Answer to Question #176449 in Differential Equations for ishtiaq

Question

(x-y)p-(x-y+z)q=z

Answer 1

given,

$(x-y)p-(x-y+z)q=z$

This equation can be written as-

$(x-y)p+(y-x-z)q=z$

The lagrange's subsidary equation are-

$\dfrac{dx}{x-y}=\dfrac{dy}{y-x-z}=\dfrac{dz}{z}~~~~~~-(1)$

Each ratio of (1) is equal to-

$\dfrac{dx+dy+dz}{x-y+y-x-z+z}=\dfrac{dx+dy+dz}{0}$

$\Rightarrow dx+dy+dz=0$

Integrating and we get-

$x+y+z=c_1$

Also each ratio is equal to-

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

Taking with 3 ratio of (1) we get

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

$\Rightarrow \dfrac{dz}{z}=\dfrac{d(x-y+z)}{2(x-y+z)}$

$\Rightarrow 2\dfrac{dz}{z}=\dfrac{d(x-y+z)}{(x-y+z)}$

Integrating and we get-

$log(x-y+z)+logc_2=2logz$

$\Rightarrow logz(x-y+z)c_2=logz^2$

$\Rightarrow \dfrac{z^2}{x-y+z}=c_2$

The solution of the above equation is -

$\phi(c_1,c_2)=0$

$\phi(x+y+z,\dfrac{z^2}{x-y+z})=0$