P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD JWDD027-Salas-v1 November 25, :21. is a solution. (e x +1) 2 + 1

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1 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: CHAPTER 9 SECTION SECTION 9.. y (x) = ex/ ; y y = ( ) e x/ e x/ =0; y is a solution. y (x) =x + e x/ ; y y = ( x + e x/) ( x +e x/) =4x x 0; y is not a solution.. y + xy = xe x / + xe x / = 0; not a solution y + xy = Cxe x / + x + Cxe x / = x; y is a solution. 3. y (x) = ex (e x +) ; y + y = ex (e x +) + e x + = (e x +) = y ; y is a solution. y (x) = Cex (Ce x +) ; y + y = Cex (Ce x +) + Ce x + = (Ce x +) = y ; is a solution. y 4. y +4y = 8 sin x + 8 sin x =0; y is a solution. y +4y = cos x + 8 cos x = 6 cos x; not a solution. 5. y (x) =e x, y =4e x ; y 4y =4e x 4e x =0; y is a solution. y (x) =C cosh x, y =4C sinh x; y 4y =4C sinh x 4C sinh x =0; is a solution. y 6. y y 3y = e x +8e 3x ( e x +6e 3x ) 3(e x +e 3x ) = 0; not a solution y y 3y = 7 [ (6+9x)e 3x (+3x)e 3x 3xe 3x] =7e 3x ; y is a solution y y =; H(x) = ( ) x = x, integrating factor: e x e x y e x y = e x x [ e x y ] = e x e x y = e x + C y = + Cex 8. y x y = ; H(x) = x, integrating factor: x x x y x 3 y = x x (x y)= x x y = x + C y = x + Cx

2 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 48 SECTION y + 5 ( ) 5 y =; H(x) = x = 5 x, integrating factor: e5x/ e 5x/ y + 5 e5x/ y = e 5x/ [ ] e 5x/ y = e 5x/ x e 5x/ y = 5 e5x/ + C y = 5 + Ce 5x/ 0. y y = e x ; H(x) =. y y = x; H(x) = x, integrating factor: e x e x y e x y = e x ( e x y ) = e x x e x y = e x + C y = e x + Ce x ( ) x = x, integrating factor: e x e x y e x y = e x xe x [ e x y ] = e x xe x x e x y = e x + xe x + e x + C = xe x + C y = x + Ce x. y + x y = cos x x ; H(x) = x =ln x, integrating factor: x x x y +xy = cos x x [x y] = cos x x y = sin x + C y = sin x x + C x 3. y 4 ( x y = n; H(x) = 4 ) x = 4 lnx =lnx 4, x integrating factor: e ln x 4 = x 4 x 4 y 4 x x 4 y = nx 4 [ x 4 y ] = nx 4 x x 4 y = 3 nx 3 + C y = nx + Cx4 3

3 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 4. y + y =+x; H(x) = x, integrating factor: e x SECTION y e x y =0; H(x) = e x y + e x y =(+x)e x x (ex y) = ( + x)e x e x y =xe x + C y =x + Ce x e x x = e x, integrating factor: e ex e ex y e x e ex y =0 [ e e x y ] =0 x e ex y = C 6. y y = e x ; H(x) = y = Ce ex x, integrating factor: e x 7. y + +e x y = +e x ; H(x) = integrating factor: e H(x) = ex +e x e x +e x y + +e x e x y e x y = x (e x y)= e x y = x + C y = xe x + Ce x +e x x =ln e x +e x, x e x +e x y = +e x ] e x = ( + e x ) [ e x +e x y e x +e x e x +e x y = +e x + C y = e x + C ( + e x ) This solution can also be written: y =+K (e x +), where K is an arbitrary constant. 8. y + x y = +x x ex ; H(x) = x, integrating factor: x x xy + y =(+x)e x (xy) =(+x)ex x xy = xe x + C y = e x + C x

4 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 484 SECTION y +xy = xe x ; H(x) = xx= x, integrating factor: e x 0. y x y =lnx; H(x) =. y + y =0; H(x) = x + e x y +xe x y = x [ ] e x y = x x e x y = x + C ( y = e x x + C ) x x, integrating factor: x x y x y = x ln x x integrating factor: e ln(x+) =(x +) ( x y ) = x ln x x y = (ln x) + C y = x(ln x) + Cx x + x = ln(x + ) = ln(x +), (x +) y +(x +)y =0 [ (x +) y ] =0 x (x +) y = C C y = (x +). y + x + y =(x +)5/ ; H(x) = x, integrating factor: (x +) x + (x +) y +(x +)y =(x +) 9/ [ (x +) y ] =(x +) 9/ x (x +) y = (x +)/ + C y = (x +)7/ + C(x +) 3. y + y = x; H(x) = x = x, integrating factor : e x e x y + e x y = xe x x [ex y]=xe x e x y = xe x e x + C y =(x ) + Ce x

5 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: SECTION y(0) = +C = = C =. Therefore, y =e x + x is the solution which satisfies the initial conition. 4. y y = e x ; H(x) = x, integrating factor: e x x (e x y)=e x e x y = e x + C y = e x + Ce x =y() = e + Ce = C = e an y = e x + e e x is the solution which e e satisfies the initial conition. 5. y + y = +e x ; H(x) = x = x, integrating factor : e x e x y + e x y = x [ex y]= ex +e x ex +e x e x y =ln(+e x )+C y = e x [ln ( + e x )+C] y(0)=ln+c = e = C = e ln. Therefore, y = e x [ln ( + e x )+e ln ] is the solution which satisfies the initial conition. 6. y + y = +e x ; H(x) = x, integrating factor: e x x (ex y)= ex +e x e x y = ln( + ex )+C [ ] y = e x ln( + ex )+C e = y(0) = ln 3 + C = C = e ln 3 an y = [ e x ln( + ex )+e ln 3] is the solution which satisfies the initial conition. 7. y x y = x e x ; ( H(x) = ) x = lnx =lnx, x integrating factor: e ln x = x x y x 3 y = e x x [ x y ] = e x x y = e x + C y = x (e x + C)

