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Solutions to Selected Exercises

0.2.101: Compute x′ = −2e−2t and x′′ = 4e−2t . Then (4e−2t) + 4 (− 2e−2t) + 4(e−2t) = 0 .

0.2.102: Yes.

0.2.103: y = xr is a solution for r = 0 and r = 2 .

0.2.104: C 1 = 10 0 , C2 = − 90

0.2.105: φ = − 9e8s

0.3.101: a) PDE, equation, second order, linear, nonhomogeneous, constant coefficient.
b) ODE, equation, first order, linear, nonhomogeneous, not constant coefficient, not autonomous.
c) ODE, equation, seventh order, linear, homogeneous, constant coefficient, autonomous.
d) ODE, equation, second order, linear, nonhomogeneous, constant coefficient, autonomous.
e) ODE, system, second order, nonlinear.
f) PDE, equation, second order, nonlinear.

0.3.102: equation: a(x)y = b(x) , solution:  b(x) y = a(x) .

1.1.101:  x x2 y = e + 2 + 9

1.1.102: x = (3t − 2)1∕3

1.1.103:  −1 x = sin (t + 1 )

1.1.104: 170

1.1.105: If n ⇔ 1 , then y = ((1 − n)x + 1 )1∕(1−n) . If n = 1 , then y = ex .

1.1.106: The equation is  ′ r = − C for some constant C . The snowball will be completely melted in 25 minutes from time t = 0 .

1.1.107: y = Ax3 + Bx2 + Cx + D , so 4 constants.

1.2.101:
PIC
y = 0 is a solution such that y(0) = 0 .

1.2.102: Yes a solution exists.  ′ y = f(x,y) where f(x,y) = xy . The function f (x,y) is continuous and ∂f -∂y = x , which is also continuous near (0,0) . So a solution exists and is unique. (In fact y = 0 is the solution).

1.2.103: No, the equation is not defined at (x,y) = (1,0) .

1.2.104: a)  ′ y = cosy , b)  ′ y = y cos(x) , c)  ′ y = sin x . Justification left to reader.

1.2.105: Picard does not apply as f is not continuous at y = 0 . The equation does not have a continuously solution. If it did notice that y′(0) = 1 , by first derivative test, y(x) > 0 for small positive x , but then for those x we would have  ′ y (x) = 0 , so clearly the derivative cannot be continuous.

1.3.101: y = Cex2

1.3.102:  t3 x = e + 1

1.3.103: x3 + x = t + 2

1.3.104: y = --1- 1−lnx

1.3.105: sin(y) = − cos(x) + C

1.3.106: The range is approximately 7.45 to 12.15 minutes.

1.3.107: a)  t x = 10e0t+02e4- . b) 102 rabbits after one month, 861 after 5 months, 999 after 10 months, 1000 after 15 months.

1.4.101:  −x3 1 y = Ce + ∕3

1.4.102: y = 2ecos(2x)+1 + 1

1.4.103: 250 grams

1.4.104:  2 P (5 ) = 1000e 2×5−0.05×5 = 1000e 8.75 ≈ 6.31 × 1 06

1.4.105:  ′ Ah = I − kh , where k is a constant with units  2 m s .

1.5.101: y = 32x−2

1.5.102:  3−x2 y = -2x-

1.5.103:  ( 3x )1∕3 y = 7e + 3x + 1

1.5.104:  ∘ ------------- y = x2 − ln(C − x)

1.6.101:
a) 0, 1, 2 are critical points.
b) x = 0 is unstable (semistable), x = 1 is stable, and x = 2 is unstable.
c) 1

1.6.102: a) There are no critical points. b) ∞

1.6.103: a) dx dt = kx(M − x) + A b)  -------- kM+ √(kM )2+4Ak ------2k------

1.6.104: a) α is a stable critical point, β is an unstable one. b) α , c) α , d) ∞ or DNE.

1.7.101: Approximately: 1.0000, 1.2397, 1.3829

1.7.102:
a) 0, 8, 12
b) x(4) = 16 , so errors are: 16, 8, 4.
c) Factors are 0.5, 0.5, 0.5.

1.7.103: a) 0, 0, 0 b) x = 0 is a solution so errors are: 0, 0, 0.

