that dP = 0:02 dt: dt P R dP R P = 0:02 dt implies that ln jP j = 0:02t + C: jP j = e0:02t+C implies that P = Ae0:02t ; where (A0:02=)0eC : 0:02t We are given P (0) = 20. Therefore, P (0) = Ae = A = 20. So the solution is P = 20e : dQ Q dQ dt 2. dt = 5 implies that Q = 5 : R dQ R dt 1 Q = 5 implies that ln jQj = 5 t + C: 1 1 1 So jQj = e 5 t+C = e 5 teC implies that Q = Ae 5 t ; where A = eC : From the initial conditions we know that ( 15 )0 Q(0) = 50, so Q(0) = Ae = A = 50. Thus Q = 50e 51 t : 3. dm dt = 3m: As in problems 1 and 2, we get m = Ae3t : Since m = 5 when t = 1; we have 5 = Ae3 ; so A = e53 : Thus m = e53 e3t = 5e3t,3 : 1. 4. 5. 6. 7. R dI R = 0:2 dx implies that 0:2I implies that dI I I = 0:2 dx implies that ln I = 0:2x + C . So I = Ae0:2x , where A = eC . According to the given boundary condition, I ( 1) = 6. Therefore, I ( 1) = Ae0:2(,1) = Ae,0:2 = 6 implies that A = 6e0:2 : Thus I = 6e0:2 e0:2x = 6e0:2(x+1) : R dy R 1 dy y dy y dx + 3 = 0 implies dx = 3 implies y = 3 dx: dI dx = , , , , 1 1 Integrating and moving terms, we have y = Ae, 3 x : Since y(0) = A = 10; we have y = 10e, 3 x : 1 dz z dt 5 implies dz z = 5 dt: Integrating and moving terms, we have z = Ae5t . Using the fact that z (1) = 5; we have z (1) = Therefore, z = e55 e5t = 5e5t,5 : dP dt = = R dP Ae5 = 5; so A = e5 : 5 P + 4 implies that PdP+4 = dt: P +4 = R dt implies that ln jP + 4j = t + C: + 4 = Aet implies that Therefore P = 104et 4: P 8. jj , dy = 2y , 4 = 2(y , 2). dx P = Aet , 4: P = 100 when t = 0; so P (0 ) = Ae0 , 4 = 100; and A = 104: R dy R Factoring out a 2 makes the integration easier: ydy ,2 = 2 dx implies that y,2 = 2 dx implies that ln y 2 = 2x + C: y 2 = e2x+C implies that y 2 = Ae2x where A = eC : The curve passes through (2,5), which means 3 = Ae4; so A = e34 : Thus, y = 2 + e34 e2x = 2 + 3e2x,4 : j , j 9. 10. , Factoring out the 0.1 gives dm :1m + 200 dt R= 0dm R = 0:1(m + 2000). 0:1t dm m+2000 = 0:1 dt implies that m+2000 = 0:1 dt, so ln m + 2000 = 0:1t + C: So m = Ae initial condition, m(0) = Ae(0:1)0 2000 = 1000; gives A = 3000: Thus m = 3000e0:1t 2000: dB + 2B = 50 implies dB = 2B + 50 = 2(B 25) implies R dB = R 2 dt: dt dt B,25 , , j , j , , After integrating and doing some algebra, we have B 25 = so A = 75e2 : Thus B = 25 + 75e2 e,2t = 25 + 75e2,2t : 11. j ,j , Ae,2t : Using the initial condition, we have 75 = Ae,2; We know that the general solution to a differential equation of the form dy = k(y , A) dt is , , 2000: Using the y = Cekt + A: 10.4 SOLUTIONS Thus, in our case, we get y = Cet=2 + 200: We know that at t = 0 we have y = 50, so solving for C we get y = Cet=2 + 200 50 = Ce0=2 + 200 ,150 = Ce0 C = ,150: Thus we get 12. y = 200 , 150et=2: We know that the general solution to a differential equation of the form dQ = k(Q , A) dt is H = Cekt + A: To get our equation in this form, we factor out a 0:3 to get dQ = 0:3 Q 120 = 0:3(Q dt 0:3 , , 400): Thus, in our case, we get Q = Ce0:3t + 400: We know that at t = 0 we have Q = 50 so solving for C we get Q = Ce0:3t + 400 50 = Ce0:3(0) + 400 ,350 = Ce0 C = ,350: 13. Thus we get Q = 400 , 350e0:3t : Rearrange and write Z or Z dr , ln j1 , Rj = r + C which can be written as 1 or dR = 1,R 1 , R = e,C ,r = Ae,r R(r) = 1 , Ae,r : The initial condition R(1) = 0:1 gives 0:1 = 1 , Ae,1 and so A = 0:9e: Therefore R(r) = 1 , 0:9e1,r : 521 522 14. CHAPTER TEN /SOLUTIONS Z Write and so ln y dy = 1 1 dt 3+t jyj = ln j3 + tj + C or ln where ln D = C . Therefore The initial condition y(0) = 1 gives D = Z jyj = ln Dj3 + tj y = D(3 + t): 1 3 and so y(t) = 13 (3 + t): 15. R = 2 16. 