Anhang A L¨osungen zu den Ubungsaufgaben¨weth0002/buecher/mathe/downloads/loesungen.pdf · 4 A....

AAnhang A

Losungen zu den Ubungsaufgaben

A

A Losungen zu den Ubungsaufgaben

A.1A.1 Losungen zu Zahlen, Gleichungen undGleichungssystemen

1.1 a) 2, 3, 5, 7, 11, 13, 17, 19 b) ∅1.2 (i) A ∩B = x : 1 ≤ x < 2,

(ii) A ∪B = x : 0 < x ≤ 3,(iii) A×B = (x, y) : 0 < x < 2 und 1 ≤ y ≤ 3,(iv) A\B = x : 0 < x < 1.

1.3 a) M1 ∪M2 = 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, . . .M1 ∩M2 = 6, 12, 18, . . .,M1\M2 = 2, 4, 8, 10, 14, 16, . . .M2\M1 = 3, 9, 15, 21, . . .b) M1 = 1,−2, M2 = 1, 2, M1 ∩M2 = 1 , M1 ∪M2 = 1, 2,−1,−2,M1\M2 = −2, M2\M1 = 2

1.7 a) 1, 1, 3, 3, 1, 4, 6, 4, 1 b) 108 243 216

1.9

(n

k

)1

nk = n!(n−k)! k!

1nk = 1

k!1·2·...·(n−k) (n−k+1)·...·(n)

1 · 2 · . . . · (n− k)︸︷︷︸ ·(n−k) Zahlen

nk︸︷︷︸k Zahlen

≤ 1k!· 1

1.10 x5 + 20x4 + 160x3 + 640x2 + 1280x+ 1024

625 y4 − 500 y3 + 150 y2 − 20 y + 1

a6 − 6 a4 b+ 12 a2 b2 − 8 b3

1.11 sum(kˆ2+1, k=71..125);

1.12 sum(kˆ3, k=1..n)=normal(sum(kˆ3, k=1..n));

1.13 a) ( 92)2x4a−3y b) a3 + a2b+ ab2

1.14 a)2x(2x2−3r2)√

r2−x2b) k2/

√(x− k)2 + x2 c) 3(x− 1)

1.15 a) ab2 b) a2/3 c) ab4/3 d) a13/8 e) a15/32

1.16 a) 1/2, 3 b) 3/2, −1/3, 12 c) nn+1

log a− 1m(n+1)

log b

1.18 a) IL = −5, 3 b) IL = ∅ c) IL =

53, 7

3

d) IL = −2 e) IL = −1

1.19 c = −2

1.20 a) IL = 0, 2 b) IL = ±2, ± 3 c) IL =−3, ±

√2, ± 5

1.21 a) IL = 3.5 b) IL = ∅ c) IL = ∅ d) IL = −11.22 a) IL = −4.424, 5.424 b) IL = −2, 11.23 a) IL = (8, ∞) b) IL = IR c) IL = ∅ d) IL = (−2.562, 1.562)

1.24 a) x1 = x2 = x3 = 1

b) x1 = 1, x2 = 3, x3 = 2

c) IL = ∅

1.25 a) IL =

(x1, x2, x3) ∈ IR3 : x =

− 32

12

0

+ λ

1212

1

b) IL =

x∈ IR3 :

x=

5

1

0

+ λ

−1

0

1

, λ ∈ IR

4 A. Losungen zu den Ubungsaufgaben

c) IL = ∅

1.26 a) IL =

x∈ IR3 :

x=

2

0

0

+ λ

−2

0

1

+ µ

− 32

1

0

; λ, µ ∈ IR

b) IL =

x∈ IR3 :

x=

1

0

0

+ λ

−1

0

1

+ µ

1

1

0

; λ, µ ∈ IR

c) IL = ∅

1.27 Die homogenen Systeme sind immer losbar.

A.2 A.2 Losungen zur Vektorrechnung

2.1 a) −→s 1 =

1

20

−18

; |−→s 1| = 26.92 b) −→s 2 =

5

−24

2

; |−→s 2| = 24.59

c) −→s 3 =

−19

36

−22

; |−→s 3| = 46.27 d) −→s 4 =

170

−60

−40

; |−→s 4| = 184.66

2.2 F = −(F1 + F2 + F3 + F4) =

−200

−175

10

N

2.3 −→e a = 1√14

2

3

1

; −→e b = 1√38

3

−5

2

; −→e c = 1√2

−1

0

−1

2.4 −→e = − −→a

|−→a | = 15

4

3

0

2.5 −→r (Q) = −→r (P ) + 10

−→a|−→a | =

−7.16

−6.08

−1.08

2.6 −→r (Q) = −→r (P1) + 1

2

−−−→P1 P2 =

0.5

3.5

2.5

2.7 a) 4 b) 96 c) 22

2.8 a) ϕ = 48.47 b) ϕ = 156.5

2.10 −→c = −→a +−→b ; −→a · −→b = 0

2.11 a) |−→a | =√

3 , α = β = γ = 54, 74

b) |−→a | =√

30 , α = 24.09 , β = 111.42 , γ = 79.48

2.12 |−→a | = BC = 2√

6∣∣∣−→b ∣∣∣ = AC = 2

√14 |−→c | = AB = 2

√14

α = 38.21 β = 70.89 γ = 70.89

2.13−→b a = 1

3

22

−22

11

−→b a = 1

9

−28

28

−14

2.14 Es ist γ = 90 , ax = 8.66 , ay = 5 , az = 0.

A.2 Losungen zur Vektorrechnung 5

2.15 a) α = 103.6 β = 76.37 γ = 19.47

b) α = 42.03 β = 68.19 γ = 123.9

2.16 a)

3

−10

8

b)

−12

−18

−3

c)

15

8

−18

d)

−10

14

12

2.17

−→F R =

∑4i=1 Fi =

(167.55

−148.68

) ∣∣∣−→F R

∣∣∣ = 224N, α = 41.6

2.18 a) |F | = 30N |a| = 3 b) ϕ = 63.61

c)−→F a = 4.444

2

1

−2

∣∣∣−→F a

∣∣∣ = 13.33 d) −→a · −→b = 0

2.19 a) ϕ = 60 b)−→M =

3

−3

−3

Nm;∣∣∣−→M ∣∣∣ = 5.2Nm c)

−→F r = 1

2

2

1

1

N

2.20−→F · −→s = 4Nm

−→F−→S 1 +

−→F−→S 2 = 4Nm⇒ Die Arbeit ist wegunabhangig.

2.22 g : −→x =

4

0

3

+ λ

−1

0

−1

;

λ = 1 : Q1 = (3, 0, 2)

λ = 2 : Q2 = (2, 0, 1)

λ = −5 : Q3 = (9, 0, 8)

2.23 g : −→x =

1

3

−2

+ λ

5

2

10

2.24 Ja: −→x =

3

0

4

+ λ

−2

1

−3

; P3 : λ = 2

2.25 d = 1, 22

2.26 g : −→x =

5

3

1

+ λ

√

32

0

− 12

2.27 a) g1 und g2 sind windschief; d = 2.04.

b) Geraden sind parallel, da −→a ‖−→b ; d = 1.79

c) Geraden schneiden sich genau in einem Punkt S = (5, 2, 10) ; α = 32.4

2.28 g1 und g2 sind windschief zueinander; d = 2.85.

2.29 E =

3

5

1

+ λ

1

1

1

+ µ

2

1

3

; −→n =

2

−1

−1

; Q = (10, 9, 11)

2.30 −→r (P ) = −→r 1 + λ (−→r 2 −−→r 1) + µ (−→r 3 −−→r 1) =

3

1

0

+ λ

−7

0

1

+ µ

2

8

3

2.31 Ja: E =

1

1

1

+ λ

2

1

−1

+ µ

3

−2

4

=

12

−4

12

⇒ λ = 1 , µ = 3.

2.32 4x+ 3y + z = 54

2.33 a) g und E schneiden sich, da −→n · −→a = 2 6= 0. Schnittpunkt λs = 4.5

⇒ S = (18.5, 5.5, 11). Schnittwinkel ϕ = 9.27

b) g‖E, da −→n · −→a = 0; Abstand d = 1.51


c) E =

1

−2

−2

+λ

−1

1

1

+µ

−2

2

1

;−→n =

−1

−1

0

, g =

2

0

3

+λ

3

6

15

.