1.7.104: a) Improved Euler: y(1) ≈ 3.38 97 for h = 1∕4 , y(1) ≈ 3.42 37 for h = 1∕8 , b) Standard Euler: y(1) ≈ 2.88 28 for h = 1∕4 , y(1) ≈ 3.131 6 for h = 1∕8 , c) y = 2ex − x − 1 , so y(2) is approximately 3.4 366 . d) Approximate errors for improved Euler: 0.04 6852 for  1 h = ∕4 , and 0.012881 for  1 h = ∕8 . For standard Euler: 0.553 75 for  1 h = ∕4 , and 0.3049 9 for h = 1∕8 . Factor is approximately 0.27 for improved Euler, and 0.55 for standard Euler.

1.8.101: a) exy + sin(x) = C , b) x2 + xy − 2y2 = C , c) ex + ey = C , d) x3 + 3xy + y3 = C .

1.8.102: a) Integrating factor is y , the equation becomes dx + 3y 2dx = 0 . b) Integrating factor is ex , the equation becomes exdx − e−ydx = 0 . c) Integrating factor is y2 , the equation becomes cos(x) + ydx + xdy = 0 . d) Integrating factor is x , the equation becomes  2 2 (2xy + y )dx + (x + 2xy )dx = 0 .

1.8.103: a) The equation is − f(x)dx + 1g(y)dy , and this is exact because M = −f(x) , N = g1(y) , so My = 0 = Nx . b)  1 − xdx + ydy = 0 , leads to potential function  x2 F(x,y) = − 2 + ln|y| , solving F (x,y) = C leads to the same solution as the example.

2.1.101: Yes. To justify try to find a constant A such that sin(x) = Aex for all x .

2.1.102: No.  x+2 2 x e = e e .

2.1.103: y = 5

2.1.104: y = C1 ln(x) + C2

2.1.105:  ′′ ′ y − 3y + 2y = 0

2.2.101:  √- √- y = C1e(−2+ 2)x + C 2e(−2− 2)x

2.2.102:  3x 3x y = C1e + C 2xe

2.2.103: y = e−x∕4 cos((√7∕4)x ) − √ 7e−x∕4 sin((√7∕4)x)

2.2.104: y = 2(a−5b)e−3x∕2 + 3a+52bex

2.2.105:  −t z(t) = 2e cos(t)

2.2.106: y = aβ−beαx + b−aαeβx β−α β−α

2.2.107: y′′ − y′ − 6y = 0

2.3.101: y = C ex + C x3 + C x2 + C x + C 1 2 3 4 5

2.3.102: a)  3 2 r − 3r + 4r − 12 = 0 , b)  ′′′ ′′ ′ y − 3y + 4y − 12y = 0 , c)  3x y = C 1e + C2sin(2x) + C 3cos(2x)

2.3.103: y = 0

2.3.104: No.  1 x x+1 e e − e = 0 .

2.3.105: Yes. (Hint: First note that sin(x) is bounded. Then note that x and xsin(x) cannot be multiples of each other.)

2.3.106: y′′′ − y′′ + y′ − y = 0

2.4.101:  8 k = ∕9 (and larger)

2.4.102:
a) 0.05I ′′ + 0.1I′ + (1∕5)I = 0
b)  -- x(t) = Ce−t cos(√ 3t − γ )
c)  −t √ -- 1√0-−t √ -- x(t) = 10e cos( 3t) + 3e sin( 3t)

2.4.103: a) k = 50 0000 , b) -1√- ≈ 0.141 5 2 , c) 45000 kg, d) 11250 kg

2.4.104: m 0 = 1 3 . If m < m 0 , then the system is overdamped and will not oscillate.

2.5.101: y = −16sin(3x7)+36cos(3x)

2.5.102: a)  2ex+3x3−9x y =----6--- , b)  √ -- √-- 2ex+3x3−9x y = C1 cos( 2x) + C 2sin( 2x) + ---6----

2.5.103: y(x) = x2 − 4x + 6 + e−x(x − 5)

2.5.104:  x x −x 2x y = 2xe−(e-+e-)log(e-+1) 4

2.5.105:  √- √- y = −sin(3x+c)+ C 1e 2x + C 2e− 2x

2.6.101:  √31 ω = 4√2 ≈ 0.984  16 C(ω ) = 3√7 ≈ 2.016

2.6.102:  (ω2−ω2)F xsp = -----02---20-22 cos(ωt) +----22ωpF02-2-2 sin (ωt ) + Ak m(2ωp)+m(ω0−ω) m(2ωp)+m(ω0−ω ) , where p = 2cm and  ∘ -- ω 0 = k- m .