6 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 486 SECTION 9. y() = e + C =0 = C = e. Therefore, y = x (e x e) is the solution which satisfies the initial conition. 8. y + x y = e x ; H(x) = x, integrating factor: x x x (x y)=x e x x y = e x (x +x +)+C y = e x x (x +x +)+ C x =y() = 5e + C = C =5e an y = e x x (x +x +)+ 5e x solution which satisfies the initial conition. is the 9. Set z = y y. Then z = y y. y y = y y = z = z = z = C e x Now, z = y y = C e x = e x y e x y = C = (e x y) = C = e x y = C x + C = y = C xe x + C e x 30. General solution: y = Ce rx. (a) y(a) =0=Ce ra = C =0 = y(x) = 0 for all x. (b) r<0 an y 0 = y as x (c) r>0 an y 0 = y 0asx () If r = 0, then y(x) = C, constant. 3. (a) Let y an y be solutions of y + p(x)y = 0, an let u = y + y. Then u + pu =(y + y ) + p (y + y ) = y + y + py + py = y + py + y + py =0+0=0 Therefore u is a solution. (b) Let u = Cy where y is a solution of y + p(x)y = 0. Then u + pu =(Cy) + p(cy)=cy + Cpy = C(y + py) =C 0=0 Therefore u is a solution.

7 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 3. (a) y + p(x)y =0 (b) y(b) =0 = Ce b e x x p(t) t a y + p(x)e p(t) t a y =0 [ x e p(t) t a y] =0 e x a p(t) t y = C y(x) =Ce x p(t) t a a p(t) t =0 = C =0 = y(x) = 0 for all x. SECTION (c) Let z = y y. Then z is a solution of y + p(x)y =0.If y (b) =y (b), then z(b) =0 = z(x) = 0 for all x. x 33. Let y(x) =e H(x) q(t) e H(t) t. a a Note first that y(a) =e H(a) q(t) e H(t) t =0 so y satisfies the initial conition. a Now, x ] x y + p(x)y = [e H(x) q(t) e H(t) t + p(x) e H(x) q(t) e H(t) t a = e H(x) q(x) e H(x) + e H(x) [ p(x)] x a a x q(t) e H(t) t + p(x) e H(x) q(t) e H(t) t = q(x) x Thus, y(x) =e H(x) q(t) e H(t) t is the solution of the initial value problem. 34. Let z = y y. Then a z = y y = q py (q py )= p (y y )= pz = z + pz =0 a 35. Accoring to Newton s Law of Cooling, the temperature T at any time t is given by T (t) = 3 + [7 3]e kt We can etermine k by applying the conition T (/) = 50 : Therefore, T (t) = 3+40e.5970t. 50=3+40e k/ e k/ = 8 40 = 9 0 k = ln(9/0) k = ln(9/0) =.5970

8 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 488 SECTION 9. Now, T () = e.5970 = 40.00; the temperature after minute is (approx.) To fin how long it will take for the temperature to reach 35, we solve for t: 3+40e.5970t = e.5970t =35 40 e.5970t =3.5970t =ln(3/40) t = ln(3/40).5970 =.6 It will take approximately.6 minutes for the thermometer to rea By (9..4) T (t) = 00 80e kt T () = = 00 80e k = = k = ln(39/40) = T (6) = 00 80e 0.066(6) = 5.85 C; T (t) =90 = 80e 0.066t = 0 = t = 64.5 secs. 37. (a) The solution of the initial value problem v =3 kv, (k >0) v(0) = 0 is: v(t) = 3 ( e kt ). k (b) At each time t, e kt <. Therefore v(t) = 3 ( e kt ) < 3 an lim v(t) = 3 k k t k (c) 38. (a) P t +(b a)p =0; H(t) = (b a) t =(b a)t, integrating factor : e (b a)t P (0) = P 0 = P (t) =P 0 e (a b)t. (b a)t P e t +(b a)e(b a)t P =0 [ ] e (b a)t P =0 t e (b a)t P = C P = Ce (a b)t

9 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: SECTION (b) (i) a>b= P 0 e (a b)t is increasing. P (t) as t. (ii) a = b = P (t) =P 0 is a constant. (iii) a<b= P 0 e (a b)t is ecreasing. P (t) 0 as t. 39. (a) i t + R L i = E R L ; H(t) = L t = R L t, integrating factor : e R L t e R L t i t + R L e R L t i = E L e R L t t [ ] e R L t i = E L e R L t e R L t i = E R e R L t + C i(0) = 0 = C = E R, so i(t) =E R i(t) = E R + Ce R L t [ e (R/L) t]. ( (b) lim i(t) = lim ) E t t R e (R/L) t = E R amps (c) i(t) =0.9 E R = e (R/L) t = 0 = R L t = ln 0 = t = L R ln 0 secons. 40. (a) i t + R L i = E R sin ωt; H(t) = L L t = R L t, integrating factor : e R L t i(0) = i 0 e R L t i t + R L e R L t i = E L e R L t sin ωt t [ e R L t i ] = E L e R L t sin ωt e R L t i = E L e R L t L [ ] R R + ω L sin ωt ω cos ωt + C L [ ] EL R i(t) = R + ω L sin ωt ω cos ωt + Ce R L t L [ ] [ ] EL R EL = i(t)= R + ω L sin ωt ω cos ωt + i 0 + ω L R + ω L e R L t. (b) lim t oes not exist because the trigonometric functions continue to oscillate.

10 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 490 SECTION 9. (c) A sample graph is: 4. (a) V (t) =kv (t) = V (t) =V 0 e kt (b) Loses 0% in 5 minutes, so V (5) = V 0 e 5k =0.8V 0 = k = 5 ln 0.8 = V (t) =V 0 e 5 (ln 0.8)t ( = V 0 e ln 0.8 ) t/5 = V0 (0.8) t/5 ( = V 4 t/5 0 5). Since V 0 = 00 liters, we get V (t) = 00 ( 4 5 ) t/5 V (t) =ktv (t) V (t) ktv (t) =0 e kt / V (t) kte kt / V (t) =0 [ ] e kt / V (t) =0 t e kt / V (t) =C V (t) =Ce kt /. V (0) = C = 00 = V (t) = 00e kt /. V (5) = 60 = 00e k(5/) = 60, e k(5/) = 4 5, ek = ( 4 /5 5). Therefore V (t) = 00 ( ) 4 t /5 5 liters. 4. Let s(t) be the number of pouns of salt present after t minutes. Since ( ) s(t) s (t) = rate in rate out = 3 (0.) 3, 00 we have Multiply by e 0.03t = e 0.03t : s (t)+0.03s(t) =0.6. e 0.03t s (t)+0.03e 0.03t s(t) =0.6e 0.03t [ e 0.03t s(t) ] =0.6e 0.03t t e 0.03t s(t) =0e 0.03t + C s(t) = 0 + Ce 0.03t. Use the initial conition s(0) = 00(0.5) = 5 to etermine C: 5=0+Ce 0 so C =5. Thus, s(t)=0+5e 0.03t lb.