⇒ Schnittpunkt S = (1,−2,−2) ; Schnittwinkel ϕ = −22.79

2.34 E1‖E2, da n1 × n2 =−→0 ; Abstand d = 3.74

2.35 E1 6 ‖E2, Schnittgerade −→r = 13

0

59

5

+ λ

3

−5

−2

, Schnittwinkel ϕ = 27.2

2.36 Nein: −→a 3 = −→a 1+−→a 2: die Vektoren sind linear abhangig.

2.37 Ja.

2.38−→b = −→a 1+

−→a 2 − 2−→a 3 −−→a 4.

2.39 Linear abhangig, da det(−→a 1,−→a 2,

−→a 3,−→a 4,

−→a 5) = 0.

2.40 a)−→b = −→a 1+

−→a 2 +−→a 3. b) Nein.

2.41−→d = −2−→a 1+

−→a 2 −−→a 3.

A.3 A.3 Losungen zu Matrizen und Determinanten

3.1 a)

6 32

13 13

−8 7

b)

−7 −3

1 −4

0 4

c)

(29 23 10

11 18 2

)d)

8 13

8 8

−1 2

e)

(−2 4 −3

9 2 4

)f)

−2 9

4 2

−3 4

3.2 a) A2 =

4 18 3

0 16 5

0 0 1

, B2 =

3 −6 2

−4 7 −4

4 −6 5

,

A ·B =

−1 6 1

−3 5 −4

1 −3 0

, B ·A =

2 3 2

−2 5 1

2 −9 −3

b) A ·B =

(1 26

0 6

)B ·A =

0 2 0 2

−2 −1 0 −2

10 18 0 23

2 7 0 8

3.3 a) A−1 = 1

9

−1 −1 5

3 3 −6

2 11 −10

b) B−1 = 13

(1 1

−1 −4

)

c) C−1 = 13

−2 1 −3 −3

−3 3 −3 −6

1 1 0 0

−1 −1 0 3

3.5 A =

3 2 −1 05 1 2 04 5 1 0

, (18, 22, 38).

3.6 5 , 0 , λ

3.8 −12 , −21 , −53

A.4 Losungen zu elementaren Funktionen 7

3.9 a) 0 , −1 b) 1 , 2 , 3

3.10 a) 142 b) 180

3.11 detA = −8 , (x1, x2, x3) = (−3, 3, 0)

3.12 A−1 = 19

−1 −1 5

3 3 −6

2 11 −10

B−1 = 17

3 −2 −2

0 0 7

−1 3 −11

3.13 Rang (A) = 3 Rang (B) = 3 Rang (C) = 3 Rang(D) = 3.

3.14 det

1 1 1

1 2 4

1 3 9

= 2 6= 0 −→x =

−3

2

1

3.15 a) Rang (A) = 2 Rang (A/b) = 3 ⇒ nicht losbar.

b) Rang (A) = 2 = Rang (A/b) ⇒ losbar, nicht eindeutig

z.B. (−2, 1,−1) ist Losung.

3.16 a) det(−→a ,−→b ,−→c

)= 0 ⇒ linear abhangig

b) det(−→a ,−→b ,−→c

)6= 0 ⇒ linear unabhangig,

−→d = −3−→a +

−→b + 2−→c .

A.4A.4 Losungen zu elementaren Funktionen

4.1 a) ID = x : |x| ≥ 1 VW = IR≥0

b) ID = IR \ 0 VW = IR

c) ID = IR \ −2, 2 VW = (−∞, 0] ∪(

14, ∞

)d) ID = IR \ −1 VW = IR \ 1e) ID = IR VW = IR≥1

f) ID = IR VW =[− 1

2, + 1

2

]4.2 a) gerade b) ungerade c) ungerade d) gerade e) gerade f) -

4.3 a) streng monoton fallend in IR≤0 ; streng monoton wachsend in IR≥0

b) streng monoton wachsend

c) streng monoton wachsend

e) streng monoton wachsend

4.4 a) y = 12 x

ID = IR>0

b) y = 13x2 ID = IR≥0

c) y = lnx+ 0, 5− ln 2 ID = IR>0

d) y = − x+1x−1

ID = (−∞, 1)

4.5 y = −2x+ 5

4.6 a) 1 2 − 5 b) −1 c) 0 2 5 − 5

4.7 a) f (2) = −5 b) f (3) = 49.1

4.8 Ja: z.B. x2 + 1

4.9 y = x3 − 2x+ 1

4.10 a) −1 (doppelt), 1 b) ±2, ±3

4.11 f (x) = 12x3 + 1

2x+ 1

4.12 factor(”,x)

4.13 fsolve(”,x)


4.14 unapply

4.15 plot

4.16 factor, convert(”, ’horner’), degree

4.17 a) NS : −2, 1

Pol : 2

b) NS : 3, 4

Pol : −1, 0

c) NS : 1

Pol : −1

d) NS : −Pol : ±1

4.18 a) NS : ±2

Pol : −Asymptote : y = 1

b) NS : 2 doppelt

Pol : −2

Asymptote : x− 6

c) NS : 1

Pol : 2

Asymptote : y = 1

d) NS : 1 doppelt

Pol : −1 doppelt

Asymptote : y = 14.19 plot, numer, denom, factor, normal, asympt, solve

4.21 t = 2.3RC

4.22 t = 1.5 s

4.23 a = 8 b = 0.4159

4.24 a) x1 = −0.3012

x2 = 2.3012

b) Subtitution t = ex. x1 = 0, x2 = 0.693 .

4.25 x = 2.

4.26 γ = 1T

ln x(t)x(t+T )

= 100 ln 2.

4.27 Grad 40, 36 81, 19 −322, 08 278, 19

Bogen 0, 7044 1.4171 −5.6213 4.85534.28 cos (x1 − x2) = cosx1 cosx2 + sinx1 sinx2

x1 = x2 = x⇒ cos (0) = 1 = cos2 x+ sin2 x

4.30 Amplitude Phasenverschiebung Periode

a) 2 − π18

23π

b) 5 2.1 π

c) 10 −3 2

d) 2.4 −π8

12π

4.32 π/2, π/4, −π/3, 0.5018, π/3, π/6, π, 0.5489, π/4, −π/3, 2π/3, π/3

4.33 0.7071, 0.9792, 0.5225, −4.455, 0.8776

4.34 y = arccos (x) → x = cos y√1− x2 =

√1− cos2 y = sin y = sin (arccos (x))

4.35 analog 6.8.

4.36 x, x,√

1− x2,√

1− x2, x/√

1 + x2,√

1− x2/x

4.37 a) ID=[−1, 1], VW=[1, π − 1),

b) ID=[0, 1], VW=[0, π/2 + 1],

c) ID=[0, 2], VW=[0, π]

A.5 Losungen zu komplexen Zahlen 9

A.5A.5 Losungen zu komplexen Zahlen

5.1 a) 6 ei π6 b) 2

√2 ei 5

4 π c) 2 ei 53 π d) 5 e0 i e) 5 ei 3

2 π f) ei π

5.2 a) 3√

2(cos π

4+ i sin π

4

)= 3 + 3 i

b) 2(cos 2

3π + i sin 2

3π)

= −1 +√

3 i

c) 1 (cosπ + i sinπ) = −1

d) 4(cos 4

3π + i sin 4

3π)

= −2− 2√

3 i

5.3 a) 3−√

2 i b) 4 (cos 125 − i sin 125) c) 5 e−i 32 π d)

√3 e−i 0.734

5.4 a) 2(cos 2

3π + i sin 2

3π)

b)√

2 (cos 135 + i sin 135)

c) 2(cos 45 + i sin 45)

d) 5(cos 233.13 + i sin 233.13)

5.5 a) 1− 4 i b) −9− 46 i c) 115− 2

5i d) −1 e) 10

37f) 16

5− 2

5i

5.6 a) −1− 4 i b) 170 c) −1024 i d) 12 e) 35

f) − 17

g) −7 + 3√

3 +√

3 i h) 765 + 128√

3 i)(6√

3+4)7

5.7 a) −512 + 512√

3 i b) 8 (cos 135 + i sin 135) c) −46 656

d) 27 ei 1.66 π = 2 ei 5.21

5.8 a) 3 eiϕ mit ϕ = π4, 3

4π, 5

4π, 7

4π

3 (cosϕ+ i sinϕ) mit ϕ = 45, 135, 225, 315

b) 6√

2 eiϕ mit ϕ = π9, 4

9π, 7

9π, 10

9π, 13

9π, 16

9π

6√

2 (cosϕ+ i sinϕ) mit ϕ = 20, 80, 140, 200, 260, 320

5.9 a) 1, 1, 2,−1± i

b) 1,−2, 12i,− 1

2i

5.11 evalc (Re((-2+7∗I ) / (15∗I))); etc.

evalc (Im((-2+7∗I) / (15∗I))); etc.