17. R tez implies e,z dz = tdt implies e,z dz = t dt implies ,e,z = t2 + C: Since the solution passes through the origin, z = 0 when t = 0, we must have ,e,0 = 02 + C; so C = ,1: Thus ,e,z = t2 , 1; or z = , ln(1 , t2 ): R R dy=dx = 5y=x implies dy=y = 5dx=x. So ln jyj = 5 ln jxj + C = 5 ln x + C implies jyj = e5 ln x eC , and thus y = Ax5 where A = eC . Since y = 3 when x = 1, so A = 3. Thus y = 3x5: dy = y2 (1 + t) implies that R dy = R (1 + t) dt implies that , 1 = t + t + C implies that y = , 1 . dt y 2 y t+ t +C 2 = , t +2t,4 : Since y = 2 when t = 1, then 2 = , 1+ 1+C : So 2C + 3 = ,1; and C = ,2: Thus y = , t 1 +t,2 dz dt 2 2 2 2 2 2 1 2 18. 19. R 2 2 R 2 z + zt2 = z (1 + t2) implies that dzz = (1 + t2 )dt implies that ln jz j = t + t3 + C implies that z = Aet+ t : z = 5 when t = 0; so A = 5 and z = 5et+ t : dw = w2 sin 2 implies that R dw = R sin 2 d implies that , 1 = , 1 cos 2 + C . According to the initial d w 2 w conditions, w(0) = 1; so ,1 = , 12 + C and C = , 12 : Thus , w1 = , 12 cos 2 , 12 implies that w1 = cos 2 +1 implies that w = cos 2 +1 : du = u2 implies R du = R dx = R ( 1 , 1 )dx implies , 1 = ln jxj , ln jx + 1j + C: x(x + 1) dx x(x+1) x 1+x u u u(1) = 1; so , 11 = ln j1j, ln j1 + 1j + C: So C = ln 2 , 1: Solving for u yields , u1 = ln jxj, ln jx + 1j + ln 2 , 1 = ln jx2j+xj1j , 1, so u = ln j ,x1 j,1 . x Separating variables and integrating with respect to r gives dz dt = 3 3 3 3 3 2 2 2 20. 2 2 +1 21. Z z dz = 1 so that ln Z (1 + r2 )dr jz j = r + 13 r3 + C: The initial condition z (0) = 1 gives C = 0 so that z (r) = er+(1=3)r : 3 22. Separating variables and integrating with respect to Z 1 gives Z w2 dw = Now set 2 = t, then this becomes Z 1 w2 dw = 2 1 cos Z 2 d : cos tdt 10.4 SOLUTIONS and so , w1 = 12 sin t + C 2 w = sin(, t) + D or Using the initial conditions give D = 23. 24. 25. 26. 27. 28. 29. 30. 32. w = sin ,2 2+ D : ,2, so the solution is w = sin ,22, 2 : dR = kR implies that dR = k dt which implies that R dR = R k dt. Integrating gives ln jRj = kt + C; so jRj = dt R R ekt+C = ekt eC . R = Aekt ; where A = eC : dQ , Q = 0 so dQ = Q : This is now the same problem as Problem 25, except the constant factor on the right is 1 dt k dt k k 1 instead of k: Thus the solution is Q = Ae k t for any constant A: dP = P , a, implying that dP = dt so R dP = R dt. Integrating yields ln jP , aj = t + C; so jP , aj = et+C = dt P ,a P ,a et eC . P = a + Aet; where A = eC or A = 0. dQ = b , Q implies that dQ = dt which, in turn, implies R dQ = R dt. Integrating yields , ln jb , Qj = t + C; so dt b,Q b,Q jb , Qj = e,(t+C ) = e,te,C . Q = b , Ae,tR; where A = e,C or A = 0. dP = k(P , a),so dP = k dt, so R dP = k dt. Integrating yields ln jP , aj = kt + C so P = a + Aekt where dt P ,a P ,a A = eC or A = 0. dR = aR + b: If a = 0; then this is just dR = b; where b is a constant. Thus in this case R = bt + C is a solution for dt dt any constant C: b If a 6= 0; then dR dt = a(R + a ): Now this is just the same as Problem 27, except here we have a in place of k and , ab in place of a, so the solutions are R = , ab + Aeat where A can be any constant. dy = y(2,y) which implies that dy = , dt, implying that R dy = , R dt, so , 1 R ( 1 , 1 )dy = , R dt. dt y(y,2) (y ,2)(y ) 2 y y,2 2j Integrating yields 12 (ln jy , 2j , ln jyj) = ,t + C , so ln jyj, = , 2 t + 2 C . yj 2 Exponentiating both sides yields j1 , y2 j = e,2t+2C ) y2 = 1 , Ae,2t ; where A = e2C . Hence y = 1,Ae ,2t . But 2 2 y(0) = 1,A = 1; so A = ,1, and y = 1+e,2t : R cos x R 1 2 ln t dx 1+2 ln t t dx dt = (1 + 2 ln t) tan x 2implies that tan x = ( t ) dt which implies that sin x dx = ( t + t ) dt. j j ln sin x = ln t + (ln t) + C: 2 sin x = eln t+(ln t) +C = t(eln t )ln t eC = t(tln t )eC . So sin x = At(ln t+1) , where A = eC . Therefore x = ln t+1 arcsin(At ). dx = x ln x , so R dx = R dt and thus ln ln x = ln t + C , so ln x = eC eln jtj = eC t . Therefore ln x = At, where dt t x ln x t A = eC , so x = eAt : dy = y ln( y ) implies that dyy = dt implies that R dyy = R ( dt): dt 2 y ln( ) y ln( ) j 31. 523 j j , j jj j , j jj , 2 2 Substituting w = ln( y2 ); dw = y1 dy gives: R R dw that ln jwj = ln j ln( y2 )j = ,t + C . Since y(0) = 1; we have C w = (, dt) implies y ln(ln 2): Thus ln j ln( )j = ,t + ln(ln 2), or 2 j ln( y2 )j = e,t +ln(ln 2) Again, since y(0) = 1, we see that not satisfy y(0) = 1.) = j j ln ln 12 = ln j , ln 2j = e,t = (ln 2) , ln(y=2) = (ln 2)e,t and thus y = 2(2,e,t ). (Note that ln(y=2) = (ln 2)e,t does 524 33. CHAPTER TEN /SOLUTIONS (a), (b) y x (c) dy dx = (b). 34. R x y , so R y dy = x dx and thus y2 2 = x2 + 2 C , or y2 , x2 = 2C: This is the equation of the hyperbolas in (a), (b) y x (c) 35. , xy , which implies that A y = x ; where A = eC : = R dy y = , R dx , ln jxj+C x ; so ln jyj = , ln jxj+C implies that jyj = e jxj),1eC : =( By looking at the slope fields, we see that any solution curve of y0 = yx intersects any solution curve to y0 = xy : Now if the two curves intersect at (x; y); then the two slopes at (x; y) are negative reciprocals of each other, because y 1 x=y = x . Hence, the two curves intersect at right angles. , 36. dy dx , , If y(x) = xv(x), then y0 = v + xv0 and the differential equation becomes Solving for v0 gives + xv 1+v v + xv0 = xx , xv = 1 , v : dv = 1 + v , v = 1 + v2 x dx 1,v 1,v 10.4 SOLUTIONS 525 which is a separable equation. Rearranging and integrating with respect to x gives Z Z 1 v 1 dv = dx 1 + v2 x Z Z Z 1 dv 12 1 +2vv2 dv = x1 dx 1 + v2 1 ln(1 + v2 ) = ln x + C: arctan(v) 2 , , , Now put v = y x to find arctan 37. jj Starting off with the homogeneous equation we replace y(x) with xv(x) and get y 1 2 2 x , 2 ln(x + y ) = C: f xy = dy dx ; y dy d(xv(x)) dv f (v) = f xv =f x x = dx = dx = v + x dx : Thus we have dv f (v) , v = x dx which is clearly separable. 38. Making the substitution s = x + a, t = y + b. the differential equation becomes dy = x + y + a + b + 5 : dx x , y + a , b , 1 Now select a and b so that a + b + 5 = 0 and a , b , 1 = 0, that is, a = ,2, b = ,3. The differential equation becomes dy = x + y : dx x , y Now set y(x) = xv(x), then y0 = v + xv0 and the differential equation becomes + xv 1+v v + xv0 = xx , xv = 1 , v : Solving for v0 gives dv = 1 + v , v = 1 + v2 x dx 1,v 1,v which is a separable equation. Rearranging and integrating with respect to x gives Z Z 1,v 1 dv = 1 + v2 x dx Z Z Z 1 1 2v 1 dv , dv = 1 + v2 2 1 + v2 x dx 1 arctan(v) , ln(1 + v2 ) = ln jxj + C: 2 Now put v = xy to find y 1 2 2 arctan x , 2 ln(x + y ) = C: Transforming back to the original variables x = s + 2 and y = t + 3 gives t + 3 , 1 ln((s + 2)2 + (t + 3)2) = C: arctan s+2 2

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# R dP RR dQ R dt R dI RR dy y = ,R 1 R dP RR dy RR dm RR dB