11 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 43. (a) P t = k(m P ) (b) P t + kp = km; H(t) = kt= kt, integrating factor : e kt kt P e t + kekt P = km e kt [ e kt P ] = km e kt t e kt P = Me kt + C P = M + Ce kt SECTION P (0) = M + C =0 = C = M an P (t) =M ( e kt) P (0) = M ( e 0k) =0.3M = k = an P (t) =M ( e t) (c) P (t) =M ( e t) =0.9M = e t =0. = t = 65 Therefore, it will take approximately 65 ays for 90 % of the population to be aware of the prouct. 44. (a) Q = rate in rate out = r kq, k > 0 t (b) Q t + kq = r, Q(0) = 0 = Q(t) = r ( e kt ) k (c) lim t Q(t) = r k. 45. (a) P t cos(πt)p =0 = P = Ce π sin πt. P (0) = C = 000 = P = 000e π sin πt. (b) P t cos(πt)p = 000 cos πt = P = Ce π sin πt 000. P (0) = 000 = C = 000 = P = 000e π sin πt (a) Let Q =lnp. Then Q t = P P t = a bq. Solving the ifferential equation Q t + bq = a P (0) = P 0 = e c = P 0 e a b. Thus P = e a b = Q = a b + Ce bt, so P = e a b +Ce bt. [ ] P0 e a e bt b.

12 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 49 SECTION 9. (b) e bt 0ast, so P e a b. ( ) b (c) P = P (a b ln P ) = P = P P + P (a b ln P )=P(a bln P )(a b b ln P ). P If 0 <P <e a/b, then P is increasing an the graph is concave up; if e a/b <P <e a/b, then P is increasing an the graph is concave own; if e a/b <P, then P is ecreasing an the graph is concave own. () SECTION 9.. y = y sin(x +3) y = sin(x +3)x y y y = sin(x +3)x ln y = cos(x +3)+C This solution can also be written: y = Ce (/) cos(x+3).. y =(x + )(y + y) y y + y = (x +)x ln y y + = x3 3 + x + C This solution can also be written: y = Ke x x3 /3 (K = ec ). 3. y =(xy) 3 y 3 y = x3 x, y 3 y = x 3 x y 0 y = 4 x4 + C This solution can also be written: x 4 + y = C, or y = C x 4 ;

13 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 4. y =3x ( + y ) y +y = 3x x tan y = x 3 + C y = tan(x 3 + C) 5. y = sin(/x) x y cos y y cos yy = sin(/x) x x y cos yy = sin(/x) x x y sin y + cos y = cos(/x)+c SECTION y +y y = y = y + y + yx +x x ln +y =ln +x + C +y = K( + x) (K =lnc) 7. y = xe y+x e y y = xe x x e y y = xe x x e y = xe x e x + C e y = e x xe x + C This solution can also be written: y = ln(e x xe x + C). 8. This solution can also be written: y = xy x y +=(x )(y ) y y = x x ln y y + =ln x + C y = +Kex x Ke x x (K = e C ). 9. (y ln x)y (y +) = x y (y +) y = x ln x x y (y +) y = x ln x x ln y + + =ln ln x + C y +

14 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 494 SECTION e y sin xx+ cos x(e y y) y =0 sin x cos x x + (e y ye y ) y = C cos x + e y + e y ( + y) =C (y ln x)y = y + x y y + y = x ln x x y y + y = x ln x x ln (y +)=ln ln x + K =ln Cln x (K =ln C ) ln (y +)=ln C ln x =ln(cln x) y = C(ln x). y = +y y sin x y y = csc xx +y 4 ln( + y )=ln csc x cot x + C the integral curves can be written as: ln( + y )=ln [ C(csc x cot x) 4], or as y = K(csc x cot x) y = x y y = y = y y x, y(0) = 0 x x x x x x sin y = x + C Thus, arcsin y = x. y(0) = 0 = arcsin 0 = +C = C = 4. y = ex y +e x e y e x y = +e x x e y =ln(+e x )+C y()=0 = = ln( + e)+c = C = ln( + e) an e y =ln(+e x )+ ln( + e)

15 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 5. y = x y y y +, y(3) = y + y =(x ) x, y 0 y y + y = (x ) x y y +ln y = 3 x3 x + C y(3) = = +ln= 3 (3)3 3+C = C = 5. Thus, y +ln y = 3 x3 x 5. SECTION x y = y xy y y = ( x)x x ln y = ln x + C x =y( ) = C = an ln xy + x = 7. (xy + y + x +)x +(y ) y =0, y() = 0 (x + )(y +)x +(y ) y =0 (x +)x + y y y =0 + y (x +)x + y + y = C x + x + ln (y +) tan y = C y()=0 = C =4. Thus, x + x + ln (y +) tan y =4 8. cos yx+(+e x ) sin yy =0 x sin y +e x + cos y y = C 9. y =6e x y, y(0) = 0 y =6e x y e y y =6e x x e y =3e x + C y(0) = 0 = =3+C = C = Thus, e y =3e x = y =ln [ 3 e x ] ln(e x +)+ln sec y = C; π 4 = y(0) = ln+ln =C ln(e x +)+ln sec y = 3 ln

16 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 496 SECTION xy y =x y, y() = xy = y ( +x ) +x y = x = y x ln y =ln x + x + C ( ) x +x x y() = = 0=+C = C = Thus, ln y =ln x + x or y = xe x. We assume that C = 0 at time t =0. (a) Let A 0 = B 0. Then Integrating, we get C t = k(a 0 C) an (A 0 C) C = kt A 0 C = kt + M C (A 0 C) = k t. M a constant. Since C(0) = 0, M = A 0 an Solving this equation for C gives A 0 C = kt + A 0. C(t) = ka 0t +ka 0 t. (b) Suppose that A 0 B 0. Then C t = k(a C 0 C)(B 0 C) an (A 0 C)(B 0 C) = k t. Integrating, we get (A 0 C)(B 0 C) C = kt ( ) B 0 A 0 A 0 C C = kt B 0 C [ ln (A 0 C)+ln(B 0 C)] = kt + M B 0 A 0 ( ) B0 C ln = kt + M M an arbitrary constant B 0 A 0 A 0 C Since C(0) = 0, M = B 0 A 0 ( ln B0 A 0 ) B 0 A 0 ln Solving this equation for C, gives an ( B0 C A 0 C ) = kt + ln(b 0/A 0 ) B 0 A 0. C(t) = A 0B 0 ( e ka 0 t e kb 0t ) A 0 e ka 0t B 0 e kb 0t.