5.12 abs ( ” ) → Betrag

argument ( ” ) → Winkel

5.13 evalc

5.14 evalc

5.15 solve (z ˆ3 = I, z) bzw. fsolve (. . ., z)

5.16 fsolve ( ” , z, complex)

5.17 a) R = 100Ω + i (199 999.95) Ω ⇒ R =∣∣∣R∣∣∣ = 199 999.98 Ω

b) R = 86.21Ω + i 34.48Ω ⇒ R =∣∣∣R∣∣∣ = 92.85Ω

5.18 a) R = R1 + (ω L)2·R2(ω L)2+R2

2+ i

ω L R22

(ω L)2+R22

b) Rg = 609.36Ω + i 497.11Ω

5.19 u = 231.77V · sin (ωt+ 0.48)

5.20 y = 22.37 cm · cos (ωt+ 5.74)


A.6 A.6 Losungen zu Grenzwert und Stetigkeit

6.1 a) n > 103 b) n > 104 c) n ≥ 1010

6.2 a) 12

b) ∞ c) 1 d) 5 e) 12

f) 43

g) 45

h) 3 i) sin π2

= 1

6.3∣∣an − 1

2

∣∣ < ε⇔ n > 14 ε

6.4 a) 12

b) Mit q :=∣∣∣an+1

an

∣∣∣ folgt |an| ≤ qn−1 |a1| → 0 fur n→∞.

6.5 L1 : |an + bn − (a+ b)| ≤ |an − a|+ |bn − b| n→∞−→ 0

L2 : |an · bn − a · b| = |an (bn − b) + (an − a) b|≤ |an| |bn − b|+ |b| |an − a| n→∞−→ 0

6.6 a) limit (1 / n ∗ sum(1 / i, i = 1..n), n = infinity);

b) limit (n / sqrt[n] (n!), n = infinity);

6.7 a) 7 b) − 23

c) 0

6.8 a) 0 b) −7 c) 2 d) 74

e) 12

f) 1

6.9 limx→1

f (x) = 2

6.10 limh→0

f (x− h) = 0 6=limh→0

f (x+ h) = −2

6.11 limh→0

f (x0 + h) = 2 = f (x0) = limh→0

f (x0 − h)

6.12 Ja: f (1) := 12

A.7 A.7 Losungen zur Differenzialrechnung

7.1 a) y′ = 56x6 − 30x2 − 30x−4 + 56x−8

b) y′ = 9x−14 − 4x−

37 + 11 + 12x−

52

c) y′ = 160l−

5960

d) y′ = −13·92

a−152

e) y′ = 2 b x (c+ e x)3 +(a+ b x2

)· 3 (c+ e x)2 · e

f) y′ = 72x

52 + 5

2x

32

g) y′ = (α+ β) xα+β−1

7.2 a) 10 cos(x)

x3 − 30 sin(x)

x4

b) cos2(x)− sin2(x)

c) nxn−1ex + xnex

d) −7(x−10)2

e) −11−cos(ϕ)

f) −4 2t2−t+1(t+1)2(t−1)(t2−1)

g) − x2−2(x2+2)2

h) −xex(−2ex+2+x)

(ex−1)2

i) - ln(x)+1−x+x ln(x)

(x−1)3

7.3 a) y′ (x) = − sin (3x+ 2) · 3b) y′ (x) = 3 (3x− 2)2 · 3c) y′ (x) = 15 cos (5x)

d) y′ (x) = (8x− 3) e4 x2−3 x+2

e) y′ (x) = 20 x1+x2

A.7 Losungen zur Differenzialrechnung 11

f) x′ (t) = Aω cos (ωt+ ϕ)

g) y′ (x) = cos(2 x−3)sin(2 x−3)

· 2h) y′ (x) = 1

21√

ln(x2−1)· 2 x

x2−1

7.4 a) y′ (x) = xx (lnx+ 1)

b) y′ (x) = xsin x · x cos x ln x+sin xx

7.5 a) y′ = x(xx) · xx((lnx+ 1) lnx+ 1

x

)b) y′ = (xx)x · x (2 lnx+ 1)

c) y′ = x(xa)+a−1 (a lnx+ 1)

d) y′ = x(ax) ax(ln a lnx+ 1

x

)e) y′ = a(xx) · xx (lnx+ 1) ln a

7.6 a) tt2−a2 b) 2 x

x4−1c) x+13

x2−4 x−5d) aln(x−3) ln a

x−3e)

ex (2−x2)(1−x)

√1−x2

f) 1

7.7 sinh′ (x) = cosh (x) , cosh′ (x) = sinh (x) , tanh′ (x) = 1cosh2(x)

7.8 arcsin′ (x) = 1√1−x2

, arccos′ (x) = − 1√1−x2

arctan′ (x) = 1x2+1

, arccot′ (x) = − 1x2+1

7.9 ar sinh′ (x) = 1√x2+1

, ar cosh′ (x) = 1√x2−1

7.10 y (x) = xn → ln y = n lnx → y′ = y · n 1x

= nxn−1

7.11 a)−2 sin(2 x)−ex y·y− y3

xex y·x+3 y2 ln x

b)y·ln y·ex y− 4·ln a

y3

e−x y

y−x·ln y·e−x y− 12 x+6

y4

c) 2 yy−2

√y

d) 2 x ycos y−x2

7.12 y′ (4) = −0.436

7.13 f√

1 + x4 3 ln(1 + 3x5

)2 cosx

f ′ 2 x3√1+x4

45 x4

1+3 x5 −2 sinx

i) df = f ′ (x) dx 2 x3√1+x4

dx 45 x4

1+3 x5 dx −2 sinx dx

ii) df (x0) = f ′ (x0) dx√

2 dx 4.993 dx√

2 dx

iii) yt = f (x0) + df (x0)√

2x 4.993x+ 4.8 1.414x− 0.3034

iv) Linearisierung√

2x 4.993x+ 4.8 1.414x+ 0.3034

v) f (x0 + ε) ≈ 1.4283 19.82893 1.428

exakt 1.4142 19.82898 1.400

Punkt (x0, y0)(1,√

2)

(3, 19.779)(

π4,√

2)

7.14 a) x (t) = Ae−γ t cos (ωt)

x (t) = −γ A e−γ t cos (ωt)− ωAe−γ t sin (ωt)

x (t) = Aγ2 e−γ t cos (ωt) +Aγ ω e−γ t sin (ωt)−Aω2 e−γ t cos (ωt) +Aγ ω e−γ t sin (ωt)

b) x (t) = 0 ⇒ −γ cos (ωt)− ω sin (ωt) = 0 ⇒ tan (ωt) = − γω

7.15 V ′ (a) = 0 mit V ′′ (a) = 2 Da2 > 0

7.16 relativer Fehler dαα

= 3.9 · 10−3 ≈ 4 0/00

7.17 a) Minimum(− 1

2;−5

)Maximum (1.5; 27)

b) Maximum (0; 16) Minimum (±2; 0)

c) Maximum (0; 2)

d) Maximum (1; 0.368)

e) Maxima xk = π4

+ k · π yk = 0.5 k ∈ ZZ


Minima xk = 34π + k · π yk = −0.5 k ∈ ZZ

f) Minimum (0, 5;−0.08)

7.18 a) y = x2+1x−3

: ID = IR \ 3, VW = (−∞,−0.325] ∪ 12.325, ∞),

Pol: x = 3, Vertikale Asymptote x = 3, Asymptote im Unendlichen y =

x+ 3,

Extremwerte: Max (−0.162,−0.325) Min (6.162, 12.325) .

b) y = (x−1)2

x+1: ID = IR \ 1, VW = (−∞,−8] ∪ [0, ∞),

Pol: x = −1, Vertikale Asymptote x = −1; Asymptote im Unendlichen y =

x− 3,

Extremwerte: Max (−3,−8) Min (1, 0) .

c) y = ln xx

: ID = (0, ∞), VW = (−∞, 0.368) ,

Nullstellen: x = 1, Pol: x = 0, Asymptote fur x→∞: y = 0,

Extremwert: Max (2.71, 0.368) , Wendepunkt: (4.48, 0.335) .

d) y = sin2 x : ID = IR, VW = [0, 1] ,

Periodizitat π, Nullstellen: xk = k π,

Extrema: Max(xk = π

2+ k π; yk = 1

)Min (xn = k π; yk = 0) ,

Wendepunkte xk = π4

+ k · π2

yk = 12

.

