history "TutoriumMathe3L5":Projekt Einsparzähler

Revision history for TutoriumMathe3L5

Revision [73269]

Last edited on 2016-10-19 12:25:26 by Jorina Lossau

Additions:

**Aufgabe 1.2.5**
y'=-3y+1
dy/dx=-3y+1
dy/(-3y+1)=dx I ∫(...)
∫dy/(-3y+1)=∫dx
-1/3ln(-3y+1)=x+K
lnI-3y+1I=-3x-3K I e^(...)
I-3y+1I=e^(-3x-3K)=e^(-3x)e^(-3K)
mit e^(-3K)=K*; K*єR; K*>0
I-3y+1I=K*e^(-3x)
-3y+1=+-K*e^(-3x)
-3y=-1+-K*e^(-3x)
y=1/3-+1/3K*e^(-3x)
mit K^=-+1/3K*; K^єR\{0}
y=1/3+K^e^(-3x) -> allgemeine Lösung
**Aufgabe 1.2.6**
y'=-3x+1
dy/dx=-3x+1
dy=(-3x+1)dx I ∫(...)
∫dy=∫(-3x+1)dx
y=-3/2x^2+x+K -> allgemeine Lösung

Revision [73241]

Edited on 2016-10-18 13:36:30 by Jorina Lossau

Additions:

**Aufgabe 1.2.3**
du/dt=3ucos(3t+4)
Lösung:
du/u=3cos(3t+4)dt I ∫(...)
∫du/u=∫3cos(3t+4)dt
lnIuI=sin(3t+4)+K Ie^(...)
IuI=e^(sin(3t+4)+K)=e^(sin(3t+4)e^K
u=+-e^K*e^(sin(3t+4))
e^K>0;K~=+-e^K; K~єR\{0}
u=K~e^(sin(3t+4)) -> allgemeine Lösung
**Aufgabe 1.2.4**
di(t)/dt=(3t+1)i
Lösung:
di(t)/i=(3t+1)dt I ∫(...)
∫di(t)/i=∫(3t+1)dt
ln(i)=3/2t^2+t+K I e^(...)
IiI=e^(3/2t^2+t+K)=e^(3/2t^2+t)e^K; K~=+-e^K; K~єR\{0}
i=+-e^Ke^(3/2t^2+t)
K~=+-e^K; K~єR\{0}
i=i(t)=K~e^(3/2t^2+t) -> allgemeine Lösung

Revision [73219]

Edited on 2016-10-18 12:27:24 by Jorina Lossau

Additions:

||**Aufgabe 1.2.1**
**Aufgabe 1.2.2**
x.=(e^(-3t))/(x+2); x(0)=2
dx/dt=(e^(-3t))/(x+2)
(x+2)dx=e^(-3t)dt I ∫(...)
∫(x+2)dx=∫e^(-3t)dt
1/2x^2+2x=-1/3e^(-3t)+K
0=x^2+4x+2/3e^(-3t)-2K
x1,2=-2+-√(4-2/3e^(-3t)+2K)
mit K^=2K+4
x1,2=-2+-√(K^-2/3e^(-2t))
Da x(0)=2>-2, muss für spezielle Lösung "+" voranstehen:
x(0)=-2+√(K^-2/3e^(-3t))=2
√(K^-2/3e^(-3t))=4
K^-2/3e^(-3t)=16
K^=16+2/3=50/3
x=x(t)=-2+√(50/3-2/3e^(-3t)) (spezielle Lösung)
||

Deletions:

~~**Aufgabe 1.2.1**~~

Revision [73033]

Edited on 2016-10-12 12:00:04 by Jorina Lossau

Additions:

**Aufgabe 1.2.1**
y'=(2+sin(x)/(3+4y)
y'=(2+sin(x))/(3+4y)
dy/dx=(2+sin(x)/(3+4y)
dy/dx=(2+sin(x)/(3+4y)
(3+4y)dy=(2+sin(x))dx I ∫(...)
∫(3+4y)dy=∫(2+sin(x))dx
3y+2y^2=2x-cos(x)+K
0=2y^2+3y-2x+cos(x)-K
0=y^2+3/2y-x+1/2cos(x)-1/2K
y1,2=-3/4+-√(9/16+x-1/2cos(x)+1/2K)
mit K~=9/16+1/2K
y1,2=-3/4+-√(K~+x-1/2cos(x))
Das Vorzeichen + oder - ergibt sich erst beim Einsetzen von Anfangsbedingungen eindeutig.

Revision [43800]

Edited on 2014-08-27 20:14:46 by Jorina Lossau

Additions:

{{image url="Mathe351.jpg" width="650" class="center"}}
{{image url="Mathe352.jpg" width="650" class="center"}}
{{image url="Mathe353.jpg" width="650" class="center"}}
{{image url="Mathe354.jpg" width="650" class="center"}}

Deletions:

{{image url="Mathe351.jpg" width="600" class="center"}}
{{image url="Mathe352.jpg" width="600" class="center"}}
{{image url="Mathe353.jpg" width="600" class="center"}}
{{image url="Mathe354.jpg" width="600" class="center"}}

Revision [43799]

Edited on 2014-08-27 20:14:16 by Jorina Lossau

Additions:

||**{{files download="Mathe3L5.pdf"text="PDF Dokument Lösungen Trennen der Variablen"}}**||

Deletions:

{{files}}
||**{{files download="Mathe35.pdf"text="PDF Dokument Lösungen Trennen der Variablen"}}**||

Revision [43798]

Edited on 2014-08-27 20:11:36 by Jorina Lossau

Additions:

{{files}}
{{image url="Mathe351.jpg" width="600" class="center"}}
{{image url="Mathe352.jpg" width="600" class="center"}}
{{image url="Mathe353.jpg" width="600" class="center"}}
{{image url="Mathe354.jpg" width="600" class="center"}}
||**{{files download="Mathe35.pdf"text="PDF Dokument Lösungen Trennen der Variablen"}}**||

Deletions:

{{image url="Mathe3Aufgaben13.jpg" width="600" class="center"}}
||**{{files download="Mathe3A1.pdf"text="PDF Dokument Aufgaben Integralrechnung"}}**||

Revision [43796]

The oldest known version of this page was created on 2014-08-27 20:02:57 by Jorina Lossau