2.6.103: a) ω = 2 , b) 25

3.1.101: y = C e3x 1 1 , y = y(x ) = C ex + C1e3x 2 2 2 , y = y(x) = C ex + C1e3x 3 3 2

3.1.102: x = 53e2t − 23e−t , y = 53e2t + 43e−t

3.1.103:  ′ x1 = x2 ,  ′ x2 = x3 ,  ′ x3 = x1 + t

3.1.104: y′ + y 1 + y2 = t 3 , y′ + y1 − y2 = t2 4 , y ′= y3 1 , y′ = y4 2

3.1.105: x1 = x2 = at . Explanation of the intuition is left to reader.

3.2.101: − 15

3.2.102: − 2

3.2.103: ⃗x = [15] −5

3.2.104: a) [1∕a 0] 0 1∕b b) [ ] 1∕0a 01∕b 00 0 0 1∕c

3.3.101: Yes.

3.3.102: No.  [cosh(t)] [et] [e−t] 2 1 − 1 − 1 = ⃗0

3.3.103: [x]′ [3 −1][x] [et] y = t 0 y + 0

3.3.104: a)  [ ] ⃗x′ = 00 2 2ttx⃗ b)  [Cet2+C ] ⃗x = 2C et21 2

3.4.101:
a) Eigenvalues: 4,0,−1 Eigenvectors: [ ] 10 1 , [ ] 01 0 , [ ] 35 −2
b)  [ ] [] [ ] ⃗x = C1 10 e4t + C2 01 + C3 35 e−t 1 0 −2

3.4.102:
a) Eigenvalues:  √- √- 1+-3i,1−--3i 2 2 , Eigenvectors: [ ] −2√- 1− 3i , [ ] −√2 1+ 3i
b)  [ (√3t) ] [ ( √3t) ] ⃗x = C1et∕2 (√−2)cos√-2-(√-) + C 2et∕2 (√−2)sin√-2 (√-) cos-32t+ 3sin -32t sin 23t − 3cos -32t

3.4.103:  [1] t [1 ] −t ⃗x = C 1 1e + C 2 −1 e

3.4.104:  [ ] [ ] ⃗x = C cos(t) + C sin(t) 1 −sin(t) 2 cos(t)

3.5.101: a) Two eigenvalues:  √-- ± 2 so the behavior is a saddle. b) Two eigenvalues: 1 and 2 , so the behavior is a source. c) Two eigenvalues: ± 2i , so the behavior is a center (ellipses). d) Two eigenvalues: − 1 and − 2 , so the behavior is a sink. e) Two eigenvalues: 5 and − 3 , so the behavior is a saddle.

3.5.102: Spiral source.

3.5.103:
PIC
The solution will not move anywhere if y = 0 . When y is positive, then the solution moves (with constant speed) in the positive x direction. When y is negative, then the solution moves (with constant speed) in the negative x direction. It is not one of the behaviors we have seen.
Note that the matrix has a double eigenvalue 0 and the general solution is x = C1t + C 2 and y = C1 , which agrees with the above description.

3.6.101:  [1 ]( √ -- √ --) [ 0]( √ -- √ --) [0] ( ) [−1] ⃗x = −1 a1cos( 3t) + b1sin( 3t) + 1 a2cos( 2t) + b2sin( 2t) + 0 a3 cos(t) + b 3sin 1 −2 1 ∕3

3.6.102: [ ] [ ] m00 m 00 ⃗x′′ = −kk−k2k 0k ⃗x 0 0 m 0 k −k . Solution:  [ ] [ ] [] ⃗x = 1−2 (a cos(√ 3k∕mt) + b sin(√ 3k∕mt)) + 10 (a cos(√k∕mt) + b sin(√k∕mt)) +

3.6.103:  2 √ 1-- 2 x2 = (∕5)cos( ∕6t) − (∕5)cos(t)

3.7.101: a) 3,0,0 b) No defects. c)  [] [ ] [ ] ⃗x = C 11 e3t + C 10 + C 01 1 1 2 −1 3 −1