7.19 a) 2 a b) 2 c) 2 d) 112

e) 0 f) 1−2

g) 1 h) ea

A.8 A.8 Losungen zur Integralrechnung

8.1 rightbox (sqrt(x), x = 0..2, 10); rightsum (sqrt(x), x = 0..2, 10);

8.2 a) x6

6+ C

b) − 1x

+ C

c) 34z

43 + C

d) 3x13 + C

e) 23x3 − 5

2x2 + 3x+ C

f) 23x

32 − 2

5x

52 + C

8.3 a) 1 b) 2π2 + 2 c) ln a

8.4 a) x sinx+ cosx+ C

b) 12

sin2 x+ C

c) −x2 cosx+ 2x sinx+ 2 cosx+ C

d) 13x3 lnx− 1

9x3 + C

e) x ex − ex + C

f) x2 ex − 2x ex + 2 ex + C

8.5 a) ln |x+ 2|+ C b) 12

ln∣∣x2 − 1

∣∣+ C c) − 16

ln∣∣1− 2x3

∣∣+ C

d) 19

13

(3s+ 4)9 + C e) − 1ω

cos (ωt+ ϕ) + C f) 13

sin (3t) + C

g) −e−x + C h) − ln |cos t|+ C i) ln |x ex|+ C

j) 12

sin2 x+ C k) 23

13

(4 + 3x)32 + C

8.6 Nachrechnen durch Differenzieren der rechten Seite

8.7 a) − 23x√x+ x+ 2

√x− 2 ln (1 +

√x) + C b) − 1

3

√1− x2

3+ C

A.8 Losungen zur Integralrechnung 13

8.8 a) 23

√1 + x3 + C

(u = 1 + x3

)b) 2

15

√5x+ 12

3+ C (u = 5x+ 12)

c) − 34

3√

(1− t)4 + C (u = 1− t)

d) 0 (u = cosx)

e) 12

arctan2 (z) + C (u = arctan z)

f) ln∣∣x2 + 6x− 12

∣∣+ C(u = x2 + 6x− 12

)g) ln |ln (x)|+ C (u = lnx)

h) − 12

cos(x2)

+ C(u = x2

)i) 1

2ln∣∣2x3 − 4x+ 2

∣∣+ C(u = 2x3 − 4x+ 2

)j) 0

(u = 1 + t2

)k) 0.47

(u = 3 t− π

4

)l) 2.055 (u = 5− x)

m) 13ex3−2 + C

(u = x3 − 2

)n) 1

2tan2 (z + 5) + C (u = tan (z + 5))

o) −√

4−x2

x− arcsin

(x2

)+ C (x = 2 sinu)

8.9 a) 12x2 lnx− 1

4x2 + C

b) x · sinx+ cosx+ C

c) t ln t− t+ C

d) − 13x cos (3x) + 1

9sin (3x) + C

e) x arctanx− 12

ln(1 + x2

)+ C

f) 12

(t− 1

ωsin (ωt) cos (ωt)

)g) 1

2ex (sinx+ cosx) + C

h) −x2 e−x − 2x e−x − 2 e−x + C

8.10 a) 12 a

(ln |x− a| − ln |x+ a|) + C

b) 2 ln |x+ 1|+ 23

ln |x− 1| − 323

ln |x+ 2|+ C + 4x

c) 13

ln∣∣∣ z−1

z+2

∣∣∣− 2 1z+2

+ C

d) 178

ln |x− 9|+ 158

ln |x+ 7|+ C

e) 19

ln∣∣∣ x

x−3

∣∣∣− 73 (x−3)

+ C

8.11 a) 23

(lnx)32 + C

b) ln |sinx|+ C

c) x sinh (x)− cosh (x) + C

d) −ecos x + C

e) x+ 14

ln |x− 1| − 54

ln |x+ 1| − 12

1x+1

+ C

f) x− 5 ln |x+ 1|g) 1

4(lnx)4 + C

h) 2 ln∣∣2x3 − 1

∣∣+ C i) 12

(x2 + 1) arctanx− 12x+ C

8.12 ln(x+√

1 + x2)x−√

1 + x2

8.13 83 −

43

√1 +

√x(2−

√x) +C

8.14 convert ( ” , x, parfrac)

8.15 a) 0 b) 0 c) 0 fur n 6= m ; π fur n = m

d) 0 fur n 6= m ; π fur n = m e) 0

8.16 a) 12

b) π2

c) 12

d) 1−a+s

e) ss2+a2 f) n!

sn+1


A.9 A.9 Losungen zu Funktionenreihen

9.1 a)∑∞

n=11√n

=∑∞

n=11

n12⇒ Satz: Divergenz

b) Quotientenkriterium ⇒ Konvergenz

c) |sin n|n2 ≤ 1

n2 ⇒ Majorantenkriterium: Konvergenz

d) n2 n+1

−→n→∞

12⇒ Koeffizienten keine Nullfolge ⇒ Divergenz

e) Quotientenkriterium ⇒ Konvergenz

f) Quotientenkriterium ⇒ Konvergenz

g) Quotientenkriterium ⇒ Konvergenz

h) Leibniz-Kriterium ⇒ Konvergenz

i) Leibniz-Kriterium ⇒ Konvergent

j) Quotientenkriterium ⇒ Konvergenz

k) Quotientenkriterium ⇒ Divergenz

l) Majorantenkriterium ≤ 1n2

9.2 a) 6 b) e c) 6

9.4 5, 1, 1e, 4

9.5 a) K = (−2, 2) b) K = [−1, 1] c) K = (−1, 1) d) K = (−1, 1]

e) K = (−2, 2) f) K = (−1, 1) g) K = IR h) K =[− 1

2, 1

2

)9.6 a) K = (−e+ 4, e+ 4) b) (−1, 3) c) K = IR d) K = IR

9.7 siehe §9.3 Tabelle 9.1

9.8 f (x) = −1 + (x− 1)2 − 2 (x− 1)3 + 3 (x− 1)4 − . . .± (n− 1) (x− 1)n ± . . .

= −1 +∑∞

n=2 (n− 1) (x− 1)n (−1)n+1 ; K = (0, 2]




9.12 f (x) = 12

∑∞n=0

(−1)n

(2 n)!

(x− π

3

)2 n+ 1

2

√3∑∞

n=0(−1)n+1

(2 n+1)!

(x− π

3

)2n+1

K = IR

9.14 f (x) = x− x2 + 12x3 +R3 (x) mit |R3 (x)| ≤ 1

6|x|4

9.15 Rn (x) = 1n!

12

1·3·...·(2 n−3)2n (1− ξ)−

2 n−12 xn ≤ 10−4 fur ξ ∈ (0, 0.05)

⇒ n = 5

9.18 F (x) =∑∞

n=0

(−1)n

2 n+1x2 n+1

9.19 taylor (m / k ∗ ln(cosh(sqrt(k ∗ g / m) ∗ t)), k = 0, 3);12g t2 − 1

12g2 t4

mk + 1

45g3 t6

m2 k2 +O(k3)

9.20 sinc(x) = x− x3

3! 3+ x5

5! 5− x7

7! 7± . . .

erf (x) = 2√π

(x1− x3

1! 3+ x5

2! 5− x7

3! 7± . . .

