history "TutoriumMathe3L7":Projekt Einsparzähler

Revision history for TutoriumMathe3L7

Revision [60567]

Last edited on 2015-10-27 09:37:52 by Jorina Lossau

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~~{{files}}~~

Revision [60566]

Edited on 2015-10-27 09:24:55 by Jorina Lossau

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~~{{image url="KonstanteL47.jpg" width="300" class="left"}}~~

Revision [60565]

Edited on 2015-10-27 09:23:12 by Jorina Lossau

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C1 = -19/18
C2 = 701/3430
spezielle Lösung:
{{image url="KonstanteL47.jpg" width="300" class="left"}}

Revision [60564]

Edited on 2015-10-27 08:35:16 by Jorina Lossau

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Revision [60563]

Edited on 2015-10-27 08:32:34 by Jorina Lossau

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Anfangsbedingungen:
{{image url="KonstanteL46.jpg" width="300" class="left"}}

Revision [60397]

Edited on 2015-10-21 08:01:59 by Jorina Lossau

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~~{{image url="KonstanteL45.jpg" width="200" class="left"}}~~

Revision [60396]

Edited on 2015-10-21 08:00:52 by Jorina Lossau

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x = xh + xp
allgemeine Lösung:
{{image url="KonstanteL45.jpg" width="200" class="left"}}

Revision [60395]

Edited on 2015-10-21 07:34:11 by Jorina Lossau

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~~{{image url="KonstanteL44.jpg" width="550" class="left"}}~~

Revision [60394]

Edited on 2015-10-21 07:32:56 by Jorina Lossau

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t²: 3 = 21a --> a = 1/7
t: 0 = 20a + 21b --> b = -20/147
t^0: 0 = 2a + 10b +21c --> c = -158/3087
e^-2t: 4 = 5d --> d = 4/5
{{image url="KonstanteL44.jpg" width="550" class="left"}}

Revision [60372]

Edited on 2015-10-17 21:19:18 by Jorina Lossau

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Revision [60371]

Edited on 2015-10-17 21:17:18 by Jorina Lossau

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Revision [60370]

Edited on 2015-10-17 20:58:15 by Jorina Lossau

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~~{{image url="KonstanteL42.jpg" width="150" class="left"}}~~

Revision [60369]

Edited on 2015-10-17 20:53:15 by Jorina Lossau

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Revision [60368]

Edited on 2015-10-17 20:44:03 by Jorina Lossau

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Revision [60367]

Edited on 2015-10-17 20:41:19 by Jorina Lossau

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Revision [60362]

Edited on 2015-10-16 21:45:51 by Jorina Lossau

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~~{{image url="KonstanteL40.jpg" width="250" class="left"}}~~

Revision [60361]

Edited on 2015-10-16 21:44:31 by Jorina Lossau

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~~{{image url="DGL4.jpg" width="350" class="left"}}~~

Revision [60360]

Edited on 2015-10-16 21:33:46 by Jorina Lossau

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**Aufgabe 1.4.3**
b) Anfangswerte: y(0) = y'(0) = 0
-----

Revision [60359]

Edited on 2015-10-16 21:23:26 by Jorina Lossau

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Revision [60358]

Edited on 2015-10-16 21:18:11 by Jorina Lossau

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Revision [60357]

Edited on 2015-10-16 21:15:56 by Jorina Lossau

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{{image url="DGL3.jpg" width="150" class="left"}}
a) Anfangswerte: y(0) = 2, y'(0) = -3

Revision [60356]

Edited on 2015-10-16 20:47:42 by Jorina Lossau

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Revision [60355]

Edited on 2015-10-16 20:46:27 by Jorina Lossau

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Revision [60354]

Edited on 2015-10-16 20:45:13 by Jorina Lossau

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Lösung des LGS:
{{image url="KonstanteL29.jpg" width="250" class="left"}}

Revision [60342]

Edited on 2015-10-16 15:51:22 by Jorina Lossau

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Revision [60341]

Edited on 2015-10-16 15:49:25 by Jorina Lossau

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Revision [60340]

Edited on 2015-10-16 15:43:59 by Jorina Lossau

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Revision [60339]

Edited on 2015-10-16 15:42:03 by Jorina Lossau

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Koeffizientenvergleich:
sin(3x) = 0 = -13b --> b = 0
cos(3x) = 5 = -13a --> a = -5/13
{{image url="KonstanteL27.jpg" width="250" class="left"}}

Revision [60338]

Edited on 2015-10-16 15:15:48 by Jorina Lossau

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~~{{image url="KonstanteL26.jpg" width="250" class="left"}}~~

Revision [60337]

Edited on 2015-10-16 15:12:46 by Jorina Lossau

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Revision [60336]

Edited on 2015-10-16 15:11:40 by Jorina Lossau

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yp, y°°p in DGL:
{{image url="KonstanteL26.jpg" width="250" class="left"}}

