ich war hier: TutoriumMathe3Variationen

Revision history for TutoriumMathe3Variationen


Revision [71970]

Last edited on 2016-09-26 19:01:52 by Jorina Lossau
Additions:
-C(x)/x^2 + C'(x)/x + 1/x* C(x)/x = 5e^(-2x)
C'(x)=5xe^(-2x) integrieren !
C(x)=-5/2xe^(-2x)-5/4e^(-2x)+k
y=-5/2e^(-2x)-5/(4x)e^(-2x)+k/x=allgemeine Lösung
y(-0,5)=-5/2e^1-5/-2e^1+k/-0,5=-2k=-8
-> k=4
y=-5/2e^(-2x)-5/(4x)e^(-2x)+4/x


Revision [71969]

Edited on 2016-09-26 18:52:21 by Jorina Lossau
Additions:
y=C(x)/x -> y' = -C(x)/x^2 + C'(x)/x


Revision [71968]

Edited on 2016-09-26 18:50:34 by Jorina Lossau
Additions:
y'+(1/x)y=5e^(-2x)
y'+(1/x)y=0->(dy)/(dx)+(1/x)y=0->(dy)/y=-(dx)/x
ln Betrag von y = -ln Betrag von x + Cgestrichen -> ln Betrag von xy = Cgestrichen -> xy = +-e^C gestrichen = C


Revision [58985]

Edited on 2015-09-07 10:42:53 by Jorina Lossau
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Revision [58984]

Edited on 2015-09-07 10:41:03 by Jorina Lossau
Additions:
spezielle Lösung:
{{image url="Konstante8.jpg" width="400" class="left"}}


Revision [58983]

Edited on 2015-09-07 10:33:42 by Jorina Lossau
Additions:
{{image url="Konstante7.jpg" width="400" class="left"}}
Deletions:
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Revision [58980]

Edited on 2015-09-07 10:26:50 by Jorina Lossau
Additions:
Anfangsbedingung:
{{image url="Konstante7.jpg" width="200" class="left"}}


Revision [58978]

Edited on 2015-09-07 10:19:25 by Jorina Lossau
Additions:
{{image url="Konstante6.jpg" width="200" class="left"}} = allgemeine Lösung
Deletions:
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Revision [58976]

Edited on 2015-09-07 10:18:10 by Jorina Lossau
Additions:
in Ansatz einsetzen:
{{image url="Konstante6.jpg" width="250" class="left"}}


Revision [58975]

Edited on 2015-09-07 10:15:54 by Jorina Lossau

No Differences

Revision [58974]

Edited on 2015-09-07 10:14:21 by Jorina Lossau
Additions:
einfache DGL für C(x)
{{image url="Konstante5.jpg" width="250" class="left"}}


Revision [58972]

Edited on 2015-09-07 10:00:45 by Jorina Lossau
Additions:
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Deletions:
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Revision [58971]

Edited on 2015-09-07 09:59:04 by Jorina Lossau
Additions:
einsetzen:
{{image url="Konstante4.jpg" width="350" class="left"}}


Revision [58970]

Edited on 2015-09-07 09:54:59 by Jorina Lossau
Additions:
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Deletions:
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Revision [58969]

Edited on 2015-09-07 09:53:29 by Jorina Lossau
Additions:
homogene Lösung: yH = C/X
Ansatz:
{{image url="Konstante3.jpg" width="450" class="left"}}


Revision [58961]

Edited on 2015-09-07 09:11:23 by Jorina Lossau
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Deletions:
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Revision [58960]

Edited on 2015-09-07 09:09:42 by Jorina Lossau
Additions:
zug. hom. DGL:
{{image url="Konstante2.jpg" width="130" class="left"}}


Revision [58959]

Edited on 2015-09-07 08:59:38 by Jorina Lossau
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Deletions:
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Revision [58958]

Edited on 2015-09-07 08:58:17 by Jorina Lossau
Additions:
**Beispiel: **
Anfangsbedingung: y(-5) = -8
{{image url="Konstante1.jpg" width="300" class="center"}}
Deletions:
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Revision [58957]

Edited on 2015-09-07 08:54:22 by Jorina Lossau
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Revision [58343]

Edited on 2015-08-27 10:44:31 by Jorina Lossau
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Revision [58340]

Edited on 2015-08-27 10:23:46 by Jorina Lossau
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Revision [58339]

Edited on 2015-08-27 09:46:56 by Jorina Lossau
Additions:
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|=|{color:#00386a; background-color: #E0E0E0; width: 700px} Vorgehensweise||
|=|{background-color: #FFFFFF; width: 700px}
Diese Methode bietet sich für lineare inhomogene DGL mit konstanten Koeffizienten an, wenn die Störfunktion nicht auf einen einfachen Ansatz für eine partikuläre Lösung hinweist. Darüber hinaus ist die Methode auch oft für lineare homogene DGL mit konstanten Koeffizienten geeignet.
Im einfachsten Fall einer inhomogenen DGL erster Ordnung wird wie folgt vorgegangen:
Die freie Konstante der allgemeinen Lösung C der zugehörigen homogenen DGL wird als Funktion C (x) von x betrachtet. Durch Einsetzen in die inhomogene DGL erhält man eine einfachere DGL für C (x), welches dann bestimmt werden kann.
||


Revision [43698]

Edited on 2014-08-27 10:06:15 by Jorina Lossau
Additions:
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Deletions:
{{files}}


Revision [43697]

The oldest known version of this page was created on 2014-08-27 09:42:50 by Jorina Lossau
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