)9.21 a) z3 = x3 − 3x y2 + i

(3x2 y − y3

)b) 1

1−z= 1−x

(1−x)2+y2 + i y

(1−x)2+y2

c) e3 z = e3 x cos 3 y + i e3 x sin 3 y

9.22∣∣ei z

∣∣ = 1 e−3√

3

9.23 a) f′i (x) = 3 (1 + i)3 x2; f

′ii (x) = 3 (1 + i) e3 (1+i) x

b)∫fi (x) dx = (1 + i)3 1

4x4 + c;

∫fii (x) dx = 1

3 (1+i)e3 (1+i) x + c

9.24 RL = iwL

A.10 Losungen zur Differenzialrechnung bei Funktionen mit mehreren Variablen 15

A.10A.10 Losungen zur Differenzialrechnung bei Funktionenmit mehreren Variablen

10.1 plot3d(z, x=a..b, y=c..d);

10.2 plot3d(z, x=a..b, y=c..d, style=contour, contours=20);

10.3 a) fx = 3x2 + y; fy = x+ 2 y−3

b) fa = 3x; ft = 2 y t−1

c) fu = (u+v)−(u+w)

(u+v)2; fv = − u+w

(u+v)2

d) fx = 2 x√1+(x2+z2)2

; fy = 0; fz = 2 z√1+(x2+z2)2

e) fx1 = fx3 = 0; fx2 = 1

f) fa = −(a x+ b x2

)−2x+ y b ea b; fb = −

(a x+ b x2

)−2x2 + y a ea b

10.4 a) fx = 6x+ 4 y; fy = 4x− 4 y; fxx = 6; fxy = fyx = 4; fyy = −4

b) fx = −6 y sin (3x y) ; fy = −6x sin (3x y)

fxx = −18 y2 cos (3x y) ; fxy = fyx = −6 sin (3x y)− 18x y cos (3x y) ;

fyy = −18x2 cos (3x y)

c) fx = 12 (3x− 5 y)3 ; fy = −20 (3x− 5 y)3

fxx = 108 (3x− 5 y)2 ; fxy = fyx = −180 (3x− 5 y)2 ; fyy = 300 (3x− 5 y)2

d) f (x, y) = x− y; fx = 1; fy = −1; fxx = 0; fxy = fyx = 0; fyy = 0

e) fx = 3 ex y + 3x y ex y; fy = 3x2 ex y

fxx = 6 y ex y + 3x y2 ex y; fxy = fyx = 6x ex y + 3x2 y ex y; fyy = 3x3 ex y

f) fx = x−y

(x2−2 x y)1/2 ; fy = −x

(x2−2 x y)1/2

fxx = −y2

(x2−2 x y)3/2 ; fxy = fyx = x y

(x2−2 x y)3/2 ; fyy = −x2

(x2−2 x y)3/2

10.5 ∂∂xf (x, y) = 2x cos

(x2 + 2 y

); ∂

∂yf (x, y) = 2 cos

(x2 + 2 y

)∂

∂y∂

∂xf (x, y) = 2x·[− sin

(x2 + 2 y

)]·

∂∂y

(x2 + 2 y

)= −4x sin

(x2 + 2 y

)∂

∂x∂

∂yf (x, y) = −2 sin

(x2 + 2 y

)·

∂∂x

(x2 + 2 y

)= −4x sin

(x2 + 2 y

)⇒ ∂

∂y∂

∂xf (x, y) = ∂

∂x∂

∂yf (x, y)

10.7 fxx = 108 (3x− 5 y)2 ; fxy = fyx = −180 (3x− 5 y)2 ; fyy = 300 (3x− 5 y)2

10.8 fxx =(x2 + y2 + z2

)−5/2 3 a x2 − a

(x2 + y2 + z2

)fyy =

(x2 + y2 + z2

)−5/2 3 a y2 − a

(x2 + y2 + z2

)fzz =

(x2 + y2 + z2

)−5/2 3 a z2 − a

(x2 + y2 + z2

)⇒ fxx + fyy + fzz = 0

10.10 fxx = −x2+y2

(x2+y2)2; fyy = x2−y2

(x2+y2)2⇒ fxx + fyy = 0

10.11 zt = −9 + 18x+ 6 y

10.12 df = 0.5 dx+ 0.826 dy

10.13 grad f =

(2 (3x+ x y) (3 + y)

2x (3x+ x y)

); f~a = 2√

5x (3 + y)2 + 4√

5x2 (3 + y) ;

→ gradplot

10.14 grad f =

2 x

(1+x2) y

cos (z)− ln(1+x2)y2

−y sin (z)

; f~a = 2√5

x

(1+x2 ) y− 2√

5y sin (z) ;

→ gradplot3d


10.15 a) dz =(12x2 y − 3 ey

)dx+

(4x3 − 3x ey

)dy

b) dz = x2−2 x y−y2

(x−y)2dx+ x2+2 x y−y2

(x−y)2dy

c) df = xx2+y2+z2 dx+ y

x2+y2+z2 dy + zx2+y2+z2 dz

10.16 a) ∂∂x1

h (x1, x2) = c1 a1;∂

∂x2h (x1, x2) = c2 a

22

b) ∂∂x1

h (x1, x2) = (2 sin (x1) + cos (x2)) cos (x1) ;∂

∂x2h (x1, x2) = − sin (x1) sin (x2)

10.17 a) f (x, y) = 14

(y2 − x2

)+ (x− y) +R2 (x, y)

b) f (x, y) = e+ 2 e (x− 1) + 3 e (x− 1)2 + e y2 +R2 (x, y)

10.18 a) df = 2x cos(x2 + 2 y

)dx+ 2 cos

(x2 + 2 y

)dy

b) df = (6x+ 4 y) dx+ (4x− 4 y) dy

c) df = −y sin (x− 2 y) dx+ (cos (x− 2 y) + 2 y sin (x− 2 y)) dy

d) df =(2x z + 4x3

)dx− z3 dy +

(x2 − 3 y z2

)dz

10.19 dV = ∂V∂ddd+ ∂V

∂hdh; |dV | ≤ 0.6786 (absoluter Fehler);∣∣ dV

V

∣∣ ≤ 0.6% (relativer Fehler)

10.20 dRR

= 3.8 %

10.21 E = (212 206± 6030) Nmm2 ,

da dE = kπ r2 z

dl − 2 l kπ r3 z

dr + lπ r2 z

dk − l kπ r2 z2 dz

10.22 1.1 ist der relative Fehler.

10.23 2 + 12

ln 2 + 12

(x− 1) +(1− 1

4ln 2)

(y − 2)

10.24 a) Stationare Punkte: (0, n π) n ∈ ZZ, fur n ungerade liegen lokale Mini-

ma vor.

b) Stationarer Punkt: (0, 0), Sattelpunkt

c) Stationare Punkte: (0, 0) ist Sattelpunkt; (1,−1) ist lokales Minimum.

10.25 d = |P − z| =

∣∣∣∣∣∣∣ 1

−2

0

−

x

y√1 + (x− 2 y)2

∣∣∣∣∣∣∣

=√

6− 2x+ 2x2 − 4x y + 4 y + 5 y2

d!= minimal→ dx = 0 und dy = 0 ⇒ (x, y) =

(16, − 1

3

)10.27 (1, 2): relatives Minimum;

(−1, −2): relatives Maximum;

(−1, 2) und (1,−2): Sattelpunkte

10.28 a) (0, 0): Sattelpunkt ; (1, 1): relatives Maximum

b)(− 1

2, 1

2

): relatives Minimum

c)(

12, 0): Sattelpunkt

d) (0, 0): Sattelpunkt ;(0, ±

√ln 2): lokales Minimum

10.29 Stationare Punkte:

P1 (0, 0, 0) ; P2 (0, 0, 1) ;P3 (0, 0, −1) ;P4 (0, λ, 0) ;

P5 (1, 0, 0) ;P6 (−1, 0, 0).

