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

0.2.104: C 1 = 10 0 , C2 = - 90

0.2.105:  8s φ = - 9e

1.1.101: y = ex + x2 + 9 2

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  1∕(1-n) y = ((1 - n)x + 1 ) . If n = 1 , then  x y = e .

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.

1.3.101:  x2 y = Ce

1.3.102: x = et3 + 1

1.3.103:  3 x + x = t + 2

1.3.104:  --1- y = 1-lnx

1.3.105: sin(y) = - cos(x) + C

1.3.106: The range is approximately 7.45 to 12.15 minutes.

1.4.101: y = Ce-x3 + 1∕3

1.4.102:  cos(2x)+1 y = 2e + 1

1.4.103: 250 grams

1.4.104:  2 6 P (5 ) = 1000e 2×5-0.05×5 = 1000e 8.75 ≈ 6.31 × 1 0

1.5.101: y = -2-- 3x-2

1.5.102: y = 3-x2 2x

1.5.103:  ( ) y = 7e3x + 3x + 1 1∕3

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= kx(M - x) + A dt b)  √-------- kM+--(kM-)2+4Ak 2k

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.

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

2.1.102: No. ex+2 = e2ex .

2.1.103: y = 5

2.1.104: y = C1 ln(x) + C2

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

2.2.102: y = C1e3x + C 2xe3x

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

2.2.104:  2(a-b)-3x∕2 3a+2b-x y(x) = 5 e + 5 e

2.2.105: z(t) = 2e-t cos(t)

2.3.101:  x 3 2 y = C1e + C2x + C 3x + C 4x + C 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.)

2.4.101: k = 8∕9 (and larger)

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

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

2.5.101: y = -16sin(3x7)+36cos(3x)

2.5.102: a)  x 3 y = 2e-+3x6--9x , b)  √ -- √-- x 3 y = C1 cos( 2x) + C 2sin( 2x) + 2e+36x-9x

2.5.103:  2 -x y(x) = x - 4x + 6 + e (x - 5)

2.5.104: y = 2xex-(ex+e-x)log(e2x+1) 4

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

2.6.101:  √-- ω = 43√12 ≈ 0.984 C(ω ) = 31√67 ≈ 2.016

2.6.102:  2 2 xsp = ---(ω0-2ω-)F20--2 cos(ωt) +----22ωpF02---2 sin (ωt ) + A m(2ωp)+m(ω0-ω2) m(2ωp)+m(ω0-ω2) k , where p = -c 2m and  ∘ -- ω 0 = k- m .

2.6.103: a) ω = 2 b) 25

3.1.101:  3x y1 = C1e ,  x C1 3x y2 = y(x ) = C 2e + 2 e ,  x C1 3x y3 = y(x) = C3e + 2 e

3.1.102: x = 5e2t - 2e-t 3 3 , y = 5e2t + 4e-t 3 3

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

3.1.104: y′ + y + y = t 3 1 2 , y′ + y - y = t2 4 1 2 , y ′= y 1 3 , y′ = y 2 4

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

6.2.103: H (s) = s1+1

6.3.101: 1 -t 2(cos t + sin t - e )

6.3.102: 5t - 5 sin t

6.3.103: 1(sint - tco st) 2

6.3.104: ∫ ( ) tf(τ) 1 - cos(t - τ) d τ 0

6.4.101: x(t) = t

6.4.102: x(t) = e-at

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

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

7.1.104: ∞∑ -1--xn n=7 (n-7)!

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

7.2.101: a = 0 2 , a = 0 3 , a = 0 4 , recurrence relation (for k ≥ 5 ): a = - 2a k k-5 , so:
 5 6 10 11 15 16 y(x) = a0 + a1x - 2a0x - 2a 1x + 4a0x + 4a 1x - 8a0x - 8a1x + ⋅⋅⋅

7.2.102: a) a2 = 12 , and for k ≥ 1 we have ak = ak-3 + 1 , so
y(x) = a0 + a1x + 1x2 + (a0 + 1)x3 + (a1 + 1)x4 + 3x5 + (a0 + 2)x 6 + (a1 + 2)x7 + 5x8 + (a0 + 3
b)  1 2 3 4 3 5 6 7 5 8 9 10 y(x) = 2x + x + x + 2x + 2x + 2x + 2x + 3x + 3x + ⋅⋅⋅

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:  1+√25 1-2√5 y = Ax + Bx

7.3.103:  3∕2 ∞∑ (-1)-1-k y = x k=0 k!(k+2)!x (Note that for convenience we did not pick a 0 = 1 )

7.3.104: y = Ax + Bx ln(x )