17 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9:. (a) From (9..4) with K =0.000, M = 800, R = P (0) = 00, we have SECTION (b) P (t) = 80, e.6 t (c) P is maximal at t t =.6 Maximum value = (a) m v = αv βv t v v(α + βv) = m t v(α + βv) v = m t α v v β α α + βv v = m t α ln v α ln(α + βv) = m t + M, M a constant ( ) v ln = α α + βv m t + αm v α + βv = [ Ke αt/m K = e ] αm Solving this equation for v we get v(t) = αk e αt/m βk = α Ce αt/m β [C =/K]. (b) Setting v(0) = v 0, we get C = α + βv 0 an v 0 αv 0 v(t) = (α + βv 0 )e αt/m βv 0 (c) lim t v(t) =0 4. F = ma = m v t (a) m v t = mg βv t = mv mg βv = m ( ) v β v c v t = m β v c v v = m v c β = m [ ( )] vc + v ln + C v c β v c v ( v c + v + v c v ) v

18 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 498 SECTION 9. (b) At t =0, v(0) = v 0. Therefore C = m [ ( )] vc + v 0 ln v c β v c v 0 Thus t = v c + v v c + v 0 vc v 0 v c v = etg/v c v c + v = v c + v 0 e tg/v c (v c v) v c v 0 [ ( ) ] [( vc + v 0 v + e tg/v vc + v c 0 = v c v c v 0 v c v 0 ]. v = v c [ (vc + v 0 )e tg/v c (v c v 0 ) (v c + v 0 )e tg/v c +(vc v 0 ) = m [ ln v c β ( vc v 0 v c + v 0 )]. m [ ( )] vc + v ln vc v 0 = v [ ( )] c vc + v ln vc v 0. v c β v c + v 0 v c v g v c + v 0 v c v v c = mg/β ) ] e tg/v c We can bring the hyperbolic functions into play by writing [ (vc + v 0 )e gt/v c (v c v 0 )e gt/v ] c v = v c (v c + v 0 )e gt/v c (vc v 0 )e gt/v c [ ] v0 cosh(gt/v c )+v c sinh(gt/v c ) = v c v 0 sinh(gt/v c )+v c cosh(gt/v c ) { } v c v 0 (c) a = g [v 0 sinh(gt/v c )+v c cosh(gt/v c )] The acceleration can not change sign since the enominator is always positive an the numerator is constant. As t, the enominator, an the fraction 0. () We can write [ (vc + v 0 ) (v c v 0 )e tg/v ] c v = v c. (v c + v 0 )+(v c v 0 )e tg/v c As t, gt/v c an e tg/v c 0. Thus v v c 5. (a) Let P = P (t) enote the number of people who have the isease at time t. Then, substituting into (9..4) with M =5, 000 an R = 00, we get 5, 000(00) P (t) = 00 + (49, 00)e 5,000kt = 5, e 5,000kt. 5, 000 P (0) + 49e = 400 = 5, 000k 5,000(0k) = Therefore, P (t) = 5, e 0.398t.

19 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: SECTION (b) 5, e t =, 500 = t = 40; It will take 40 ays for half the population to have the isease. (c) 6. y t = ky(m y) =kmy ky = y t y t > 0 for 0 <y<m y, so t y t < 0 for M <y<m, so y t =(km ky)y t = k (M y)(m y)y is increasing is ecreasing Therefore infecte. y t is maximal at y = M. The isease is spreaing fastest when half the population is 7. Assume that the package is roppe from rest. (a) Let v = v(t) be the velocity at time t, 0 t 0. Then 00 v t = 00g v or v t + 50 v = g (g =9.8 m/sec ) This is a linear ifferential equation; e t/50 is an integrating factor. t/50 v e t + 50 et/50 v = ge t/50 t [ ] e t/50 v = ge t/50 e t/50 v =50ge t/50 + C v =50g + Ce t/50 Now, v(0) = 0 = C = 50g an v(t) =50g ( e t/50). At the instant the parachute opens, v(0) = 50g ( e /5) = 50g(0.83) = 88.8 m/sec. (b) Now let v = v(t) enote the velocity of the package t secons after the parachute opens. Then 00 v t = 00g 4v or v t = g 5 v

20 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 500 SECTION 9. This is a separable ifferential equation: v t = g 5 v u g u = 5 t g ln u + g u g = t 5 + K ln u + g u g = g 5 t + M u + g u g = gt/5 = Ce Ce.5 t set u = v/5, u=(/5)v u = g Ce.5 t + Ce.5t v =5 g Ce.5 t + Ce.5t Now, v(0) = 88.8 = 5 g C + C =88.8 = C =.43. Therefore, v(t) =5 g.43e.5 t +.43e.5t = 5.65 ( ) +0.70e.5 t 0.70e.5 t (c) From part (b), lim v(t) =5.65 m/sec. t 8. (a) By the hint C (A 0 ) = C kt A 0 C = kt + K. First, C(0) = 0 = K =/A 0. Then, C() = A 0 = k =/A 0. Thus, A 0 C = ( ) t (t +), which gives C(t) =A 0. A 0 t + (b) By the hint C (A 0 )(A C 0 ) = C kt A 0 A 0 C A 0 C = C kt

21 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: A 0 [ ln A 0 C +ln A 0 ] C = kt + K A 0 ln First, C(0) = 0 = K = A 0 ln. Then, A 0 C A 0 = kt + K. C SECTION Thus, C() = A 0 = A 0 ln 3 = k + A 0 ln = k = A 0 ln 3. A 0 ln A 0 C [ A 0 C = t ln 3 A 0 + ln = ln A 0 A 0 ( ) ] t 3 so that A 0 C ( ) t 3 A 0 C = 3 t t an therefore C(t) =4A 0 (3 t ) t. (c) By the hint [ (m + n)ln A 0 (m n) A 0 (m n) ( A 0 m A 0 First, C(0) = 0 = K = m + n A 0 (m n) ln C() = A 0 = k = m + n A 0 (m n) ln m m + n C m m + n C C )( A 0 n ) = kt m + n C n A 0 n m + n C C = kt A 0 m m + n C +(m + n)ln A 0 n ] m + n C = kt + K A 0 A 0 m + n A 0 (m n) ln =0. Then, A 0 n m + n A 0 A 0 m m + n A 0 A 0 n m + n C A 0 m m + n C = kt + K. = m + n A 0 (m n) ln m n.