Lokale Minima: P2 , P3 , P5 , P6

10.30 a) −1.3150x+ 2.0607 b) 0.9619x+ 0.4769

A.11 Losungen zur Integralrechnung bei Funktionen mit mehreren Variablen 17

A.11A.11 Losungen zur Integralrechnung bei Funktionen mitmehreren Variablen

11.1 a) 13

ln(l) b) −25 c) −2 d) 32π

11.2 16

11.3 I1 = I2 = 323

11.4 2π

11.5 a) 1256

b) xs = 2.5 ys = −2.5

11.6 Ix = Iy = π16R4 Ip = π

8R4

11.7 xs = 23a ys = 1

3b

11.8 xs = 0 ys = 43πR

11.9 a) 12

b) 427l9 c) 4

3R3 d) 2

3R4

11.10 zs = 23R2 Ix = Iy = 1

6M R2

(3R2 + 1

)Iz = 1

3M R2

11.11 Ix = Iy = 25M R2 Iz = 2

5M R2

11.12 I = 115

11.13 xs = ys = 43π, zs = 1

2

A.12A.12 Losungen zu Differenzialgleichungen

Differenzialgleichungen 1. Ordnung

12.1 a) c e−4 x b) c e−2 x c) c e−83 x d) c e

ba

x e) c e6 x f) c e−RL

t

12.2 a) v (t) = 10 ms· e−2 t , t > 2.3 sec

b) y (x) = e−λ x , λ = ln 2

c) N (t) = N0 e− t

τ , τ = 8321.5 Jahre!

12.3 a) y0 e− 1

2 x2+ 4

b) 14

11+x

(4 y0 + (2x+ 1)) e2 x

c) 1x

(y0 − x cosx+ sinx)

d) 1cos x

(y0 + x)

e) y0 e2 sin x − 1

2

f) x(y0 + x− 4

x

)12.4 a) y (t) = const · e−

1R C

t + U0 R C

1+(ω R C)2sin (ωt) + U0 ω R2 C2

1+(ω R C)2cos (ωt)

b) y (t) = const · e−RL

t + U0L

R−2 Le−2 t

12.5 a) x1+C x

b) c√

1 + x2

c) e2x +2 c−1e2 x+2 c+1

d) 1− 1x+C

e) arccos(

12x2 + C

)f) − ln (− sinx+ C)

12.6 a) y (x) = 2π e− sin x+1

b) y (x) = xx+1

c) y (x) = 3√

3 + 3x− x3


d) y (x) = x−1−ln x

e) y (x) =√

2 + 2 e2 x

12.7 a) y (x) = 4x lnx+ C x b) y (x) = 12x− x

ln C x

12.8 a) y (x) = 3√

12x6 + C b) y (x) = C x−1

x+1c) y (x) = C

√x2 + 1

d) y (x) =√C · ex − 1 e) y (x) = a tan (x) f) y (x) =

(C −

√x3)− 1

3

12.9 a) y (x) = 1cos(x)

b) y (x) = cosh (x) c) y (x) = π2

sin (x)

d) y (x) = 2(x2 + 1

)Anwendungen DG 1. Ordnung

12.11 a) x (t) =a b (e(a−b) k t−1)

b e(a−b) k t−a

b) limt→∞

x (t) = a ba

= b, d.h. wenn alle Molekule vom Typ B reagiert haben.

12.12 a) v (t) = m gk

(1− e−

km

t)

+ v0 e− k

mt

s (t) = m gkt+(

mk

)2g(e−

km

t − 1)− v0

mk

(e−

km

t − 1)

b) vmax = limt→∞

v (t) = m gk

12.13 v (t) =√

m gk

tanh√

m gkt

Fur t→∞ wird die Endgeschwindigkeit erreicht vE = limt→∞

v (t) =√

m gk

s (t) =v2

Eg

ln cosh gvE

t

12.14 T (t) = e−a t (T0 − TL) + TL ⇒ limt→∞

T (t) = TL

12.15 vz (r) = − 14 η

∆p∆z

(R2 − r2

)(Gesetz von Hagen-Poiseuille)

12.16 a) v (t) = 1R·m · g · sinϕ ·

(1− e−

Rm

t)

b) v (t) =√

m g sin ϕD

tanh

(√D g sin ϕ

mt

)=(

52

tanh (2 t))

c) v (t) = 18

(25 tanh

(ln 2 + 5

2t)− 15

)Lineare Differenzialgleichungssysteme

12.17 a)

2

0

1

et ,

0

1

0

e4 t ,

1

0

−2

e6 t

b)

2

1

3

et ,

1

1

2

e−t ,

1

−1

−1

e2 t

12.18

(2

1

)e√

0 t ,

(1

−2

)e±

√5 t

12.19 a)

(1

−i

)eiωt ,

(1

i

)e−iωt komplexes Fundamentalsystem(

cos (ωt)

sin (ωt)

),

(sin (ωt)

− cos (ωt)

)reelles Fundamentalsystem

b) ψ (t) = E0

(− 1

ωt

1ω2

)spezielle Losung

A.12 Losungen zu Differenzialgleichungen 19

12.20 a) −2 EW →

(1

1

)EV − 4 EW →

(1

−1

)EV

b)

(1

1

)e−2 t ,

(1

−1

)e−4 t FS

c)

(1

1

)e√

2 i t ,

(1

1

)e−

√2 i t ,

(1

−1

)e2 i t ,

(1

−1

)e−2 i t kompl. FS

d)

(1

1

)cos

√2 t ,

(1

1

)sin

√2 t ,

(1

−1

)cos 2 t ,

(1

−1

)sin 2t reelles FS

e) ~y ′ (t) =

0 1 0 0

−3 0 1 0

0 0 0 1

1 0 −3 0

~y (t)

12.21 a) det (A− λ I) = λ2 + 2λ+ 5 = (−1 + 2 i− λ) (−1− 2 i− λ)

~ϕ1 (x) =

(− 4

5− 2

5i

1

)e(−1+2 i) x , ~ϕ2 (x) =

(− 4

5+ 2

5i

1

)e(−1−2 i) x

b) det (A− λ I) = (3− λ) (2− λ) (6− λ)

~ϕ1 (x) =

−1

0

1

e2 x , ~ϕ2 (x) =

−1

1

−1

e3 x , ~ϕ3 (x) =

1

2

1

e6 x

c) det (A− λ I) = (2− λ)2 (5− λ)

~ϕ1 (x) =

1

1

0

e2 x , ~ϕ2 (x) =

0

1

1

e2 x , ~ϕ3 (x) =

1

−1

1

e5 x

12.22 ~y ′ (x) =

3 2 −1

2 3 −1

−1 −1 4

~y (x) , det (A− λ I) = (1− λ) (3− λ) (6− λ)

~ϕ1 (x) =

−1

1

0

ex , ~ϕ2 (x) =

1

1

2

e3 x , ~ϕ3 (x) =

1

1

−1

e6 x

AWP: ~y (x) = ~ϕ1 (x) + ϕ2 (x) + 2 ~ϕ3 (x)

12.23

(y1

y2

)′

=(

0 1−6 5

) (y1

y2

), det (A− λ I) = (2− λ) (3− λ)

~ϕ1 (x) =(

12

)e2 x , ~ϕ2 (x) =

(13

)e3 x , ~y (x) = α ~ϕ1 (x) + β ~ϕ2 (x) ,

y (x) = y1 (x) = α e2 x + β e3 x AWP: y (x) = 3 e2 x − 2 e3 x

12.24 ~ϕ1 (x) =

(−1

1

)sin (x) , ~ϕ2 (x) =

(−1

1

)cos (x) ,

~ϕ3 (x) =

(2

3

)e−2 x , ~ϕ4 (x) =

(2

3

)e2 x

Differenzialgleichungen hoherer Ordnung

12.25 Kein Fundamentalsystem!

12.26 Fundamentalsystem

12.27 a) λ2 + 13λ+ 40 = (λ+ 5) (λ+ 8) = 0 , ϕ1 (t) = e−5 t , ϕ2 (t) = e−8 t

b) λ2 − 12λ+ 36 = (λ− 6)2 = 0 , ϕ1 (t) = e6 t , ϕ2 (t) = t e6 t


c) λ2 + 6λ+ 34 = (λ+ 3− 5 i) (λ+ 3 + 5 i) = 0

ϕ1 (x) = e−3 x cos (5x) , ϕ2 (x) = e−3 x sin (5x)

d) λ2+16 = (λ+ 4 i) (λ− 4 i) = 0 , ϕ1 (x) = cos (4x) , ϕ2 (x) = sin (4x)

