DGL 2Ordnung AWP < gewöhnliche < Differentialgl. < Analysis < Hochschule < Mathe < Vorhilfe
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| Aufgabe | ÜB12 A2 NR.2
Man löse die Anfangswertprobleme:
y'' - 10*y' + [mm] x^{2} [/mm] = 0
y'' - 10*y' = - [mm] x^{2}
[/mm]
[mm] x_{0}=0
[/mm]
[mm] y_{0}=0
[/mm]
[mm] y_{0}'=0
[/mm]
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y'' - 10*y' = - [mm] x^{2}
[/mm]
1. Schritt: Charakteristisches homogenes Polynom aufstellen:
[mm] \lambda^{2} [/mm] - 10 * [mm] \lambda [/mm] + 0 = 0
PQ-Formel:
[mm] \lambda_{1,2} [/mm] = 5 [mm] \pm \wurzel{5^{2}}
[/mm]
[mm] \lambda_{1} [/mm] = 10
[mm] \lambda_{2} [/mm] = 0
Daraus ergibt sich:
[mm] \lambda_{1} \not= \lambda_{2}
[/mm]
[mm] \Rightarrow y_{1}=e^{\lambda_{1}*x} [/mm] = [mm] y_{1}=e^{10*x}
[/mm]
[mm] \Rightarrow y_{2}=e^{\lambda_{2}*x} [/mm] = [mm] y_{2}=e^{0*x}=e^{0*x}=1
[/mm]
Daraus ergibt sich die homogene Lösung:
[mm] y_{h}(x)=c_{1}*e^{10*x}+c_{2}
[/mm]
Soweit sogut!
Nun können wir noch folgendes aus der gegebenen Ausgangsgleichung ableiten:
Störfkt.: g(x) = [mm] -x^{2}
[/mm]
Vorfaktoren: a = -10; b = 0
y(x) = [mm] y_{h}(x) [/mm] + [mm] y_{p}(x)
[/mm]
Ermitteln von [mm] y_{p}:
[/mm]
Aus dem oben gegebenen [mm] \Rightarrow [/mm] Polynom 2. Grades ist die Störefkt.!
g(x) = [mm] P_{n}(x)
[/mm]
da b = 0 & a [mm] \not= [/mm] 0
[mm] \Rightarrow y_{p} [/mm] = x * [mm] Q_{n}
[/mm]
Ermitteln des Polynomes & der Koeffizienten:
[mm] Q_{n} [/mm] = [mm] a_{2}*x^{2} [/mm] + [mm] a_{1}*x [/mm] + [mm] a_{0}
[/mm]
[mm] y_{p} [/mm] = x * [mm] Q_{n}
[/mm]
[mm] y_{p} [/mm] = x * [mm] (a_{2}*x^{2} [/mm] + [mm] a_{1}*x [/mm] + [mm] a_{0})
[/mm]
[mm] y_{p} [/mm] = [mm] a_{2}*x^{3} [/mm] + [mm] a_{1}*x^{2} [/mm] + [mm] a_{0}*x
[/mm]
[mm] y_{p}' [/mm] = [mm] 3*a_{2}*x^{2} [/mm] + [mm] 2*a_{1}*x [/mm] + [mm] a_{0}
[/mm]
[mm] y_{p}'' [/mm] = [mm] 6*a_{2}*x [/mm] + [mm] 2*a_{1}
[/mm]
Einsetzen für den Koeffizientenvergleich!