22 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 50 SECTION 9. Thus, so that m + n A A 0 (m n) ln 0 n m + n C A 0 m m + n C = m + n ( m ) A 0 (m n) ln (t)+0 n A 0 n m + n C ( m ) [ t A 0 m m + n C = an therefore C(t) =A 0 (m + n) n m t n t m t+ n t+ ]. PROJECT 9.. (a) x +3y = C = +3y =0 = y = 3 The orthogonal trajectories are the solutions of: y = 3. y = 3 = y = 3 x + C (b) Curves: y = Cx, y = C = y x orthogonal trajectories: y = x y yy+ xx= K ; x + y = K (= K )

23 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: SECTION (c) xy = C = y + xy =0 = y = y x The orthogonal trajectories are the solutions of: y = x y. y = x y xx = yy x = y + C or x y = C () y = Cx 3, y =3Cx = 3y x orthogonal trajectories: y = x 3y 3yy+ xx= K ; 3y + x = K (= K ) (e) y = Ce x = y = Ce x = y The orthogonal trajectories are the solutions of: y = y. y = y yy = x y = x + K or y = x + C (f) x = Cy 4, =4Cy 3 y ; y = y 4x orthogonal trajectories: yy+ y x = 4x y 4xx= K ; y +4x = K (= K )

24 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 504 SECTION 9.3. (a) Curves: y x = C; yy x =0 = y = x y orthogonal trajectories: y = y x ; y y = x x; y = C x (b) Curves: y = Cx 3, yy =3Cx ; y = xyy = y = 3y 3 x orthogonal trajectories: y = x 3y ; 3yy+ xx= C; x + 3 y = C or x +3y = C (c) Curves: y = Cex x, xy = Cex ; xy + y = Ce x = y = orthogonal trajectories: y = x y( x) ; yy= y(x ) x ( ) x x x = x y = ln x x + C or y +x +ln( x) = C () Curves: e x sin y = C; e x cos yy + e x sin y =0 = y = sin y cos y orthogonal trajectories: y = cos y sin y ; 3. (a) A ifferential equation for the given family is: sin y cos y y = y =xyy + y (y ) x; x; ln sec y = x + C or sec y = Ce x A ifferential equation for the family of orthogonal trajectories is foun by replacing y by /y. The result is: y = xy y + y (y ) which simplifies to y =xyy + y (y ) Thus, the given family is self-orthogonal. (b) x C + y C = = x 4 C + yy C 4 =0 = C = A ifferential equation for the given family is: 4x x + yy x + xyy xy y y =4 A ifferential equation for the family of orthogonal trajectories is foun by replacing y by /y. The result is: Thus, the given family is self-orthogonal. x xy y + xyy y =4

25 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: SECTION 9.3 SECTION The characteristic equation is: r +r 8=0 or (r + 4)(r ) = 0. The roots are: r = 4,. The general solution is: y = C e 4x + C e x.. r 3r +4=0 = r =6, 7; y = C e 6x + C e 7x 3. The characteristic equation is: r +8r +6 = 0 or (r +4) =0. There is only one root: r = 4. By Theorem II, the general solution is: y = C e 4x + C xe 4x. 4. r +7r +3=0 = r = 7 37 ± ; y = C e x + C e The characteristic equation is: r +r +5=0. x. The roots are complex: r = ± i. By Theorem III, the general solution is: 6. r 3r +8=0 = r = 3 ± 3 7. The characteristic equation is: y = e x (C cos x + C sin x). ( i; y = e3x/ C cos 3 3 ) x + C sin x r +5r 3=0 or (r )(r +3)=0. The roots are: r =, 3. The general solution is: y = C e x/ + C e 3x. 8. r = 0 = r = ± 3; y = C e 3x + C e 3x. 9. The characteristic equation is: r +=0. The roots are complex: r = ± 3 i. The general solution is: y = C cos 3 x + C sin 3 x.

26 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 506 SECTION r 3r =0 = r = 3 ; y = C e 3 x + C xe 3 x.. The characteristic equation is: 5r + 4 r 3 4 = 0 or 0r +r 3=(5r )(4r +3)=0. The roots are: r = 5, 3 4. The general solution is: y = C e x/5 + C e 3x/4.. r +3r =0 = r =0, 3 ; y = C + C e 3 x. 3. The characteristic equation is: r +9=0. The roots are complex: r = ±3i. The general solution is: y = C cos 3x + C sin 3x. 4. r r 30 = 0 = r =6, 5; y = C e 6x + C e 5x. 5. The characteristic equation is: r +r +=0. The roots are complex: r = ± i. The general solution is: y = e x/ [C cos(x/) + C sin(x/)]. 6. r 4r +4=0 = r =; y = C e x + C xe x. 7. The characteristic equation is: 8r +r =0 or (4r )(r +)=0. The roots are: r = 4,. The general solution is: y = C e x/4 + C e x/. 8. 5r r +=0 = r = 5 ± ( 5 i; y = ex/5 C cos x 5 + C sin x ) The characteristic equation is: r 5r + 6 = 0 or (r 3)(r ) = 0. The roots are: r = 3,. The general solution an its erivative are: y = C e 3x + C e x, y =3C e 3x +C e x. The conitions: y(0) =, y (0) = require that C + C = an 3C +C =. Solving these equations simultaneously gives C =, C =. The solution of the initial value problem is: y =e x e 3x.