12.28 a) λ4 − 10λ2 + 9 = (λ+ 1) (λ− 1) (λ+ 3) (λ− 3) = 0

ϕ1 (x) = ex , ϕ2 (x) = e−x , ϕ3 (x) = e3 x , ϕ4 (x) = e−3 x

b) λ3 − 2λ2 + λ = λ (λ− 1)2 = 0

ϕ1 (t) = 1 , ϕ2 (t) = et , ϕ3 (t) = t et

c) λ6 − 1 = 0 = (λ− 1) (λ+ 1)(λ− 1

2− 1

2

√3 i)(

λ+ 12− 1

2

√3 i) (λ− 1

2+ 1

2

√3 i) (λ+ 1

2+ 1

2

√3 i)

ϕ1/2 (x) = e±x, ϕ3/4 (x) = e±12 x sin

(12

√3x), ϕ5/6 (x) = e±

12 x cos

(12

√3x)

12.29 λ2 − 3λ+ 2 = (λ− 1) (λ− 2) = 0

a) yp (x) = 3

b) yp (x) = 34

+ 12x

c) yp (x) = x e2 x

d) yp (x) = 110

(cosx− 3 sinx)

e) yp (x) = 3 + 4x+ cosx− 3 sinx

f) yp (x) =(

12x2 − x

)e2 x

g) yp (x) = − 12

(sinx+ cosx) ex

12.30 a) λ2 + 16 = (λ− 4 i) (λ+ 4 i) = 0 , x (t) = 3 cos (4 t) + sin (4 t)

b) λ2 + 2λ + 2 = (λ+ 1 + i) (λ+ 1− i) = 0 , x (t) = 2 sin (t) e−t +

2 cos (t) e−t

c) λ2 + 13λ+ 40 = (λ+ 5) (λ+ 8) = 0 , x (t) = 8 e−5 t − 5 e−8 t

12.31 a) λ4 − 10λ2 + 9 = (λ+ 1) (λ− 1) (λ+ 3) (λ− 3) = 0

y (x) = 120

sin (x) + C1 e−x + C2 e

x + C3 e−3 x + C4 e

3 x

b) λ3 − 7λ− 6 = (λ+ 1) (λ+ 2) (λ− 3) = 0

y (x) = −ex + C1 e−x + C2 e

−2 x + C3 e3 x

c) λ3 − 2λ2 + λ− 2 = (λ− 2) (λ+ i) (λ− i) = 0

y (x) = − 15

(x sin (x) + 2x cos (x)) + C1 e2 x + C2 cos (x) + C3 sin (x)

d) λ3 − 6λ2 + 12λ− 8 = (λ− 2)3 = 0

y (x) = x3 e2 x + C1 e2 x + C2 x e

2 x + C3 x2 e2 x

A.13 Losungen zur Laplace-Transformation 21

A.13A.13 Losungen zur Laplace-Transformation

13.1 a) 3s+4

; s > −4

b) 4s3 ; s > 0

c) 4 ss2+25

; s > 0

d) πs2+π2 ; s > 0

e) −3√

πs

; s > 0

13.2 a) 72s2 − 3

√π

2 s52

+ 6s

b) 10−3 ss2+4

c)

√( 1

3 )

s43

+ 4s−2

d) nicht moglich!

13.3 a) F (s) = As

(1− e−s t0

)+ A

s+2e−s t0

b) F (s) = Ase−a s − A

se−b s

c) F (s) = 1s2

(1− e−3 s

)d) F (s) = 1

s2+1

(1 + e−π s

)13.4 a) 5 e−2 t

b) 4 cos (2 t)− 32

sin (2 t)

c) 2− 5 t

d) tk−1

Γ(k)

e) 1√π

(8 t

12 − 5 t−

12

)f) 1

2

(1− e−2 t

)13.5 a) 2 et cos (2 t) + 5

2et sin (2 t)

b) S (t− 2) (t− 2)

c) 16S (t− 5) (t− 5)3

d) 12t2 − 1

2S (t− 2) (t− 2)2

13.6 a) − 43e2 t − 1

6e−t + 7

2e3 t

b) 2 et − 2 cos t+ sin t

c) −e−t − 12t2 e2 t + 2 t e2 t + e2 t

d) et(t2 − t+ 3

)13.7 y (t) = 1

13.8 dsolve(DG, y(0)=1, (D(y))(0)=2, (D@@2)(y)(0)=3,

(D@@3)(y)(0)=0, y(t), method = laplace);

13.9 I (t) = I0 e−R

Lt

Anwendungen der Laplace-Transformation

13.10 I (t) = − 52

cos (5 t) + 52

sin (5 t) + 52e−5 t

13.11 a) mx = −Dx → x (t) = 5 cos

(√Dmt

)ω: x (2) = 5 cos (2ω) = 2.5 ⇒ ω = π

6⇒ x (t) = 5 cos

(π6t)

b) x (t0) = −5 π6

c) x (t0) = 0

13.12 x+ 4 x+ 8x = 20 cos (2 t) ; x (0) = 0 , x (0) = 0

⇒ X (s) = 20 ss2+4

· 1s2+4 s+8

⇒ x (t) = cos (2 t) + 2 sin (2 t)− e2 t (cos (2 t) + 3 sin (2 t))

13.13 −13.14 X (s) = 8 ω

s2+ω2 · 1s2+4

⇒ x (t) = − 8ω2−4

sin (ωt) + 4ω2−4

sin (ωt)

Resonanz bei ω = 2


A.14 A.14 Losungen zu Fourier-Reihen

14.1 a) f (t) =

1 −π

2< t < π

2

0 t = ±π2

−1 −π < t < −π2, π

2< t < π

f (t) gerade ⇒ bn = 0 fur n = 1, 2, 3, . . .

a0 = 0 , an = 4π n

[sin(n π

2

)]=

4

n πn = 1, 5, 9, . . .

− 4n π

n = 3, 7, 11, . . .

0 n gerade

⇒ f (t) = 4π

cos (t)− 1

3cos (3 t) + 1

5cos (5 t)− 1

7cos (7 t)± . . .

= 4

π

∑∞n=0 (−1)n 1

2n+1cos ((2n+ 1) t)

b) f (t) =(1− x

2π

)fur 0 < x < 2π

a0 = 12, an = 0 , bn = 1

π n⇒ f (t) = 1

2+ 1

π

∑∞n=1

1n

sin (n t)

Der Fourierwert an der Stelle t = 0 ist f (0) = 12.

14.2 a) f gerade, bn = 0 , f (t) = 2T

(1− e−

T2

)+4T

∑∞n=1

1−(−1)n e−T

2

T2+4 n2 π2 cos(n 2π

Tt)

b) f (t) = 34h− 2 h

π2

∑∞n=1

n ungerade

1n2 cos

(n 2π

Tt)− h

π

∑∞n=1

1n

sin(n 2π

Tt)

14.3 a) cn = 1n π

2 12i

[ei n π

2 − e−i n π2

]= 2

n πsin(n π

2

)=

2

n πfur n = 1, 5, 9, . . .

− 2n π

fur n = 3, 7, 11, . . .

0 fur n gerade

b) cn = 12π

1i n

= −i 12π n

, c0 = 12⇒ f (t) = −i

2π

∑∞n=−∞n6=0

1nei n t + 1

2

14.4 Nach den Moivreschen Formeln gilt

sin3 (t) = 14

(3 · sin (t)− sin (3 t)) = 34

sin (t)− 14

sin (3t).