3.7.102:
a) 1,1,2
b) Eigenvalue 1 has a defect of 1
c)  [ ] ([] [ ]) [ ] 0 t 1 0 t 3 2t ⃗x = C 1 −11 e + C 2 00 + t −11 e + C 3 3−2 e

3.7.103:
a) 2,2,2
b) Eigenvalue 2 has a defect of 2
c)  [ ] ([ ] [ ]) ([ ] [ ] []) ⃗x = C 03 e2t + C −01 + t 03 e2t + C 10 + t 0−1 + t2 03 e2t 1 1 2 0 1 3 0 0 2 1

3.7.104: A = [5 5] 0 5

3.8.101:  [ ] tA e3t+2e−t e−t−2e3t e = e−t−e3t e3t+e−t 2 2

3.8.102:  ⌊ t 3t ⌋ tA |||2e3t−4e2t+3et 3e2 −3e2 −e3t+4e2t−3et||| e = ||⌈ 2et−2e2t tet3t 2e2t−2et ||⌉ 2e3t−5e2t+3et 3e2 −3e2 −e3t+5e2t−3et

3.8.103: a)  tA [(t+1)e2t −te2t] e = te2t (1−t)e2t b)  [(1−t)e2t] ⃗x = (2−t)e2t

3.8.104: [1+2t+5t2 3t+6t2 ] 2t+4t2 1+2t+5t2  0.1A [1.25 0.36] e ≈ 0.24 1.25

3.9.101: The general solution is (particular solutions should agree with one of these):
x(t) = 1C (e9t + e4t) + 4C (e9t − e4t) − 18t+5 5 1 5 2 54 y(t) = 1C (e9t − e4t) + 1C (4e5t + e4t) + t+-7 5 1 5 2 6 216

3.9.102: The general solution is (particular solutions should agree with one of these):
x(t) = 12C1(et + e−t) + 12C2(et − e−t) + tet y(t) = 12C 1(et − e−t) + 12C 2(et + e−t) + tet

3.9.103:  [ ]( ) [ ] ⃗x = 11 52et − t − 1 + −11 −12 e−t

3.9.104:  [ ](( ) √- ( ) √- cos(t)) ⃗x = 19 1140-+ 1201√6- e 6t + 1140-+ 1201√-6 e− 6t − 6t0 −-70
 [1](−9 -1 9t cos(t)) + −1 80 sin(2t) + 30 cos(2t) + 40 − 30

4.1.101:  ∘ 15 ω = π 2

4.1.102:  2 2 λk = 4k π for k = 1,2,3,... xk = cos(2kπt) + B sin(2kπt) (for any B )

4.1.103: x(t) = − sin(t)

4.1.104: General solution is  −λt x = Ce . Since x(0) = 0 then C = 0 , and so x(t) = 0 . Therefore, the solution is always identically zero. One condition is always enough to guarantee a unique solution for a first order equation.

4.1.105:  √3 −33√λ √3 (√ 3 3√λ) (√3 3√λ) -3-e2 − -3-cos --2-- + sin -2--- = 0

4.2.101: sin(t)

4.2.102: ∞ ∑ (π−n)sin(πn+π2)+2(π+3n)sin(πn−π2)sin(nt) n=1 πn −π

4.2.103: 1 1 2 − 2 co s(2t)

4.2.104: π4 ∞∑ (−1)n(8π2n2−48) -5 + ----n4-----cos(nt) n=1

4.3.101: a) 8+ ∞∑ 16(−1)ncos(nπt) 6 n=1 π2n2 2 b) 8− 16-cos(πt) + 4-cos(πt) −-16-cos(3πt) + ⋅⋅⋅ 6 π2 2 π2 9π2 2

4.3.102: a) ∞∑ n+1 ( ) (−1)--2λ-sin nπt n=1 nπ λ b)  ( ) ( ) ( ) 2λ-sin πt − λsin 2πt + 2λ-sin 3πt − ⋅⋅⋅ π λ π λ 3π λ

4.3.103:  ′ ∑∞ π f (t) = n2+1 cos(nπt) n=1

4.3.104: a)  t ∞∑ 1- F(t) = 2 + C + n=1 n4 sin (nt) b) no.