27 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: SECTION r +r +=0 = r = ; y = C e x + C xe x =y() = C e +C e, =y () = C e C e = C = 5e,C =3e = y = 5e x +3xe x.. The characteristic equation is: r + 4 =0. The roots are: r = ± i. The general solution an its erivative are: y = C cos(x/) + C sin(x/) y = C sin(x/) + C cos(x/). The conitions: y(π) =, y (π) = require that C = an C =. The solution of the initial value problem is: y = cos(x/) + sin(x/).. r r +=0 = r =± i; y = e x (C cos x + C sin x). =y(0) = C, =y (0) = C + C = C =0, y = e x cos x 3. The characteristic equation is: r +4r + 4 = 0 or (r +) =0. There is only one root: r =. The general solution an its erivative are: y = C e x + C xe x y = C e x + C e x C xe x. The conitions: y( ) =, y ( ) = require that C e C e = an C e +3C e =. Solving these equations simultaneously gives C =7e, C =5e. The solution of the initial value problem is: y =7e e x +5e xe x =7e (x+) +5xe (x+). 4. r r +5=0 = r =± i; y = e x (C cos x + C sin x). 0=y(π/) = e π/ ( C ) = C =0; =y (π/) = e π/ ( C ) = C = e π/ = y = e x π/ sin x. 5. The characteristic equation is: r r =0 or (r )(r +)=0. The roots are: r =,. The general solution an its erivative are: y = C e x + C e x y =C e x C e x. (a) y(0) = = C + C = = C = C. Thus, the solutions that satisfy y(0) = are: y = Ce x +( C)e x.

28 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 508 SECTION 9.3 (b) y (0) = = C C = = C =C. Thus, the solutions that satisfy y (0) = are: y = Ce x +(C )e x. (c) To satisfy both conitions, we must have C = C = C = 3. The solution that satisfies y(0) =, y (0) = is: y = 3 ex + 3 e x. 6. r ω =0 = r = ±ω; y = A e ωx + A e ωx Since e ωx = cosh ωx + sinh ωx an e ωx = cosh ωx sinh ωx, we can write y = C cosh ωx + C sinh ωx (with C = A + A,C = A A ). 7. α = r + r, β = r r ; y = k e r x + k e r x = e αx (C cosh βx + C sinh βx), where k = C + C, k = C C. 8. r + ω =0 = r = ±ωi; y = C cos ωx + C sin ωx. Assuming that C + C > 0, we have C cos ωx + C sin ωx = ( ) C C + C C + C cos ωx + C C + C sin ωx = A (sin φ 0 cos ωx + cos φ 0 sin ωx) =A sin(ωx + φ 0 ), where A = C + C an φ 0, φ 0 [0, π), is the angle such that sin φ 0 = C C + C an cos φ 0 = C C + C 9. (a) Let y = e αx, y = xe αx. Then W (x) =y y y y = e αx [e αx + αxe αx ] xe αx [αe αx ]=e αx 0 (b) Let y = e αx cos βx, y = e αx sin βx, β 0. Then W (x) =y y y y = e αx cos βx[αe αx sin βx + βe αx cos βx] e αx sin βx[αe αx cos βx βe αx sin βx] = βe αx Characteristic equation: r +0 3 r + C = 0; roots: r = 03 ± 0 6 4/C. (a) r = 00( 5 ± 5); y = C e 00( 5+ 5)t + C e 00( 5 5)t (b) r = 500; y = C e 500t + C te 500t (c) r = 500( ± i); y = e 500t (C cos 500t + C sin 500t)

29 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: SECTION (a) The solutions y = e x, y = e 4x imply that the roots of the characteristic equation are r =, r = 4. Therefore, the characteristic equation is: (r )(r +4)=r +r 8=0 an the ifferential equation is: y +y 8y =0. (b) The solutions y =3e x, y =4e 5x imply that the roots of the characteristic equation are r =, r = 5. Therefore, the characteristic equation is (r + )(r 5) = r 4r 5=0 an the ifferential equation is: y 4y 5y =0. (c) The solutions y =e 3x, y = xe 3x imply that 3 is the only root of the characteristic equation. Therefore, the characteristic equation is (r 3) = r 6r +9=0 an the ifferential equation is: y 6y +9y =0. 3. (a) We want r = ±i, so r = 4. Differential equation: y +4y =0 (b) We want r = ± 3i, so (r +) = 9. Differential equation: y +4y +3y =0 33. (a) Let y = e αx u. Then Now, y = αe αx u + e αx u an y = α e αx u +αe αx u + e αx u y αy + α y = ( α e αx u +αe αx u + e αx u ) α (αe αx u + e αx u )+α e αx u = e αx u Therefore, y αy + α y =0 = e αx u = u =0. (b) y αy + ( α + β ) y = y αy + α y + β y. From part (a) y = e αx u = y αy + α y = e αx u. Therefore, y αy + ( α + β ) y =0 = e αx u + β e αx u =0 = u + β u = r + ar + b =0 = r,r = a ± a 4b. If a 4b >0, then a 4b <a, so y = C e r x + C e r x 0asx. a ± a 4b is negative, an the solutions: If a 4b = 0, then r = r = r = a/ < 0, an the solutions: y = C e rx + C xe rx 0asx.

30 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 50 SECTION 9.3 If a 4b <0, then y = e ax/ ( C cos b 4a x+ C sin b 4a x ) satisfies y <e ax/ = y 0 as x. 35. (a) If a = 0, b > 0, then the general solution of the ifferential equation is: y = C cos bx+ C sin ( ) bx= A cos bx+ φ where A an φ are constants. Clearly y(x) A for all x. (b) If a>0, b = 0, then the general solution of the ifferential equation is: y = C + C e ax an lim y(x) =C. x The solution which satisfies the conitions: y(0) = y 0, y (0) = y is: y = y 0 + y a y a e ax an lim x y(x) =y 0 + y a ; k = y 0 + y a. 36. Let y an y be solutions of the homogeneous equation. Suppose that y = ky for some scalar k. Then y ky W (y,y )= y ky =0 Now suppose that W (y,y ) = 0, an suppose that y which y (x) 0. Then, is not ientically 0. Let I be an interval on ( y y ) = y y y y (y ) = W (y,y ) (y ) =0 Therefore, theorem. y y = k constant on I. Finally, y = ky on I implies y = ky for all x by the uniqueness 37. Let W be the Wronskian of y an y. Then 0 0 W (a) = y (a) y (a) =0 Therefore one of the solutions is a multiple of the other (see the Supplement to this Section). 38. From the hint, y x = y z z x x + y z y x = y z x. ( x ) Differentiating with respect to x again, we have = ( y x z y ). z