14.5 a) f (t) = 13T 2 +

∑∞n=1

(T2

n2 π2 cos(n 2π

Tt)− T2

n πsin(n 2π

Tt))

b) f (t) = 43π2 +

∑∞n=1

(4

n2 cos (n t)− 4πn

sin (n t))

c) f (2π) = (2π)2+02

= 2π2 = 43π2 +

∑∞n=1

4n2 ⇒

∑∞n=1

1n2 = π2

6

14.6 a0 = u02, an = 0 , bn = −u0

π1n⇒ u (t) = u0

2− u0

π

∑∞n=1

1n

sin(n 2π

Tt)

Gleichspannungsanteil u02

Grundschwingung mit Amplitude u0π

und Frequenz ω0 = 2πT

Sinusformige Oberschwingungen mit Amplitude u02π, u0

3π, u0

4π, . . . bei den Fre-

quenzen 2ω0 , 3ω0 , 4ω0, . . ..

14.7 a) T = 4 , ω0 = 2πT

= π2. Funktion ungerade ⇒ a0 = 0 , an = 0 n ∈

IN.

bn = 16n π

(1− cos (nπ)) =

0 n gerade32n π

n ungerade

⇒ f (x) =∑∞

n=1

n ungerade

32n π

sin(n π

2x)

b) T = 8 , ω0 = 2πT

= π4. Funktion gerade ⇒ bn = 0 n ∈ IN.

a0 = 2 , an = 8n2 π2 (cos (nπ)− 1) =

0 n gerade

− 16n π

n ungerade

⇒ f (x) = 2−∑∞

n=1

n ungerade

16n2 π2 cos

(n π

4x).

14.8 34− 2

π2

∑∞n=1

n ungerade

1n2 cos

(n 2π

Tt)− 1

π

∑∞n=1

1n

sin(n 2π

Tt).

A.15 Losungen zur Fourier-Transformation 23

14.9 a) 20− 40π

∑∞n=1

1n

sin(n π

5t)

b) 32− 12

π2

∑∞n=1

n ungerade

1n2 cos

(n π

3t)− 6

π

∑∞n=1 (−1)n 1

nsin(n π

3t).

A.15A.15 Losungen zur Fourier-Transformation

15.1 a) F (f1) (ω) = 2Asin(ω T

2 )ω

= ATsin(ω T

2 )ω T

2= AT si

(ω T

2

)b) Fur A = 1

Tergibt sich F (f1) (ω) = si

(ω T

2

).

c) F (f2) (ω) = 2Ae−i ω t0sin(ω T

2 )ω

= e−i ω t0 F (f1) (ω)

15.2 F (f) (ω) = ATsin2(ω T

2 )(ω T

2 )2= AT si2

(ω T

2

)= 2 A

ω2T(cos(ωT )− 1)

15.3 a) f ungerade → F (f) (ω) = −2i∫∞0f (t) sin (ωt) dt = −i 2 ω

α2+ω2

b) F (f) (ω) = sin(ωT )

ω(1−( ωT

π )2)

15.4 > fourier (f(t), t, w);

15.5 Die Funktion setzt sich zusammen aus der Funktion f1, aus Aufgabe 15.1a)

mit der Breite 2T und der Dreiecksfunktion aus Aufgabe 15.2 . Mit der

Skalierungseigenschaft und der Linearitat der Fourier-Transformation erhalt

man

F (f) (ω) = 2AT sin(ωT )ωT

+ATsin2(ω T

2 )(ω T

2 )2= 2AT si (ωT ) +AT si2

(ω T

2

).

15.9 a) F (δ (t)) = 1

b) F (δ (t− t0)) = e−iωt0

c) F(

i2

(δ (t+ t0)− δ (t− t0)))

= − sin (ω t0)

d) F (sin (ω0t)) = πi

(δ (ω − ω0)− δ (ω − ω0))

15.10 a) F(ei a t

)(ω) = F

(ei a t · 1

)(ω) =

↑Verschiebungssatz

F (1) (ω − a) = 2π δ (ω − a)

b) δ (t− a) ∗ f (t) =∫∞−∞ δ (t− a− τ)︸︷︷︸

=0τ=t−a

f (τ) dτ = f (t− a)

15.11 rect(

2 tT

)∗ rect

(2 tT

)= tri

(tT

)=

1− |t| |t| ≤ T

0 |t| > T

15.12 a) 1iω

(1− eiωT

)2= −4

iωeiωT sin2

(ωT2

)b) −2i

[−T

ωcos (ωt) + sin(ωT )

ω2

]c) 2

ω[sin (ωT2)− sin (ωT1)]

15.13 t < 0: (f ∗ h) (t) = 0 0 ≤ t ≤ T : (f ∗ h) (t) = t− 1 + e−t

T < t: (f ∗ h) (t) = T − e−t(eT − 1

)


A.16 A.16 Losungen zu Partielle Differenzialgleichungen

16.1 u (x, t) = cos (ωt) sin (k x) ; uxx (x, t) = −k2 cos (ωt) sin (k x) ;

utt (x, t) = −ω2 cos (ωt) sin (k x)

utt − c2 uxx =(−ω2 + c2 k2

)cos (ωt) sin (k x) = 0 fur ω = c k

u (x = 0, t) = cos (ωt) sin (0) = 0; u (x = L, t) = cos (ωt) sin(n π

L

)L =

0 fur k = n πL

16.2 u (x, t) = e−ωt sin (k x) ; uxx (x, t) = −k2 e−ωt sin (k x) ;

ut (x, t) = −ω e−ωt sin (k x)

ut −Duxx =(−ω +Dk2

)e− ωt sin (k x) = 0 fur ω = Dk2

16.3 a) f (x, y) = 12

ln(x2 + y2

); fxx (x, y) = − x2−y2

(x2+y2)2; fyy (x, y) = x2−y2

(x2+y2)2;

fxx + fyy = 0

b) g (x, y, z) =(x2 + y2 + z2

)− 12

gxx (x, y, z) = 3(x2 + y2 + z2

)− 52 x2 −

(x2 + y2 + z2

)− 32 ; gyy , gzz analog

gxx +gyy +gzz = 3(x2 + y2 + z2

)− 52[x2 + y2 + z2

]−3

(x2 + y2 + z2

)− 32 =

0

16.4 a) u (t) = f1 (x+ c t) + f2 (x − c t); utt = c2 (f ′′1 (x+ c t) + f ′′2 (x− c t)) ;

uxx = f ′′1 (x+ c t) + f ′′2 (x− c t) ; utt − c2 uxx = . . . = 0

b) u (x, t) = 12

(u0 (x+ c t) + u0 (x− c t)) + 12c∫ x+ct

x−ctv0 (ζ) dζ

c) u (x, t) = 12

(sin (k (x+ c t)) + sin (k (x− c t))) = sin (k x) · sin (k c t) =

sin (k x) · sin (ωt) mit ω = c k

16.5 ∂x1R

= −x−aR3 ; ∂2

x1R

= − 1R3 + 3 (x−a)2

R5 ; . . .

16.7 kn = κ(

n πL

)2; u (x, t) =

∑∞n=1 cn sin

(n πLx)e−( n π

L )2 t

16.8 u (x, t) = T (t) ·X (x) ⇒ T ′

T= X′′

X= k2

⇒ u (x, t) = Aek2 t (B sin (k x) + C cos (k x))

16.9 a) u (x, y) = X (x) · Y (y) ⇒ X′′

X= Y ′′

Y= −k2

⇒ u (x, y) = [A cos (k x) +B sin (k x)] [C cos (k y) +D sin (k y)]

b) X (0) = 0 ⇒ A = 0; Y (0) = 0 ⇒ C = 0; Y (L) = 0 ⇒ kn = n πL

⇒ u (x, y) =∑∞

n=1 cn sin(n π

Lx)

sin(n π

Ly)

16.10 a) k = ± i√n2 +m2 2π

L

b) u1 = sin(n 2π

Lx)

sin(m 2π

Ly)

sin(√n2 +m2 2π

Lt)

u2 = sin(n 2π

Lx)

sin(m 2π

Ly)cos(√n2 +m2 2π

Lt)

16.11 k =(n2 +m2

)4π2

L2

16.12 a) uxx = −k2 sin (k x)(ek y + e−k y

)uyy = k2 sin (k x)

(ek y + e−k y

)b) k = n π

L

16.13 u (x, t) = 1− xπ

+ 2π

∑∞n=1

(−1)n+1

ne−n2 t sin (nx)

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