Solutions to Selected Exercises

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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:

$y = 0$ 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:

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: $-λt x = Ce$ and if $x (0) = 0$ then $C = 0$ and so $x(t) = 0$ , so the solution is always identically zero. Note that 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) $(| 2 ||||x - x - 0.04 if 0.2 ≤ x ≤ 0.8 |||{0.6x if x ≤ 0.2 y(x,0.1) = ||| ||||0.6 - 0.6x if x ≥ 0.8 |($
b) $y(x,1∕2) = - x + x2$ c) $y(x,1 ) = x - x2$

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

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

4.9.102: $( ) u(x,y) = 0.1 sin (πx ) sinsh(inπh(2(2-πy)))$

4.10.101: $∞ u = 1 + ∑ 12rn sin(nθ) n=1n$

4.10.102: $u = 1 - x$

4.10.103: a) $-1 2 1 u = 4 r + 4$ b) $-1 2 1 2 u = 4 r + 4 + r sin(2θ)$

4.10.104: $1 ∫ π ρ2 - r2 u(r,θ) = --- --------------------2g(α) d α 2π -π ρ - 2rρ cos(θ - α ) + r$

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

5.1.102: a) $p (x) = 1$ , $q(x) = 0$ , $r(x) = 1x$ , $α 1 = 1$ , $α2 = 0$ , $β1 = 1$ , $β2 = 0$ . The problem is not regular. b) $p(x) = 1 + x 2$ , $q(x) = x2$ , $r(x) = 1$ , $α 1 = 1$ , $α2 = 0$ , $β1 = 1$ , $β2 = 1$ . The problem is regular.

5.2.101: $y(x,t) = sin(πx)cos(4π2t)$

5.2.102: $9y + y = 0 (0 < x < 10,t > 0) xxxx tt$ , $y(0, t) = y (0,t) = 0 x$ , $y(10,t) = y (10,t) = 0 x$ , $y(x,0) = sin(πx ), yt(x,0) = x(1 0 - x)$ .

5.3.101: $∞∑ --4 ( cos(nπ)-1 ) yp(x,t) = n=1 n4π4 cos(nπx ) - sin(nπ) sin(nπx) - 1 cos(n πt). nodd$

5.3.102: Approximately 1991 centimeters

6.1.101: $-8 8- 4 s3 + s2 + s$

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

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: $∫ t ( ) 0 f(τ) 1 - cos(t - τ) d τ$

6.4.101: $x(t) = t$

6.4.102: $-at x(t) = e$

6.4.103: $x(t) = (cos* sin )(t) = 12tsin(t)$

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: $11-x = - 1-(12-x)-$ so $∑∞ 11-x-= (-1)n+1(x - 2)n n=0$ , which converges for $1 < x < 3$ .

7.1.104: $∞∑ -1---n n=7 (n-7)!x$

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:
$y(x) = a0 + a1x - 2a0x5 - 2a 1x 6 + 4a0x10 + 4a 1x 11 - 8a0x15 - 8a1x16 + ⋅⋅⋅$

7.2.102: a) $a2 = 1 2$ , and for $k ≥ 1$ we have $ak = ak-3 + 1$ , 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: $1+√5 1-√5 y = Ax 2 + Bx 2$

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: $y = Ax + Bx ln(x )$