31 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: Substituting into the ifferential equation x y + αxy + βy =0, we get ( y z y ) + α y z z + βy =0, or y z + ay + by =0, z SECTION where a = α, b = β. 39. From Exercise 38, the change of variable z = lnx transforms the equation x y xy 8y =0 into the ifferential equation with constant coefficients y z y 8y =0. z The characteristic equation is: r r 8 = 0 or (r 4)(r +)=0 The roots are: r =4,r=, an the general solution (in terms of z) is: Replacing z by ln x we get y = C e 4z + C e z. y = C e 4lnx + C e lnx = C x 4 + C x. 40. Using the result of Exercise 38, we get y z 3y z +y =0, so r 3r +=0 = r =,. = y = C e z + C e z. Substituting z =lnx, we get y = C x + C x. 4. From Exercise 38, the change of variable z = lnx transforms the equation x y 3xy +4y =0 into the ifferential equation with constant coefficients y z 4y +4y =0. z The characteristic equation is: r 4r + 4 = 0 or (r ) =0. The only root is: r =, an the general solution (in terms of z) is: Replacing z by ln x we get y = C e z + C ze z. y = C e lnx + C ln xe lnx = C x + C x ln x. y 4. From Exercise 38, we get z y +5y =0 z r r +5=0 = r =± i; an y = e z (C cos z + C sin z). Substituting z =lnx we get: y = x [C cos( ln x)+c sin( ln x)].

32 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 5 REVIEW EXERCISES REVIEW EXERCISES. First calculate the integrating factor e H(x) : H(x) = x = x an e H(x) = e x Multiplication by e x gives Integrating this equation, we get e x y + e x y =e x which is x (ex y)=e x e x y = e x + C an y = e x + Ce x. The equation can be written The equation is separable: cos(x)x + cos(x)+ y + y =0. y y + y = C an sin(x)+y +ln y = C y 3. The equation can be written The equation is separable: cos xx or sin(x)+x ln(y +)=C. cos xx y y y =0. + y y + y = C an 4 sin(x)+ x ln(y +)=C 4. The equation can be written xe x (y ln y)y =0. The equation is separable: xe x x y ln yy = C an e x (x ) y ln y + 4 y + C 5. The equation can be written y + 3 sin x y = x x.

33 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: Calculate the integrating factor e H(x) : H(x) = 3 x x =lnx3 an e H(x) = x 3. Multiplying by x 3 gives x 3 y +3x y = x sin x which is x (x3 y)=xsin x. Integrating this equation, we get x 3 y = x sin xx + C = x cos x + sin x + C. 4 REVIEW EXERCISES 53 an y = cos x + x 4x 3 sin x + C x The equation can be written Calculate the integrating factor e H(x) : Multiplication by x gives Integrating the equation, we get an H(x) = x y +xy =xe x y + x y = x ex. x x =lnx an e H(x) = x. which is x y = e x + C. y = x (ex + C). x (x y)=xe x. 7. The equation can be written The equation is separable: ( + x )x or arctan y = x + x3 3 + C. +x +y y =0. x3 y = C an x + +y 3 arctan y = C 8. The equation can be written x y + y =0 y

34 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 54 REVIEW EXERCISES The equation is separable: (x )x y + y = C y an 3 x3 x y ln y = C. 9. The equation can be written Calculate the integrating factor e H(x) : Multiplication by x gives Integrating this equation, we get H(x) = x y +xy = x 4 y + x y = x. x x =lnx an e H(x) = x. which is x (x y)=x 4. x y = 5 x5 + C an y = 5 x3 + C x. 0. The equation can be written x +x y y =0 The equation is separable: x +x x y y = C an 3 ( + x ) 3/ + y = C. Solving for y, we have y = 3 ( + x ) 3/ + C.. The equation can be written The integrating factor is Multiplication by x gives y + x y = x + e H(x) = e ln x = x. xy + y = x + x which is x (xy) = x + x. Integrating this equation, we get xy =lnx + x + C an y = x (ln x + x + C). Applying the initial conition y()=,we have ln + + C = an C = 3.

35 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: REVIEW EXERCISES 55 Therefore y = x ( ln x + x + 3 ).. The equation can be written 4x y y + y =0. The equation is separable: 4xx y y + y = C an x (y +) / = C. To fin the solution that satisfies y(0) =, we set x =0,y= an solve for C: C = (+) =. Therefore x (y +) / + = 0 is the solution. 3. The equation can be written The equation is separable: e x x + Solving for y, we get e x + y y =0. y y = C an ex + ln y = C. y = + Ce ex. To fin the solution that satisfies y(0) = + e, we set x =0,y= + e an solve for C. We have C =. Therefore y = + e ex. 4. The equation can be written tan x cos yy =0. The equation is separable: tan xx cos yy = C an ln sec x sin y = C. Applying the initial conition: y(0) = π, we have C =. Therefore sin y =ln sec x The characteristic equation is r r +=0.

36 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 56 REVIEW EXERCISES The roots are: r =± i. The general solution is y = C e x cos x + C e x sin x 6. The characteristic equation is r + r + 4 =0. The roots are: r = with multiplicity. The general solution is y = C e x/ + C xe x/ 7. The characteristic equation is r r =0. The roots are: r =,. The general solution is y = C e x + C e x 8. The characteristic equation is r 4r =0. The roots are: r = 0, 4. The general solution is y = C + C e 4x 9. The characteristic equation is r 6r +9=0. The roots are: r = 3 with multiplicity. The general solution is y = C e 3x + C xe 3x 0. The characteristic equation is The roots are: r = ±i. r +4=0

37 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: The general solution is y = C cos x + C sin x REVIEW EXERCISES 57. The characteristic equation is r +4r +3=0 The roots are: r = ± 3i. The general solution is y = e x (C cos 3x + C sin 3x). The characteristic equation is 3r 5r =0. The roots are: r =, 3. The general solution is y = C e x + C e x/3. 3. The characteristic equation is r r =0. The roots are: r =0,. The general solution is y = C + C e x. Applying the initial conitions y(0) = an y (0) = 0, we have C + C =, C =0 = C =. The solution of the initial-value problem is: y =. 4. The characteristic equation is r +7r +=0. The roots are: r = 3, 4. The general solution is y = C e 3x + C e 4x.

