u'=u+v, v'=-u+v+sinx < gewöhnliche < Differentialgl. < Analysis < Hochschule < Mathe < Vorhilfe
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| Aufgabe | | Consider the System of ODEs for $u(x)$ and $v(x)$ given by $$u' = u + [mm] v,\quad [/mm] v' = -u + v + [mm] \sin(x)$$ [/mm] and reduce this to a single ODE of second order for $u(x)$. Hence find the general solution to this system of ODEs. Find the associated solution for $v(x)$. |
hi
meine Frage bezieht sich zunächst nur auf den Ansatz dieser Aufgabe, ob meiner überhaupt richtig ist. Hier ist er:
$(I)~ u' &= u + v [mm] \Leftrightarrow [/mm] v = u' - u [mm] \Rightarrow [/mm] v' = u'' - u'$ [mm] \\
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$(II)~ v' &= -u + v + [mm] \sin(x)$ \\
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[mm] $(I)~\text{in} ~(II):\quad [/mm] u'' - u' &= -u + u' - u + [mm] \sin(x)$\\
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[mm] $\Rightarrow [/mm] u(x) &: u'' - 2u' + 2u = [mm] \sin(x)$
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So, dass ist eine ODE zweiter Ordnung für $u(x)$, und diese soll ich jetzt lösen - oder bin ich auf dem falschen Weg...?
Entsprechendes Vorgehen dann für $v(x)$.
Danke & Gruß, GB
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| Status: |
(Antwort) fertig  | | Datum: | 19:06 Di 20.10.2009 | | Autor: | leduart |
Hallo
richtig, aber fuer v' nicht entsprechend, sondern Loesung fuer u benutzen. da du ja hast: v=u'-u
Gruss leduart
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ok, vielen dank schonmal.
wenn hier noch jemand drübergucken könnte:
$u'' - 2u' + 2u = [mm] \sin(x)$\\
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Char. Gleichung: [mm] $m^2 [/mm] - 2m + 2 = 0$
[mm] $m_{1,2} [/mm] = [mm] \frac{2 \pm \sqrt{4 - 8}}{2} [/mm] = [mm] \frac{2 \pm \sqrt{4i^2}}{2} [/mm] = 1 [mm] \pm i$\\
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[mm] $\Rightarrow u_h [/mm] = [mm] e^x(A\cos(x) [/mm] + [mm] B\sin(x))$
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Partikuläre Lösung:
[mm] $u_p [/mm] = [mm] C\sin(x) [/mm] + [mm] D\cos(x); u_p' [/mm] &= [mm] C\cos(x) [/mm] - [mm] D\sin(x);u_p'' [/mm] &= [mm] -C\sin(x) [/mm] - [mm] D\cos(x)$
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[mm] $\Rightarrow -C\sin(x) [/mm] - [mm] D\cos(x) [/mm] - [mm] 2C\cos(x) [/mm] + [mm] 2D\sin(x) +2C\sin(x) [/mm] + [mm] 2D\cos(x) [/mm] = [mm] \sin(x)$\\
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[mm] $\Leftrightarrow \sin(x)(-C [/mm] + 2D + 2C) + [mm] \cos(x)(-D [/mm] - 2C + 2D) = [mm] \sin(x)$\\
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[mm] $\Leftrightarrow \sin(x)(C [/mm] + 2D) + [mm] \cos(x)(D [/mm] - 2C) = [mm] \sin(x)$\\
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Koeffizientenvergleich:
$(I):~C + 2D &= 1 [mm] \Rightarrow [/mm] C = 1 - [mm] 2D$\\
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$(II):~D - 2C &= 0 [mm] \Leftrightarrow [/mm] D - 2 + 4D = 0 [mm] \Leftrightarrow [/mm] 5D = 2$ [mm] $\Leftrightarrow [/mm] D = [mm] \frac{2}{5}$\\
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[mm] $\Rightarrow [/mm] C = [mm] \frac{1}{5}$
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[mm] $$u_p [/mm] = [mm] \frac{1}{5}\sin(x) [/mm] + [mm] \frac{2}{5}\cos(x) [/mm] = [mm] \frac{\sin(x) + 2\cos(x)}{5}$$
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[mm] \bigskip
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[mm] $$\Rightarrow [/mm] u = [mm] u_h [/mm] + [mm] u_p [/mm] = [mm] e^x(A\cos(x) [/mm] + [mm] B\sin(x)) [/mm] + [mm] \frac{\sin(x) + 2\cos(x)}{5}$$
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Lösung für $v(x)$
$u' = [mm] e^x \cdot (A\cos(x) [/mm] + [mm] B\sin(x)) [/mm] + [mm] e^x \cdot (-A\sin(x) [/mm] + [mm] B\cos(x)) [/mm] + [mm] \frac{\cos(x) - 2\sin(x)}{5}$\\
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$v = u' - u = [mm] $\\
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$= [mm] e^x \cdot (A\cos(x) [/mm] + [mm] B\sin(x)) [/mm] + [mm] e^x \cdot (-A\sin(x) [/mm] + [mm] B\cos(x)) [/mm] + [mm] \frac{\cos(x) - 2\sin(x)}{5} [/mm] - [mm] e^x(A\cos(x) [/mm] + [mm] B\sin(x)) [/mm] - [mm] \frac{\sin(x) + 2\cos(x)}{5} [/mm] =$ [mm] \\
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$= [mm] e^x \cdot (-A\sin(x) [/mm] + [mm] B\cos(x)) [/mm] + [mm] \frac{\cos(x) - 2\sin(x) - \sin(x) - 2\cos(x)}{5} =$\\
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$= [mm] e^x \cdot (-A\sin(x) [/mm] + [mm] B\cos(x)) [/mm] - [mm] \frac{\cos(x) + 3\sin(x)}{5} [/mm] $
Lieben Dank und Gruß, GB
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| Status: |
(Antwort) fertig | | Datum: | 21:50 Di 20.10.2009 | | Autor: | leduart |
Hallo
ich seh keinen Fehler.
Gruss leduart
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super - lieben dank nochmal 
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