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~~yp, y''p~~

Revision [60335]

Edited on 2015-10-16 15:07:57 by Jorina Lossau

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{{image url="KonstanteL25.jpg" width="250" class="left"}}
yp, y''p

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~~{{image url="KonstanteL25.jpg" width="400" class="left"}}~~

Revision [60131]

Edited on 2015-10-12 10:59:39 by Jorina Lossau

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Revision [60127]

Edited on 2015-10-12 10:53:07 by Jorina Lossau

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Revision [60126]

Edited on 2015-10-12 10:51:34 by Jorina Lossau

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Revision [60083]

Edited on 2015-10-12 09:43:48 by Jorina Lossau

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~~{{image url="KonstanteL23.jpg" width="200" class="left"}}~~

Revision [60081]

Edited on 2015-10-12 09:43:19 by Jorina Lossau

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~~{{image url="DLG2.jpg" width="150" class="left"}}~~

Revision [60051]

Edited on 2015-10-12 09:18:54 by Jorina Lossau

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Edited on 2015-10-12 09:18:13 by Jorina Lossau

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Revision [60049]

Edited on 2015-10-12 09:15:26 by Jorina Lossau

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~~{{image url="KonstanteL21.jpg" width="200" class="left"}}~~

Revision [60047]

Edited on 2015-10-12 09:14:03 by Jorina Lossau

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Revision [60029]

Edited on 2015-10-09 14:00:34 by Jorina Lossau

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Revision [59979]

Edited on 2015-10-09 10:39:01 by Jorina Lossau

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Revision [59976]

Edited on 2015-10-09 10:19:49 by Jorina Lossau

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Revision [59975]

Edited on 2015-10-09 09:53:58 by Jorina Lossau

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Revision [59899]

Edited on 2015-10-05 11:00:15 by Jorina Lossau

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yp, y°p, y°°p in DGL:
yp, y°p, y°°p in DGL:

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~~yp, yp, yp°° in DGL:~~

Revision [59898]

Edited on 2015-10-05 10:54:19 by Jorina Lossau

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~~{{image url="KonstanteL18.jpg" width="170" class="left"}}~~

Revision [59897]

Edited on 2015-10-05 10:52:40 by Jorina Lossau

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partikuläre Lösung: Es tritt Resonanz auf; -2 ist eine doppelte Nullstelle im Lambda - Ansatz!
{{image url="KonstanteL18.jpg" width="170" class="left"}}

Revision [59895]

Edited on 2015-10-05 10:22:34 by Jorina Lossau

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~~{{image url="KonstanteL17.jpg" width="200" class="left"}}~~

Revision [59894]

Edited on 2015-10-05 10:21:35 by Jorina Lossau

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Revision [59893]

Edited on 2015-10-05 09:53:18 by Jorina Lossau

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Revision [59892]

Edited on 2015-10-05 09:51:51 by Jorina Lossau

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Revision [59887]

Edited on 2015-10-05 09:35:58 by Jorina Lossau

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Revision [59885]

Edited on 2015-10-05 09:16:58 by Jorina Lossau

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~~{{image url="KonstanteL15.jpg" width="300" class="left"}}~~

Revision [59883]

Edited on 2015-10-05 09:14:29 by Jorina Lossau

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yp, yp, yp°° in DGL:
{{image url="KonstanteL15.jpg" width="300" class="left"}}

Revision [59777]

Edited on 2015-10-02 10:56:39 by Jorina Lossau

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~~{{image url="KonstanteL14.jpg" width="170" class="left"}}~~

Revision [59776]

Edited on 2015-10-02 10:53:30 by Jorina Lossau

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partikuläre Lösung: Es tritt Resonanz auf; 2 ist eine einfache Nullstelle im Lambda - Ansatz!
{{image url="KonstanteL14.jpg" width="170" class="left"}}

Revision [59775]

Edited on 2015-10-02 10:39:18 by Jorina Lossau

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~~{{image url="KonstanteL13.jpg" width="100" class="left"}}~~

Revision [59774]

Edited on 2015-10-02 10:37:17 by Jorina Lossau

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Revision [59773]

Edited on 2015-10-02 10:17:03 by Jorina Lossau

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Unter der Bedingung 4a + c = 2b wird sich also stets (mindestens) ein λ = −2 ergeben, so dass die Störfunktion S (x) = 3e^(-2t) zu mathematischer Resonanz führt.

Deletions:

Unter der Bedingung 4a + c = 2b wird sich also stets (mindestens) ein λ = −2
ergeben, so dass die Störfunktion S (x) = 3e^(-2t) zu mathematischer Resonanz führt.

