Homogene diferencijalne jednadžbe

Uvrštavanjem $\displaystyle y^{\prime }=\frac{dy}{ dx}$ u zadanu diferencijalnu jednadžbu dobivamo

$\displaystyle ydy$

$\displaystyle =\left( y-x\right) dx,$

odnosno,

$\displaystyle \left( y-x\right) dx-ydy=0.$

(5.2)

Ovo je diferencijalna jednadžba prvog stupnja homogenosti pa je rješavamo supstitucijom:

Uvrštavanjem u (5.2) dobivamo

$\displaystyle \left( xz-x\right) -xz\left( z+xz^{\prime }\right)$	$\displaystyle =0 \slash :x$
$\displaystyle z-1$	$\displaystyle =z\left( z+xz^{\prime }\right)$
$\displaystyle z-1-z^{2}$	$\displaystyle =xz^{\prime }\cdot z$
$\displaystyle xz\frac{dz}{ dx}$	$\displaystyle =z-1-z^{2}$
$\displaystyle \frac{zdz}{z-1-z^{2}}$	$\displaystyle =\frac{ dx}{x}$
$\displaystyle \frac{zdz}{z^{2}-z+1}$	$\displaystyle =-\frac{ dx}{x}.$

Ovo je diferencijalna jednadžba separiranih varijabli, pa vrijedi

$\displaystyle \int \frac{zdz}{z^{2}-z+1}$	$\displaystyle =-\int \frac{ dx}{x}$
$\displaystyle \int \frac{\frac{1}{2}d\left( z^{2}-z+1\right) +\frac{1}{2}}{z^{2}-z+1}$	$\displaystyle =-\ln \left\vert x\right\vert +C$
$\displaystyle \frac{1}{2}\int \frac{d\left( z^{2}-z+1\right) }{z^{2}-z+1}+\frac{1}{2}\int \frac{dz}{z^{2}-z+1}$	$\displaystyle =-\ln \left\vert x\right\vert +C$
$\displaystyle \frac{1}{2}\ln \left\vert z^{2}-z+1\right\vert +\frac{1}{2}\frac{2}{\sqrt{3}} {\mathop{\mathrm{arctg}}}\frac{z-\frac{1}{2}}{\frac{\sqrt{3}}{2}}$	$\displaystyle =-\ln \left\vert x\right\vert +C$
$\displaystyle \ln \sqrt{z^{2}-z+1}+\frac{1}{\sqrt{3}}{\mathop{\mathrm{arctg}}}\frac{2z-1}{ \sqrt{3}}$	$\displaystyle =-\ln \left\vert x\right\vert +C.$

Vraćanjem susptitucije $\displaystyle\frac{y}{x}=z$ dobivamo

$\displaystyle \ln \sqrt{\frac{y}{x}^{2}-\frac{y}{x}+1}+\frac{1}{\sqrt{3}}{ \mathop{\mathrm{arctg}}}\frac{2\frac{y}{x}-1}{\sqrt{3}}$	$\displaystyle =-\ln \left\vert x\right\vert +C$
$\displaystyle \ln \sqrt{y^{2}-xy+x^{2}}-\ln \left\vert x\right\vert +\frac{1}{\sqrt{3}}{ \mathop{\mathrm{arctg}}}\frac{2y-x}{x\sqrt{3}}$	$\displaystyle =-\ln \left\vert x\right\vert +C$
$\displaystyle \ln \sqrt{y^{2}-xy+x^{2}}+\frac{1}{\sqrt{3}}{\mathop{\mathrm{arctg}}}\frac{ 2y-x}{x\sqrt{3}}$	$\displaystyle =C$
$\displaystyle \ln C \sqrt{y^{2}-xy+x^{2}}+\frac{1}{\sqrt{3}}{\mathop{\mathrm{arctg}}} \frac{2y-x}{x\sqrt{3}}$	$\displaystyle =0.$

Ovo je homogena diferencijalna jednadžba stupnja homogenosti

koju rješavamo supstitucijom:

$\displaystyle \frac{x}{y}$	$\displaystyle =z$
$\displaystyle x$	$\displaystyle =yz$
$\displaystyle \frac{ dx}{dy}$	$\displaystyle =x^{\prime }=z+yz^{\prime }$

Uvrštavanjem u zadanu diferencijalnu jednadžbu dobivamo

$\displaystyle \left( 1+e^{z}\right) \left( z^{\prime }y+z\right) +e^{z}\left( 1-z\right)$	$\displaystyle =0$
$\displaystyle z^{\prime }y+z+yz^{\prime }e^{z}+ze^{z}+e^{z}-ze^{z}$	$\displaystyle =0$
$\displaystyle z^{\prime }y\left( 1+e^{z}\right)$	$\displaystyle =-z-e^{z}$
$\displaystyle y\left( 1+e^{z}\right) \frac{dz}{dy}$	$\displaystyle =-z-e^{z}$
$\displaystyle y\left( 1+e^{z}\right) dz$	$\displaystyle =\left( -z-e^{z}\right) dy$
$\displaystyle -\frac{1+e^{z}}{e^{z}+z}dz$	$\displaystyle =\frac{dy}{y}.$

Ovo je diferencijalna jednadžba sa separiranim varijablama pa vrijedi

$\displaystyle -\int \frac{d\left( e^{z}+z\right) }{e^{z}+z}$	$\displaystyle =\int \frac{dy}{y}$
$\displaystyle -\ln \left\vert e^{z}+z\right\vert$	$\displaystyle =\ln \left\vert y\right\vert +C$
$\displaystyle \frac{1}{\left\vert e^{z}+z\right\vert }$	$\displaystyle =Cy$
$\displaystyle e^{z}+z$	$\displaystyle =\frac{1}{Cy}.$

Vraćanjem susptitucije $\displaystyle\frac{x}{y}=z$ dobivamo

$\displaystyle e^{\frac{x}{y}}+\frac{x}{y}=\frac{1}{Cy}.$