38 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 58 REVIEW EXERCISES Applying the initial conitions y(0) =, y (0) = 8, we have C + C =, 3C 4C =8 = C =6,C = 4. The solution of the initial-value problem is: y =6e 3x 4e 4x. 5. The characteristic equation is r 6r +3=0. The roots are: r =3± i. The general solution is y = e 3x (C cos x + C sin x). Applying the initial conitions y(0) =, y (0) =, we have C =, 3C +C = = C =,C =. The solution of the initial-value problem is: y = e 3x ( cos x sin x). 6. The characteristic equation is r +4r +4=0. The roots are: r = with multiplicity. The general solution is y = C e x + C xe x. Applying the initial conitions y( ) =, y ( ) =, we have C e C e =, C e +3C e = = C =7e,C =5e. The solution of the initial-value problem is: y =7e x +5xe x. 7. Curves: y = Ce x ; y =Ce x = y =y Orthogonal trajectories: y = y ; yy= x; y = C x. 8. Curves: y = C +x ; y = xc ( + x ) = y = xy +x Orthogonal trajectories: y = +x xy ; +x yy= x = x an y =ln x + x + C ( ) x + x x;

39 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: REVIEW EXERCISES Substituting y = x r into the equation, we get r(r )x r +4rx r +x r =0 or x r (r +3r +)=0 = r +3r +=(r + )(r +)=0. The solutions are: r =,. 30. Substituting y = x r into the equation, we get r(r )x r rx r 8x r =0 or x r (r r 8)=0 = r r 8=(r + )(r 4)=0. The solutions are: r = 4,. 3. Let y(t) be the value of the business (measure in millions) at time t. Then y(t) satisfies y t = ky The general solution of this equation is y(t) = kt + C Applying the given conitions, y(0) = an y() =.5, to fin C an k, we get C = an k =/3. year from now the business will be worth: y() = = 3 million years from now the business will be worth: y(.5) = 3 ( 5 = 6 million. ) + years from now the business will be worth: y(3) = =. 3 (3)+ Obviously the business cannot continue to grow at a rate proportional to its value square. 3. Let y(t) be the value of the business (measure in millions) at the time t. Then y(t) satisfies y t = k y The general solution of this equation is ( ) k y(t) = t + C Applying the given conitions, y(0) = an y() =.44, to fin C an k, we get C = an k =0. so y(t) = ( 0 t +).

40 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 50 REVIEW EXERCISES 5 years from now the business will be worth: y(7) = (0.7+) =.89 million. Solve ( ) t y(t) = 0 + =4 t = 0. The business will be worth 4 million 8 years from now. 33. (a) The general solution of the ifferential equation is y = a b + Ce bt. Applying the initial conition y(0) = 0, we get C = a b, an (b) lim t y(t) = a b (c) Setting y =0.9 a, we have b y = a b ( e bt ) 0.9 a b = a b ( e bt ). The solution to this equation is t = ln 0 b hours. 34. Let T (t) be the temperature of the bar at time t. It follows from Newton s law of cooling that By the conitions given in the problem, we have T (t) =τ + Ce kt. τ =0, T(0) = 00, T(0) = 50. Applying these conitions, we get C = 00 an k = ln 0. Therefore (a) T (30) = 00() 3/ = (b) Solve 5 = 00() t/0 for t: T (t) = 00e (t/0) ln = 00() t/0. 5 = 00() t/0, t 0 ln 4 ln = ln(/4), t = =40. 0 ln It will take 40 minutes for the bar to reach Let T (t) be the temperature of the object at time t. It follows from Newton s law of cooling that T (t) =τ + Ce kt.

41 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: By the conitions given in the problem, we have τ =70, T(0) = 0, T(0) = 35. REVIEW EXERCISES 5 Applying these conitions, we get C = an k = 0 ln(0/7). (a) The temperature of the object at time t is (b) T (0) = = 0 7 T = e (t/0) ln(0/7) = (a) Let T be the length of time neee to fill the tank. Then an T = 300 minutes. 600+(6 4)T = 00 ( ) t/0 7 0 (b) Let S(t) be the amount of salt issolve in the tank at time t. Then S t = rate in rate out = 6 S S 4=3 600+t t The equation can be rewritten The general solution to this equation is S + S t =3. C S = t + (300 + t). Since S(0) = 40, we get C = Therefore, the amount of salt in the tank at time t is: S(t) = t (300 + t) (c) S(T ) = = 535 pouns. (300 + t) 37. (a) Let T be the length of time neee to empty the tank. Then 80 (8 4)T = 0 an T = 0 minutes. (b) Let S(t) be the amount of salt in the tank at the time t. Then S t = 4 S S 8=4 80 4t 0 t, S(0) = 8 80=0 The solution to this initial-value problem is S(t) = 4(0 t) 7 40 (0 t) ()

42 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: 5 REVIEW EXERCISES (c) Let t 0 be the time that the tank contains exactly 40 gallons. Then 80 4t 0 = 40 an t 0 =0. Substituting t = 0 into (), we get S(0) =.5 pouns. 38. (a) The ifferential equation is separable an can be written as P P (0 0 5 P ) = t. With the initial conition P (0) = 000, the solution is Then (b) Setting P = , we get P (t) = 500e0.t +0.5e 0.t. lim t P (t) = = 500e0.t +0.5e 0.t. The solution to this equation is t = = 36 months. 39. Let P (t) be the number of people that have hear the rumor at time t. Then P (t) satisfies: P t = kp(0, 000 P ) The general solution of this equation is Now, P (0) = 0, 000 +C P (t) = = 500 = C = 39; 0, 000 +Ce 0,000kt. 0, 000 P (0) = +39e = 00 = 0, 000k 0,000(0k) =.09. 0, 000 Therefore, P (t) = +39e 0.090t. 0, 000 (a) P (0) = +39e 0.09(0) = 74 (b) The rumor will be spreaing fastest when the number of people who have hear it is equal to the number of people who have not hear it: 0, 000 =0, e 0.09t The solution of this equation is: t = 40 ays.

43 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-09 JWDD07-Salas-v November 5, 006 9: REVIEW EXERCISES (a) From (), u +u = a x Integrating, we get ln u + +u x = a + C. Applying the initial conition y (0) = u(0) = 0 = C = 0. Thus, ln u + +u x = a. (b) Set u = y : ln y + +(y ) x = a, y + +(y ) = e x/a an +(y ) = e x/a y. Squaring both sies an simplyfying, we get y = ex/a e x/a = sinh (x/a) = y = a cosh (x/a)+c. Applying the initial conition y(0) = a, we get C = 0. Therefore y(x) = a cosh(x/a), a catenary.