(x * [mm] Q_{n}'' [/mm] - 10(x * [mm] Q_{n}' [/mm] + 0*(x * [mm] Q_{n} [/mm] = [mm] -x^{2})
[/mm]
[mm] 6*a_{2}*x [/mm] + [mm] 2*a_{1} [/mm] - [mm] 10*(3*a_{2}*x^{2} [/mm] + [mm] 2*a_{1}*x+a_{0})+ [/mm] 0 * [mm] (a_{2}*x^{3} [/mm] + [mm] a_{1}*x^{2} [/mm] + [mm] a_{0}*x) [/mm] = [mm] -x^{2}
[/mm]
[mm] 6*a_{2}*x [/mm] + [mm] 2*a_{1} [/mm] - [mm] (30*a_{2}*x^{2} [/mm] + [mm] 20*a_{1}*x [/mm] + [mm] 10*a_{0})= -x^{2}
[/mm]
[mm] 6*a_{2}*x [/mm] + [mm] 2*a_{1} [/mm] - [mm] 30*a_{2}*x^{2} [/mm] - [mm] 20*a_{1}*x [/mm] - [mm] 10*a_{0}= -x^{2}
[/mm]
Sortieren nach Potenzen:
- [mm] 30*a_{2}*x^{2} [/mm] + x * [mm] (6*a_{2} [/mm] - [mm] 20*a_{1}) [/mm] - [mm] 10*a_{0} [/mm] + [mm] 2*a_{1} [/mm] = [mm] -x^{2}
[/mm]
Betrachten wir nun die einzelnen Koeffizienten:
- [mm] 30*a_{2}*x^{2} [/mm] = [mm] -x^{2}
[/mm]
[mm] a_{2} [/mm] = [mm] \bruch{1}{30}
[/mm]
x * [mm] (6*a_{2} [/mm] - [mm] 20*a_{1}) [/mm] = 0
x * [mm] (6*\bruch{1}{30} [/mm] - [mm] 20*a_{1}) [/mm] = 0
x * [mm] (6*\bruch{1}{30} [/mm] - [mm] 20*a_{1}) [/mm] = 0
x * [mm] (\bruch{1}{5} [/mm] - [mm] 20*a_{1}) [/mm] = 0
[mm] \bruch{1}{5} [/mm] - [mm] 20*a_{1} [/mm] = 0
- [mm] 20*a_{1}) [/mm] = - [mm] \bruch{1}{5}
[/mm]
[mm] a_{1}) [/mm] = [mm] \bruch{1}{100}
[/mm]
- [mm] 10*a_{0} [/mm] + [mm] 2*a_{1} [/mm] = 0
- [mm] 10*a_{0} [/mm] + [mm] 2*\bruch{1}{100} [/mm] = 0
- [mm] 10*a_{0} [/mm] = [mm] -\bruch{2}{100}
[/mm]
[mm] a_{0} [/mm] = [mm] \bruch{2}{1000} [/mm] = [mm] \bruch{1}{500}
[/mm]
[mm] y_{p}(x) [/mm] = [mm] \bruch{1}{30}*x^{3} [/mm] + [mm] \bruch{1}{100}*x^{2} [/mm] + [mm] \bruch{1}{500}*x
[/mm]
y(x) = [mm] y_{h}(x) [/mm] + [mm] y_{p}(x)
[/mm]
[mm] \Rightarrow [/mm] y(x) = [mm] c_{1}*e^{10*x} [/mm] + [mm] c_{2} [/mm] + [mm] \bruch{1}{30}*x^{3} [/mm] + [mm] \bruch{1}{100}*x^{2} [/mm] + [mm] \bruch{1}{500}*x
[/mm]
So damit ist die DGL 2. Ordnung richtig gelöst!
Jedoch habe ich gerade einen Blackout, wie ich jetzt noch mal die Anfangswerte einsetzen muss um einen Endwert zu erhalten!
Mußte ich das Endergebnis zwei mal Ableiten ?
Für y' & y'' ?
Bitte um Hilfe diesbezüglich!
*Danke*
------------------------------
Ich habe diese Aufgabe niergenswo anders gepostet / hochgeladen /oder sonst auf irgend eine Art und Weise publiziert!
