Answer to Question #276480 in Differential Equations for Saniya

Question

Find the general integral of the linear p.d.e

1) Px(x+y) =qy(x+y) - (x-y) (2x+2y+z)

2) x(x^2+3y^2) p - y(3x^2+y^2) q=2z(y^2-x^2)

3) (y+zx) p - (x+yz) q=x^2-y^2

Answer 1

1)

$\frac{dx}{x(x+y)}=-\frac{dy}{y(x+y)}=-\frac{dz}{(x-y) (2x+2y+z)}$

$dx/x+dy/y=0$

$lnx+lny=lnc_1$

$xy=c_1$

$\frac{dx+dy}{x+y}=\frac{-dz}{2(x+y)+z}$

let $u=x+y,z=tu$ , then:

$z'=t'u+t=-\frac{2u+tu}{u}=-2-t$

$-\frac{dt}{2(t+1)}=du/u$

$-ln(t+1)/2=lnu+lnc$

$\frac{1}{\sqrt{t+1}}=cu$

$u\sqrt{t+1}=(x+y)\sqrt{\frac{z}{x+y}+1}=c_2$

$F(c_1,c_2)=F(xy,(x+y)\sqrt{\frac{z}{x+y}+1})=0$

2)

$\frac{dx}{ x(x^2+3y^2)}=-\frac{dy}{y(3x^2+y^2)}=\frac{dz}{2z(y^2-x^2)}$

$dx/x+dy/y=dz/z$

$lnx+lny=lnz+lnc_1$

$xy/z=c_1$

$-\frac{y(3x^2+y^2)}{ x(x^2+3y^2)}=\frac{dy}{dx}$

$y=tx,y'=t'x+t$

$t'x+t=-\frac{t(3+t^2)}{1+3t^2}$

$t'x=-\frac{3t+t^3+t+3t^3}{1+3t^2}$

$\frac{1+3t^2}{t^3+t}dt=-\frac{4dx}{x}$

$ln(t^3+t)=-4lnx+lnc_2$

$x(t^3+t)=x((y/x)^3+y/x)=c_2$

$y^3/x^2+y=c_2$

$F(c_1,c_2)=F(xy/z,y^3/x^2+y )=0$

3)

$\frac{dx}{y+zx}=-\frac{dy}{x+yz}=\frac{dz}{x^2-y^2}$

$ydx+xdy+dz=0$

$xy+z^2/2=c_1$

$\frac{xdx+ydy}{z(x^2-y^2)}=\frac{dz}{x^2-y^2}$

$x^2+y^2-z^2=c_2$

$F(c_1,c_2)=F(xy+z^2/2, x^2+y^2-z^2)=0$