4.3.105: a) ∞ ∑ (−1)n+1sin(nt) n=1 n b) f is continuous at t = π∕2 so the Fourier series converges to f (π∕2) = π∕4 . Obtain π∕4 = ∞∑ (−1)n+1 = 1 − 1∕3 + 1∕5 − 1∕7 + ⋅⋅⋅ n=1 2n−1 . c) Using the first 4 terms get 76∕105 ≈ 0.72 (quite a bad approximation, you would have to take about 50 terms to start to get to within 0.01 of π∕4 ).

4.4.101: a)  ∞ 1∕2 + ∑ -−242 cos(nπ-t) n=1 π n 3 nodd b)  ∞ ∑ 2(−1)n+1sin(nπt) n=1 πn 3

4.4.102: a) cos(2t) b)  ∞∑ --−4n- n=1πn2−4π sin(nt) n odd

4.4.103: a) f(t) b) 0

4.4.104: ∞ ∑ 2-−12-sin (nt) n=1 n(1+n )

4.4.105:  ∞∑ πt+ 2n(1π−n2) sin(nt) n=1

4.5.101: x = √--1-2 sin(2πt) + √-0.1-2 co s(10 πt) 2−4π 2−100π

4.5.102: x = ∑∞ -e−n-cos(2nt) n=13−2n

4.5.103:  ∞∑ x = -1√- + -22√−4-2-2cos(nπt) 2 3 n=1 nπ ( 3−n π) nodd

4.5.104: x = -1√- − 2-tsin(πt) + ∞∑ ---−4---cos(nπt) 2 3 π3 n=3n2π4(1−n2) nodd

4.6.101: u(x,t) = 5 sin(x)e−3t + 2sin(5x)e−75t

4.6.102:  −0.1t u(x,t) = 1 + 2co s(x)e

4.6.103:  λt λx u(x,t) = e e for some λ

4.6.104: u(x,t) = Aex + Bet

4.7.101: y(x,t) = sin(x)(sin(t) + cos(t))

4.7.102: y(x,t) = 51π sin(πx)sin(5πt) + 1010π sin(2πx)sin(10πt)

4.7.103:  ∞∑ n+1 √-- y(x,t) = 2(−1)n--sin(nx)cos(n 2t) n=1

4.7.104: y(x,t) = sin(2x) + tsin(x)

4.8.101:  sin(2π(x−3t))+sin(2π(3t+x)) cos(3π(x−3t))−cos(3π(3t+x)) y(x,t) = 2 + 18π

4.8.102: a)  (|| 2 |||{x − x − 0.04 if 0.2 ≤ x ≤ 0.8 y(x,0.1) = ||0.6x if x ≤ 0.2 |||(0.6 − 0.6x if x ≥ 0.8
b)  2 y(x,1∕2) = − x + x c)  2 y(x,1 ) = x − x

4.8.103: a) y(1,1 ) = −1∕2 b) y(4,3) = 0 c) y(3,9) = 1∕2

4.9.101:  ∞∑ ( ) u(x,y) = 1n2 sin(nπx ) sinh(sinnhπ((1nπ−)y)) n=1

4.9.102: u(x,y) = 0.1 sin (πx )(sinh(π(2−y))) sinh(2π)

4.10.101:  ∞∑ 1-n u = 1 + n=1n2r sin(n𝜃)

4.10.102: u = 1 − x

4.10.103: a) u = −41r2 + 14 b) u = −14 r2 + 14 + r2 sin(2𝜃)

4.10.104:  ∫ π 2 2 u(r,𝜃) = 1-- -------ρ--− r--------g(α) d α 2π −π ρ − 2rρ cos(𝜃 − α ) + r2

5.1.101:  (2n−1)π- λn = 2 , n = 1,2,3,... ,  ((2n−1)π-) yn = cos 2 x

5.1.102: a) p (x) = 1 , q(x) = 0 , r(x) = 1 x , α = 1 1 , α = 0 2 , β = 1 1 , β = 0 2 . The problem is not regular. b)  2 p(x) = 1 + x ,  2 q(x) = x , r(x) = 1 , α 1 = 1 , α2 = 0 , β1 = 1 , β2 = 1 . The problem is regular.

5.2.101: y(x,t) = sin(πx)cos(4π2t)

5.2.102: 9yxxxx + ytt = 0 (0 < x < 10,t > 0) , y(0, t) = yx(0,t) = 0 , y(10,t) = yx(10,t) = 0 , y(x,0) = sin(πx ), yt(x,0) = x(1 0 − x) .

