Exakte DGL II < gewöhnliche < Differentialgl. < Analysis < Hochschule < Mathe < Vorhilfe
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| Aufgabe | Show that
$yf(xy)dx+xg(xy)dy=0$
is not exact in general but becomes exact on multiplying by the integrating factor
[mm] $\bruch{1}{xyf(xy)-xyg(xy)}$. [/mm] |
Hallo,
ich suche meinen Rechenfehler:
[mm] $\bruch{yf(xy)}{xyf(xy)-xyg(xy)}dx+\bruch{xg(xy)}{xyf(xy)-xyg(xy)}dy=0$
[/mm]
[mm] $\bruch{\partial M}{\partial y}=\bruch{(xyf-xyg)(f-yf_yx)-yf*(xf+xyf_y-xg-xyg_yx)}{(xyf-xyg)^2}$
[/mm]
[mm] $\bruch{\partial M}{\partial y}=\bruch{(xyf^2-xyfg+x^2y^2ff_y-x^2y^2f_yg-xyf^2-x^2y^2ff_y+xyfg+x^2y^2fg_y}{(xyf-xyg)^2}$
[/mm]
[mm] $\bruch{\partial M}{\partial y}=\bruch{x^2y^2*(fg_y-f_yg)}{(xyf-xyg)^2}$
[/mm]
[mm] $\bruch{\partial N}{\partial x}=\bruch{(xyf-xyg)(g+xyg_x)-xg*(yf+xy^2f_x-yg-xy^2g_x)}{(xyf(xy)-xyg(xy))^2}$
[/mm]
[mm] $\bruch{\partial N}{\partial x}=\bruch{xyfg-xyg^2+x^2y^2fg_x-x^2y^2gg_x-xyfg-x^2y^2f_xg+xyg^2+x^2y^2gg_x}{(xyf-xyg)^2}$
[/mm]
[mm] $\bruch{\partial N}{\partial x}=\bruch{x^2y^2(fg_x-f_xg)}{(xyf-xyg)^2}$
[/mm]
Demnach sollte [mm] $fg_y-f_yg=fg_x-f_xg$ [/mm] sein ?
Vielen Dank fürs Nachschauen.
LG, Martinius
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Hallo Martinius,
> Show that
>
> [mm]yf(xy)dx+xg(xy)dy=0[/mm]
Hier muß man f bzw. g nach der Kettenregel ableiten.
[mm]y*f\left( \ u\left(x,y\right) \ ) dx + x*g\left( \ u\left(x,y\right) \ ) dy=0[/mm]
mit [mm]u\left(x,y\right)=xy[/mm]
Schreibe also
[mm]f_{x}=f_{u}*u_{x}=y*f_{u}[/mm]
[mm]f_{y}=f_{u}*u_{y}=x*f_{u}[/mm]
[mm]g_{x}=g_{u}*u_{x}=y*g_{u}[/mm]
[mm]g_{y}=g_{u}*u_{y}=x*g_{u}[/mm]
>
> is not exact in general but becomes exact on multiplying by
> the integrating factor
>
> [mm]\bruch{1}{xyf(xy)-xyg(xy)}[/mm].
> Hallo,
>
> ich suche meinen Rechenfehler:
>
> [mm]\bruch{yf(xy)}{xyf(xy)-xyg(xy)}dx+\bruch{xg(xy)}{xyf(xy)-xyg(xy)}dy=0[/mm]
>
> [mm]\bruch{\partial M}{\partial y}=\bruch{(xyf-xyg)(f-yf_yx)-yf*(xf+xyf_y-xg-xyg_yx)}{(xyf-xyg)^2}[/mm]
[mm]\bruch{\partial M}{\partial y}=\bruch{(xyf-xyg)(f\red{+}yf_{\red{u}}\red{u_{y}})-yf*(xf+xyf_{\red{u}}*\red{u_{y}}-xg-xyg_{\red{u}}\red{u_{y}})}{(xyf-xyg)^2}[/mm]
[mm]\gdw \bruch{\partial M}{\partial y}=\bruch{(xyf-xyg)(f\red{+}yf_{\red{u}}\red{x})-yf*(xf+xyf_{\red{u}}*\red{x}-xg-xyg_{\red{u}}\red{x})}{(xyf-xyg)^2}[/mm]
>
> [mm]\bruch{\partial M}{\partial y}=\bruch{(xyf^2-xyfg+x^2y^2ff_y-x^2y^2f_yg-xyf^2-x^2y^2ff_y+xyfg+x^2y^2fg_y}{(xyf-xyg)^2}[/mm]
>
> [mm]\bruch{\partial M}{\partial y}=\bruch{x^2y^2*(fg_y-f_yg)}{(xyf-xyg)^2}[/mm]
>
>
> [mm]\bruch{\partial N}{\partial x}=\bruch{(xyf-xyg)(g+xyg_x)-xg*(yf+xy^2f_x-yg-xy^2g_x)}{(xyf(xy)-xyg(xy))^2}[/mm]
[mm]\bruch{\partial N}{\partial x}=\bruch{(xyf-xyg)(g+xyg_{\red{u}})-xg*(yf+xy^2f_{\red{u}}-yg-xy^2g_{\red{u}})}{(xyf(xy)-xyg(xy))^2}[/mm]
>
> [mm]\bruch{\partial N}{\partial x}=\bruch{xyfg-xyg^2+x^2y^2fg_x-x^2y^2gg_x-xyfg-x^2y^2f_xg+xyg^2+x^2y^2gg_x}{(xyf-xyg)^2}[/mm]
>
> [mm]\bruch{\partial N}{\partial x}=\bruch{x^2y^2(fg_x-f_xg)}{(xyf-xyg)^2}[/mm]
>
>
> Demnach sollte [mm]fg_y-f_yg=fg_x-f_xg[/mm] sein ?
>
> Vielen Dank fürs Nachschauen.
>
> LG, Martinius
>
>
>
>
Gruß
MathePower
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