Answer to Question #211496 in Differential Equations for Siddu

Question

Solve x2p+y2q=(x+y)z

Answer 1

Question

Solve x^2p +y^2q=(x+y)z

Solution

x²$\frac {dz}{dx}$ +y²$\frac {dz}{dy}$ =(x+y)z

Characteristic equation will be

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

$\int\frac {dx} {x^2}$ =$\int\frac {dy} {y^2}$

$-\frac 1 x$ =$-\frac 1 y$ +C₁

C₁=$\frac {x-y} {xy}$

y=$\frac {x}{Cx +1}$

The characteristics equation can then be written as;

$\frac {dx} {x^2} =\frac {dz}{(x+y)z}$ =$\cfrac {dz} {(x + \cfrac {x} {C_1x +1})z}$

$\frac { C_1x^2 +2x}{x^2(C_1x +1)}$dx =$\frac {C_1x +2} {x (C_1x+1)}$ =$\frac {dz} z$

$\int\frac {C_1x+2}{x (C_1x+1)}$dx=$\int\frac {dz} z$

But

$\frac {C_1x +2}{x (C_1x+1)}=\frac 1 x +\frac {1}{x(C_1x+1)}$

However,

$\frac {1} {x (C_1x+1)}$ =$\frac A x$ +$\frac {B}{C_1x+1}$

A(C₁x+1)+Bx=1

Therefore,

A=1 and B=-C₁

$\frac {1}{x (C_1x+1)}$ =$\frac 1 x$ +$\frac {-C_1}{C_1x+1}$

$\int$ $\frac {C_1x+2}{x (C_1x+1)}$ =$\int$ $\frac 2 x$ -$\int$ $\frac {C_1}{C_1x+1}$ =$\int\frac{dz}{z}$

2ln|x| -C₁ln|C₁x+1|=ln|z|+ln|C₂|

C₂=$\frac {x^2}{z(C_1x+1)}$ =$\frac {x^2}{z(1+\frac {x-y}{y})}$ =$\frac {yx^2}{zx}$ =$\frac {xy }{z}$

Answer;

F($\frac {x-y}{xy}$,$\frac {xy} z$)=0