5.3.101:  ∞∑ ( cos(nπ)−1 ) yp(x,t) = n−44π4 cos(nπx ) −-sin(nπ)-sin(nπx) − 1 cos(n πt). nn=o1dd

5.3.102: Approximately 1991 centimeters

6.1.101: s83 + 8s2-+ 4s

6.1.102:  2 −2t 2t − 2t + 1 − e

6.1.103: --1-- (s+1)2

6.1.104: ---1-- s2+2s+2

6.2.101: f(t) = (t − 1)(u(t − 1) − u (t − 2 )) + u(t − 2)

6.2.102:  e−t 3et x(t) = (2et−1 − t2 − 1)u(t − 1) −-2 +-2

6.2.103:  -1- H (s) = s+1

6.3.101: 1(cos t + sin t − e−t) 2

6.3.102: 5t − 5 sin t

6.3.103: 1(sint − tco st) 2

6.3.104: ∫ t ( ) f(τ) 1 − cos(t − τ) d τ 0

6.4.101: x(t) = t

6.4.102: x(t) = e−at

6.4.103:  1 x(t) = (cos∗ sin )(t) = 2tsin(t)

6.4.104: δ(t) − sin (t)

6.4.105: 3δ(t − 1) + 2t

7.1.101: Yes. Radius of convergence is 10 .

7.1.102: Yes. Radius of convergence is e .

7.1.103: -1- = − --1--- 1−x 1−(2−x) so 1--= ∑∞ (−1)n+1(x − 2)n 1−x n=0 , which converges for 1 < x < 3 .

7.1.104: ∞∑ (n1−7)!xn n=7

7.1.105: f(x) − g(x ) is a polynomial. Hint: Use Taylor series.

7.2.101: a2 = 0 , a3 = 0 , a 4 = 0 , recurrence relation (for k ≥ 5 ): ak = − 2ak−5 , so:
 5 6 10 11 15 16 y(x) = a0 + a1x − 2a0x − 2a 1x + 4a0x + 4a 1x − 8a0x − 8a1x + ⋅⋅⋅

7.2.102: a) a = 1 2 2 , and for k ≥ 1 we have a = a + 1 k k−3 , so
 1 2 3 4 3 5 6 7 5 8 9 10 y(x) = a0 + a1x + 2x + (a0 + 1)x + (a1 + 1)x + 2x + (a0 + 2)x + (a1 + 2)x + 2x + (a0 + 3 )x + (a1 + 3&#
b) y(x) = 12x2 + x3 + x4 + 32x5 + 2x 6 + 2x7 + 52x8 + 3x 9 + 3x10 + ⋅⋅⋅

7.2.103: Applying the method of this section directly we obtain ak = 0 for all k and so y(x) = 0 is the only solution we find.

7.3.101: a) ordinary, b) singular but not regular singular, c) regular singular, d) regular singular, e) ordinary.

7.3.102:  √- √ y = Ax 1+25+ Bx 1−25

7.3.103:  ∞ y = x3∕2 ∑ (−1)−1xk k=0 k!(k+2)! (Note that for convenience we did not pick a 0 = 1 )

7.3.104: y = Ax + Bx ln(x )

8.1.101: a) Critical points (0,0) and (0,1) . At (0,0) using u = x , v = y the linearization is u′ = −2u − (1∕π)v , v′ = − v . At (0,1) using u = x , v = y − 1 the linearization is u′ = − 2u + (1∕π)v , v′ = v .
b) Critical point (0,0) . Using u = x , v = y the linearization is  ′ u = u + v ,  ′ v = u .
c) Critical point (1∕2,−1∕4) . Using u = x − 1∕2 , v = y + 1∕4 the linearization is u′ = −u + v , v′ = u + v .

8.1.102: 1) is c), 2) is a), 3) is b)

8.1.103: Critical points are (0,0,0) , and (− 1,1,−1) . The linearization at the origin using variables u = x , v = y , w = z is u ′ = u , v′ = −v , z′ = w . The linearization at the point (−1, 1,−1) using variables u = x + 1 , v = y − 1 , w = z + 1 is u′ = u − 2w , v′ = −v − 2w , w ′ = w − 2u .

8.1.104: u′ = f(u,v,w) , v′ = g(u,v,w) , w ′ = 1 .