Revision [59638]

Edited on 2015-10-01 10:00:37 by Jorina Lossau

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Auflösen nach und Gleichsetzen liefert schließlich:
4a + c = 2b
Unter der Bedingung 4a + c = 2b wird sich also stets (mindestens) ein λ = −2
ergeben, so dass die Störfunktion S (x) = 3e^(-2t) zu mathematischer Resonanz führt.

Revision [59637]

Edited on 2015-10-01 09:55:09 by Jorina Lossau

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~~{{image url="KonstanteL11.jpg" width="250" class="left"}}~~

Revision [59636]

Edited on 2015-10-01 09:42:26 by Jorina Lossau

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~~{{image url="KonstanteL12.jpg" width="200" class="left"}}~~

Revision [59635]

Edited on 2015-10-01 09:38:40 by Jorina Lossau

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{{image url="KonstanteL11.jpg" width="250" class="left"}}
Vergleich mit charakteristischer Gleichung liefert:
{{image url="KonstanteL12.jpg" width="200" class="left"}}

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~~{{image url="KonstanteL11.jpg" width="200" class="left"}}~~

Revision [59632]

Edited on 2015-10-01 09:34:40 by Jorina Lossau

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~~{{image url="KonstanteL11.jpg" width="250" class="left"}}~~

Revision [59628]

Edited on 2015-10-01 09:32:38 by Jorina Lossau

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Revision [59618]

Edited on 2015-10-01 09:23:27 by Jorina Lossau

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Man spricht von „mathematischer Resonanz“, wenn dieStörfunktion S (x) selbst eine Lösung der zugehörigen homogenen DGL ist. Eine Teillösung der zugehörigen homogenen DGL muss also eine ebensolche Exponentialfunktion mit dem Koeffizienten α = − 2 im Exponent sein. Das ist der Fall, wenn die charakteristische Gleichung mindestens eine reelle Lösung λ1 = −2 aufweist.
Damit lässt sich die charakteristische Gleichung inder Produktform aufschreiben nd in die Summenform bringen und dann mit der zug.hom. DGL vergleichen:

Revision [59617]

Edited on 2015-10-01 09:11:07 by Jorina Lossau

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= spezielle Lösung

Revision [59616]

Edited on 2015-10-01 09:01:44 by Jorina Lossau

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Revision [59615]

Edited on 2015-10-01 08:59:15 by Jorina Lossau

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nach Lösung des LGS:

Revision [59614]

Edited on 2015-10-01 08:57:47 by Jorina Lossau

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~~{{image url="KonstanteL9.jpg" width="250" class="left"}}~~

Revision [59613]

Edited on 2015-10-01 08:56:12 by Jorina Lossau

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= allgemeine Lösung
{{image url="KonstanteL9.jpg" width="250" class="left"}}

Revision [59612]

Edited on 2015-10-01 08:36:40 by Jorina Lossau

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Revision [59611]

Edited on 2015-10-01 08:29:47 by Jorina Lossau

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~~{{image url="KonstanteL7.jpg" width="100" class="left"}}~~

Revision [59610]

Edited on 2015-10-01 08:27:28 by Jorina Lossau

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Revision [59609]

Edited on 2015-10-01 08:16:34 by Jorina Lossau

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~~{{image url="KonstanteL6.jpg" width="180" class="left"}}~~

Revision [59608]

Edited on 2015-10-01 08:15:22 by Jorina Lossau

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partikuläre Lösung:
{{image url="KonstanteL6.jpg" width="180" class="left"}}

Revision [59607]

Edited on 2015-10-01 08:13:02 by Jorina Lossau

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~~{{image url="KonstanteL5.jpg" width="200" class="left"}}~~

Revision [59594]

Edited on 2015-09-30 09:00:56 by Jorina Lossau

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~~{{image url="KonstanteL5.jpg" width="350" class="left"}}~~

Revision [59593]

Edited on 2015-09-30 08:59:09 by Jorina Lossau

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Revision [59592]

Edited on 2015-09-30 08:43:13 by Jorina Lossau

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{{image url="KonstanteL4.jpg" width="300" class="left"}}
homogene Lösung:

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~~{{image url="KonstanteL4.jpg" width="250" class="left"}}~~

Revision [59591]

Edited on 2015-09-30 08:41:33 by Jorina Lossau

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Revision [59590]

Edited on 2015-09-30 08:26:55 by Jorina Lossau

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**Lösung:**
{{image url="KonstanteL3.jpg" width="550" class="left"}}

Revision [59589]

Edited on 2015-09-30 08:25:16 by Jorina Lossau

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Auch hier müssen die Lösungen der charakteristischen Gleichung konjugiert komplexe λ – Werte sein, was unter der Wurzel einen negativen Ausdruck erfordert. Außerdem muss der Realteil negativ sein, was wegen a > 0 auch b > 0 erfordert.
{{image url="KonstanteL3.jpg" width="550" class="left"}}

Revision [59588]