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Hallo Kerberos2008,
> ÜB12 A2 NR.2
> Man löse die Anfangswertprobleme:
> y'' - 10*y' + [mm]x^{2}[/mm] = 0
>
> y'' - 10*y' = - [mm]x^{2}[/mm]
>
> [mm]x_{0}=0[/mm]
> [mm]y_{0}=0[/mm]
> [mm]y_{0}'=0[/mm]
>
>
> y'' - 10*y' = - [mm]x^{2}[/mm]
>
> 1. Schritt: Charakteristisches homogenes Polynom
> aufstellen:
>
> [mm]\lambda^{2}[/mm] - 10 * [mm]\lambda[/mm] + 0 = 0
>
> PQ-Formel:
>
> [mm]\lambda_{1,2}[/mm] = 5 [mm]\pm \wurzel{5^{2}}[/mm]
> [mm]\lambda_{1}[/mm] = 10
> [mm]\lambda_{2}[/mm] = 0
>
> Daraus ergibt sich:
> [mm]\lambda_{1} \not= \lambda_{2}[/mm]
>
> [mm]\Rightarrow y_{1}=e^{\lambda_{1}*x}[/mm] = [mm]y_{1}=e^{10*x}[/mm]
> [mm]\Rightarrow y_{2}=e^{\lambda_{2}*x}[/mm] =
> [mm]y_{2}=e^{0*x}=e^{0*x}=1[/mm]
>
> Daraus ergibt sich die homogene Lösung:
>
> [mm]y_{h}(x)=c_{1}*e^{10*x}+c_{2}[/mm]
>
> Soweit sogut!
>
> Nun können wir noch folgendes aus der gegebenen
> Ausgangsgleichung ableiten:
>
> Störfkt.: g(x) = [mm]-x^{2}[/mm]
> Vorfaktoren: a = -10; b = 0
>
> y(x) = [mm]y_{h}(x)[/mm] + [mm]y_{p}(x)[/mm]
>
> Ermitteln von [mm]y_{p}:[/mm]
>
> Aus dem oben gegebenen [mm]\Rightarrow[/mm] Polynom 2. Grades ist
> die Störefkt.!
>
> g(x) = [mm]P_{n}(x)[/mm]
>
> da b = 0 & a [mm]\not=[/mm] 0
> [mm]\Rightarrow y_{p}[/mm] = x * [mm]Q_{n}[/mm]
>
> Ermitteln des Polynomes & der Koeffizienten:
> [mm]Q_{n}[/mm] = [mm]a_{2}*x^{2}[/mm] + [mm]a_{1}*x[/mm] + [mm]a_{0}[/mm]
>
> [mm]y_{p}[/mm] = x * [mm]Q_{n}[/mm]
> [mm]y_{p}[/mm] = x * [mm](a_{2}*x^{2}[/mm] + [mm]a_{1}*x[/mm] + [mm]a_{0})[/mm]
> [mm]y_{p}[/mm] = [mm]a_{2}*x^{3}[/mm] + [mm]a_{1}*x^{2}[/mm] + [mm]a_{0}*x[/mm]
> [mm]y_{p}'[/mm] = [mm]3*a_{2}*x^{2}[/mm] + [mm]2*a_{1}*x[/mm] + [mm]a_{0}[/mm]
> [mm]y_{p}''[/mm] = [mm]6*a_{2}*x[/mm] + [mm]2*a_{1}[/mm]
>
> Einsetzen für den Koeffizientenvergleich!
> (x * [mm]Q_{n}''[/mm] - 10(x * [mm]Q_{n}'[/mm] + 0*(x * [mm]Q_{n}[/mm] = [mm]-x^{2})[/mm]
>
> [mm]6*a_{2}*x[/mm] + [mm]2*a_{1}[/mm] - [mm]10*(3*a_{2}*x^{2}[/mm] + [mm]2*a_{1}*x+a_{0})+[/mm]
> 0 * [mm](a_{2}*x^{3}[/mm] + [mm]a_{1}*x^{2}[/mm] + [mm]a_{0}*x)[/mm] = [mm]-x^{2}[/mm]
>
> [mm]6*a_{2}*x[/mm] + [mm]2*a_{1}[/mm] - [mm](30*a_{2}*x^{2}[/mm] + [mm]20*a_{1}*x[/mm] +