8.2.101: a) (0,0 ) : saddle (unstable), (1,0) : source (unstable), b) (0,0 ) : spiral sink (asymptotically stable), (0,1) : saddle (unstable), c) (1,0) : saddle (unstable), (0,1) : saddle (unstable)

8.2.102: a) 1 2 1 3 2y + 3x − 4x = C , critical points (− 2,0) : an unstable saddle, and (2,0) : a stable center. b) 12y2 + ex = C , no critical points. c) 12y2 + xex = C , critical point at (− 1,0) is a stable center.

8.2.103: Critical point at (0,0) . Trajectories are  ∘ ----------4 y = ± 2C + (1∕2)x , for C > 0 , these give closed curves around the origin, so the critical point is a stable center.

8.2.104: A critical point x0 is stable if f ′(x0) < 0 and unstable when f′(x0) > 0 .

8.3.101: a) Critical points are ω = 0 , 𝜃 = kπ for any integer k . When k is odd, we have a saddle point. When k is even we get a sink. b) The findings mean the pendulum will simply go to one of the sinks, for example (0,0 ) and it will not swing back and forth. The friction is too high for it to oscillate, just like an overdamped mass-spring system.

8.3.102: a) Solving for the critical points we get (0,− h∕d) and (bha+cad, ab) . The Jacobian at (0,− h∕d) is [ ] a+bh∕d 0 −ch∕d −d whose eigenvalues are a + bh∕d and − d . So the eigenvalues are always real of opposite signs and we get a saddle (In the application however we are only looking at the positive quadrant so this critical point is not relevant). At (bha+cad,ab) we get Jacobian matrix [ b(bh+ad)] 0ac−bh+aacd b a −d . b) For the specific numbers given, the second critical point is  550 (-3-,40) the matrix is [ 0 −11∕6] 3∕25 1∕4 , which has eigenvalues  √--- 5±i-327 40 . Therefore there is a spiral source. This means the solution will spiral outwards. The solution will eventually hit one of the axis x = 0 or y = 0 so something will die out in the forest.

8.3.103: The critical points are on the line x = 0 . In the positive quadrant the y′ is always positive and so the fox population always grows. The constant of motion is C = yae−cx−by , for any C this curve must hit the y axis (why?), so the trajectory will simply approach a point on the y axis somewhere and the number of hares will go to zero.

8.4.101: Use Bendixson-Dulac Theorem. a) fx + gy = 1 + 1 > 0 , so no closed trajectories. b)  2 fx + gy = − sin (y) + 0 < 0 for all x,y except the lines given by y = kπ (where we get zero), so no closed trajectories. c) fx + gy = y + 0 > 0 for all x,y except the line given by y = 0 (where we get zero), so no closed trajectories.

8.4.102: Using Poincarè-Bendixson Theorem, the system has a limit cycle, which is the unit circle centered at the origin as x = co s(t) + e−t , y = sin(t) + e−t gets closer and closer to the unit circle. Thus we also have that x = co s(t) , y = sin(t) is the periodic solution.

8.4.103: f(x,y) = y ,  2 g(x,y) = μ(1 − x )y − x . So  2 2 fx + gy = 1 + μ (1 − x ) = 1 + μ − μx . The Bendixson-Dulac Theorem says there is no closed trajectory lying in the set 1+μ -μ- < x2 .

8.4.104: The closed trajectories are those where sin(r) = 0 , therefore, all the circles with radius a multiple of π are closed trajectories.

8.5.101: Critical points: (0,0,0) , (3 √8,3 √ 8,27) , (−3 √8,− 3√ 8,27) . Linearization at (0,0,0 ) using u = x , v = y , w = z is  ′ u = −10u + 10v ,  ′ v = 28u − v ,  ′ 8 w = − (∕3)w . Linearization at  √-- √ -- (3 8, 3 8,27) using  √ -- u = x − 3 8 ,  √ -- v = y − 3 8 , w = z − 27 is u′ = − 10u + 10v ,  √ -- v′ = u − v − 3 8w ,  √ -- √ -- w′ = 3 8u + 3 8v − (8∕3)w . Linearization at  √-- √ -- (−3 8,− 3 8,27) using  √ -- u = x + 3 8 ,  √ -- v = y + 3 8 , w = z − 27 is u′ = −10u + 10v ,  √-- v′ = u − v + 3 8w , w ′ = − 3√ 8u − 3 √ 8v − (8∕3)w .