Overleg:Integrerende factor

Check

Als

$P\mathrm {d} x+Q\mathrm {d} y=0$

herschreven kan worden als

$yA(xy)\mathrm {d} x+xB(xy)\mathrm {d} y=0$

is

${\frac {1}{xy(A-B)}}$

een integrerende factor dus

${\frac {yA\mathrm {d} x+xB\mathrm {d} y}{xy(A-B)}}$

is totaal

Controle

${\frac {\partial }{\partial y}}{\frac {A}{x(A-B)}}=$

$={\frac {xA'}{x(A-B)}}-{\frac {Ax(xA'-xB')}{(x(A-B))^{2}}}$

$={\frac {A'(A-B)}{(A-B)^{2}}}-{\frac {A(A'-B')}{(A-B)^{2}}}$

$={\frac {-A'B-AB'}{(A-B)^{2}}}$

$={\frac {-(AB)'}{(A-B)^{2}}}$

en dus ook

${\frac {\partial }{\partial x}}{\frac {B}{y(A-B)}}={\frac {-(AB)'}{(A-B)^{2}}}$

eigenlijk

${\frac {\mathrm {d} y}{\mathrm {d} x}}+R(x,y)=0$

als

${\frac {\mathrm {d} y}{\mathrm {d} x}}+{\frac {A(x)}{B(y)}}=0$

Als

$yA(x)\mathrm {d} x+xB(y)\mathrm {d} y=0$

dan equiv met

$a(x)\mathrm {d} x+b(y)\mathrm {d} y=0$

dus scheiding vars,

$a(x)\mathrm {d} x=-b(y)\mathrm {d} y$