> [mm]10*a_{0})= -x^{2}[/mm]
>
> [mm]6*a_{2}*x[/mm] + [mm]2*a_{1}[/mm] - [mm]30*a_{2}*x^{2}[/mm] - [mm]20*a_{1}*x[/mm] -
> [mm]10*a_{0}= -x^{2}[/mm]
>
> Sortieren nach Potenzen:
>
> - [mm]30*a_{2}*x^{2}[/mm] + x * [mm](6*a_{2}[/mm] - [mm]20*a_{1})[/mm] - [mm]10*a_{0}[/mm] +
> [mm]2*a_{1}[/mm] = [mm]-x^{2}[/mm]
>
> Betrachten wir nun die einzelnen Koeffizienten:
>
> - [mm]30*a_{2}*x^{2}[/mm] = [mm]-x^{2}[/mm]
> [mm]a_{2}[/mm] = [mm]\bruch{1}{30}[/mm]
>
> x * [mm](6*a_{2}[/mm] - [mm]20*a_{1})[/mm] = 0
> x * [mm](6*\bruch{1}{30}[/mm] - [mm]20*a_{1})[/mm] = 0
> x * [mm](6*\bruch{1}{30}[/mm] - [mm]20*a_{1})[/mm] = 0
> x * [mm](\bruch{1}{5}[/mm] - [mm]20*a_{1})[/mm] = 0
> [mm]\bruch{1}{5}[/mm] - [mm]20*a_{1}[/mm] = 0
> - [mm]20*a_{1})[/mm] = - [mm]\bruch{1}{5}[/mm]
> [mm]a_{1})[/mm] = [mm]\bruch{1}{100}[/mm]
>
> - [mm]10*a_{0}[/mm] + [mm]2*a_{1}[/mm] = 0
> - [mm]10*a_{0}[/mm] + [mm]2*\bruch{1}{100}[/mm] = 0
> - [mm]10*a_{0}[/mm] = [mm]-\bruch{2}{100}[/mm]
> [mm]a_{0}[/mm] = [mm]\bruch{2}{1000}[/mm] = [mm]\bruch{1}{500}[/mm]
>
> [mm]y_{p}(x)[/mm] = [mm]\bruch{1}{30}*x^{3}[/mm] + [mm]\bruch{1}{100}*x^{2}[/mm] +
> [mm]\bruch{1}{500}*x[/mm]
>
> y(x) = [mm]y_{h}(x)[/mm] + [mm]y_{p}(x)[/mm]
> [mm]\Rightarrow[/mm] y(x) = [mm]c_{1}*e^{10*x}[/mm] + [mm]c_{2}[/mm] +
> [mm]\bruch{1}{30}*x^{3}[/mm] + [mm]\bruch{1}{100}*x^{2}[/mm] +
> [mm]\bruch{1}{500}*x[/mm]
>
> So damit ist die DGL 2. Ordnung richtig gelöst!
> Jedoch habe ich gerade einen Blackout, wie ich jetzt noch
> mal die Anfangswerte einsetzen muss um einen Endwert zu
> erhalten!
>
> Mußte ich das Endergebnis zwei mal Ableiten ?
> Für y' & y'' ?
Nein.
Du brauchst hier nur
[mm]y\left(0\right)= \ ... [/mm]
[mm]y'\left(0\right)= \ ... [/mm]
Dieses GLeichungsystem löst Du dann nach den Konstanten auf.
>
> Bitte um Hilfe diesbezüglich!
>
> *Danke*
>
>
>
> ------------------------------
> Ich habe diese Aufgabe niergenswo anders gepostet /
> hochgeladen /oder sonst auf irgend eine Art und Weise
> publiziert!
>
>
Gruss
MathePower
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Ah stimmt ich mußte dabei ja nur [mm] c_{1} [/mm] & [mm] c_{2} [/mm] ermitteln Dankeschön!