Edited on 2015-09-30 08:12:59 by Jorina Lossau

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Die zugehörige homogene DGL beschreibt eine ungedämpfte Schwingung für den Fall, dass die Lösung ihrer charakteristischen Gleichung rein imaginäre konjugiert komplexe λ enthält. Dies ist der Fall wenn sowohl b = 0 ist (es liegt dann keine 1. Ableitung von y vor) als auch c dasselbe Vorzeichen wie a hat, also c > 0 ist
b = 0 und c > 0

Revision [59587]

Edited on 2015-09-30 07:40:41 by Jorina Lossau

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~~{{image url="DGL1.jpg" width="350" class="left"}}~~

Revision [59586]

Edited on 2015-09-30 07:20:10 by Jorina Lossau

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Revision [59585]

Edited on 2015-09-30 07:18:03 by Jorina Lossau

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{{files}}
**Lösung:**
{{image url="KonstanteL1.jpg" width="150" class="left"}}
Die charakteristische Gleichung mit ihren Lösungen lautet:

Revision [59479]

Edited on 2015-09-25 15:50:36 by Jorina Lossau

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||**1.4 Lineare inhomogene Differentialgleichungen mit konstanten Koeffizienten**
**Aufgabe 1.4.1**
{{image url="DGL1.jpg" width="150" class="left"}}
**a) Unter welcher Bedingung beschreibt die zugehörige homogene DGL eine ungedämpfte Schwingung wenn a > 0 bekannt ist ?**
**b) Unter welcher Bedingung beschreibt die zugehörige homogene DGL eine gedämpfte Schwingung wenn a > 0 bekannt ist?**
**c) Lösen Sie die DGL für den Fall**
a = 2, b = 16, c = 30 und
y(0) = 2, y°(0) = 5
**d) Unter welchen Bedingungen würde man bei der inhomogenen DGL ay°° + by° + cy = 3e^(-2t) von „mathematischer Resonanz“ sprechen?**
**e) Finden Sie eine allgemeine Lösung für den Fall**
a = 1, b = 5 und c = 6
**f) Finden Sie die spezielle Lösung für den Fall**
a = 1, b = 4, c = 4 und
y(0) = 2, y°(0) = 5
------
**Aufgabe 1.4.2**
{{image url="DLG2.jpg" width="150" class="left"}}
a) Anfangswerte:y(0) = 2, y'(0) = -3
b) Anfangswerte:y(0) = y'(0) = 0
--------
**Aufgabe 1.4.3**
{{image url="DGL3.jpg" width="150" class="left"}}
a) Anfangswerte: y(0) = 2, y'(0) = -3
b) Anfangswerte: y(0) = y'(0) = 0
-----
**Aufgabe 1.4.4**
{{image url="DGL4.jpg" width="350" class="left"}}
||

Deletions:

{{image url="Mathe3L71.jpg" width="650" class="center"}}
{{image url="Mathe3L72.jpg" width="650" class="center"}}
{{image url="Mathe3L73.jpg" width="650" class="center"}}
{{image url="Mathe3L74.jpg" width="650" class="center"}}
{{image url="Mathe3L75.jpg" width="650" class="center"}}
{{image url="Mathe3L76.jpg" width="650" class="center"}}
{{image url="Mathe3L77.jpg" width="650" class="center"}}
{{image url="Mathe3L78.jpg" width="650" class="center"}}
{{image url="Mathe3L79.jpg" width="650" class="center"}}

Revision [43774]

Edited on 2014-08-27 17:26:35 by Jorina Lossau

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~~{{files}}~~

Revision [43773]

Edited on 2014-08-27 17:25:28 by Jorina Lossau

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====Tutorium Mathematik 3====
===Lineare Inhomogene Differentialgleichungen mit konstanten Koeffizienten - Lösungen===
{{image url="Mathe3L71.jpg" width="650" class="center"}}
{{image url="Mathe3L72.jpg" width="650" class="center"}}
{{image url="Mathe3L73.jpg" width="650" class="center"}}
{{image url="Mathe3L74.jpg" width="650" class="center"}}
{{image url="Mathe3L75.jpg" width="650" class="center"}}
{{image url="Mathe3L76.jpg" width="650" class="center"}}
{{image url="Mathe3L77.jpg" width="650" class="center"}}
||**{{files download="Mathe3L7.pdf"text="PDF Dokument Inhomogene Differentialgleichungen"}}**||

Deletions:

Tutorium Mathematik 3
Lineare Inhomogene Differentialgleichungen mit konstanten Koeffizienten - Lösungen
{{image url="Mathe3L7.jpg" width="600" class="center"}}
||**{{files download="Mathe3L7.pdf"text="PDF Dokument Aufgaben Integralrechnung"}}**||

Revision [43771]

The oldest known version of this page was created on 2014-08-27 17:02:01 by Jorina Lossau