Painevg

2021-12-31

Solve the following homogeneous differential equations
$\frac{dy}{dx}=\frac{x-y}{x+y}$

$\frac{dy}{dx}=\frac{x-y}{x+y}$ (1)
We have to solve the equation
Rearrange the equation
$\left(x+y\right)dy=\left(x-y\right)dx$
$xdy+ydy=xdx-ydx$
$xdy+ydx+ydy-ydx=0$
$d\left(xy\right)+ydy-xdx=0$
Integrating we get
$xy+\frac{{y}^{2}}{2}-\frac{{x}^{2}+c}{2}=0$
$xy+\frac{{y}^{2}}{2}-\frac{{x}^{2}}{2}+c=0$
It is solution of differential equation.

Hector Roberts

Put, $y=vx$
$⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$
$⇒v+x\frac{dv}{dx}=\frac{1-v}{1+v}$
$⇒x\frac{dv}{dx}=\frac{1-2v-{v}^{2}}{1+v}$
$⇒\int \frac{v+1}{{\left(v+1\right)}^{2}-2}dx=-\int \frac{1}{x}dx$
$⇒\frac{1}{2}\mathrm{ln}|{\left(v+1\right)}^{2}-2|=-\mathrm{ln}|x|+\mathrm{ln}|{c}_{1}|$
$⇒\mathrm{ln}|\left[{\left(v+1\right)}^{2}-2\right]|=2\mathrm{ln}|\frac{{c}_{1}}{x}|$
$⇒{\left(v+1\right)}^{2}-2=\frac{{c}_{1}^{2}}{{x}^{2}}$
$⇒{x}^{2}\left({v}^{2}+2v-1\right)=C$
Where $C={c}_{1}^{2}$
Since, $v=\frac{y}{x}$
$⇒{y}^{2}+2xy-{x}^{2}=C$

karton