Dann sollte das Ergebnis folgendermaßen aussehen:
y(x) = [mm] c_{1} [/mm] * [mm] e^{10x} [/mm] + [mm] c_{2} [/mm] + [mm] \bruch{1}{30} [/mm] * [mm] x^{3} [/mm] + [mm] \bruch{1}{100} [/mm] * [mm] x^{2} [/mm] + [mm] \bruch{1}{500} [/mm] * x
y(0) = [mm] c_{1} [/mm] * [mm] e^{10*0} [/mm] + [mm] c_{2} [/mm] + [mm] \bruch{1}{30} [/mm] * [mm] 0^{3} [/mm] + [mm] \bruch{1}{100} [/mm] * [mm] 0^{2} [/mm] + [mm] \bruch{1}{500} [/mm] * 0
0 = [mm] c_{1} [/mm] * 1 + [mm] c_{2}
[/mm]
[mm] -c_{1} [/mm] = [mm] c_{2}
[/mm]
y(x)' = 10 * [mm] c_{1} [/mm] * [mm] e^{10x} [/mm] + [mm] \bruch{1}{10} [/mm] * [mm] x^{2} [/mm] + [mm] \bruch{1}{50} [/mm] * x + [mm] \bruch{1}{500}
[/mm]
y(0)' = 10 * [mm] c_{1} [/mm] * [mm] e^{10*0} [/mm] + [mm] \bruch{1}{10} [/mm] * [mm] 0^{2} [/mm] + [mm] \bruch{1}{50} [/mm] * 0 + [mm] \bruch{1}{500}
[/mm]
y(0)' = 10 * [mm] c_{1} [/mm] * 1 + [mm] \bruch{1}{500}
[/mm]
0 = 10 * [mm] c_{1} [/mm] * 1 + [mm] \bruch{1}{500}
[/mm]
- [mm] \bruch{1}{500} [/mm] = 10 * [mm] c_{1}
[/mm]
[mm] \Rightarrow c_{1} [/mm] = - [mm] \bruch{1}{5000}
[/mm]
und [mm] -c_{1} [/mm] = [mm] c_{2} \Rightarrow c_{2} [/mm] = [mm] \bruch{1}{5000}
[/mm]
Sollte so stimmen oder ?
Habe dafür keine Teillösungen!
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Hallo Kerberos2008,
> Ah stimmt ich mußte dabei ja nur [mm]c_{1}[/mm] & [mm]c_{2}[/mm] ermitteln
> Dankeschön!
>
> Dann sollte das Ergebnis folgendermaßen aussehen:
>
> y(x) = [mm]c_{1}[/mm] * [mm]e^{10x}[/mm] + [mm]c_{2}[/mm] + [mm]\bruch{1}{30}[/mm] * [mm]x^{3}[/mm] +
> [mm]\bruch{1}{100}[/mm] * [mm]x^{2}[/mm] + [mm]\bruch{1}{500}[/mm] * x
>
> y(0) = [mm]c_{1}[/mm] * [mm]e^{10*0}[/mm] + [mm]c_{2}[/mm] + [mm]\bruch{1}{30}[/mm] * [mm]0^{3}[/mm] +
> [mm]\bruch{1}{100}[/mm] * [mm]0^{2}[/mm] + [mm]\bruch{1}{500}[/mm] * 0
>
> 0 = [mm]c_{1}[/mm] * 1 + [mm]c_{2}[/mm]
> [mm]-c_{1}[/mm] = [mm]c_{2}[/mm]
>
> y(x)' = 10 * [mm]c_{1}[/mm] * [mm]e^{10x}[/mm] + [mm]\bruch{1}{10}[/mm] * [mm]x^{2}[/mm] +
> [mm]\bruch{1}{50}[/mm] * x + [mm]\bruch{1}{500}[/mm]
>
> y(0)' = 10 * [mm]c_{1}[/mm] * [mm]e^{10*0}[/mm] + [mm]\bruch{1}{10}[/mm] * [mm]0^{2}[/mm] +
> [mm]\bruch{1}{50}[/mm] * 0 + [mm]\bruch{1}{500}[/mm]
>
> y(0)' = 10 * [mm]c_{1}[/mm] * 1 + [mm]\bruch{1}{500}[/mm]
>
> 0 = 10 * [mm]c_{1}[/mm] * 1 + [mm]\bruch{1}{500}[/mm]
>
> - [mm]\bruch{1}{500}[/mm] = 10 * [mm]c_{1}[/mm]
>
> [mm]\Rightarrow c_{1}[/mm] = - [mm]\bruch{1}{5000}[/mm]
>
> und [mm]-c_{1}[/mm] = [mm]c_{2} \Rightarrow c_{2}[/mm] = [mm]\bruch{1}{5000}[/mm]
>
>
> Sollte so stimmen oder ?
Das stimmt so. ![[ok]](/images/smileys/ok.gif "[ok]")
> Habe dafür keine Teillösungen!
>
Gruss
MathePower
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Danke für die Hilfe, die Wärme macht mir zu schaffen (Schunt Dachgeschoss) - da macht man schnell mal Flüchtigkeitsfehler oder vergisst die leichtesten Dinge :(
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