Answer to Question #164217 in Differential Equations for ripunjoy sharma

Question

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

Answer 1

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

This is Lagrange's linear equation $Pp+Qq=R$

∴ The auxiliary equation is

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

Taking the fraction as:-

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

Taking first two terms-

 

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

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

Integrating Both side and we get,

 

$\dfrac{1}{2}log(x+y+z)=-log(x-y)+C$

Now taking last two terms-

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

$\dfrac{d(x-y)}{y-x}=\dfrac{d(y-z)}{z-y}$

Integrating Both sides and we get-

 

 $log(x-y)=log(y-2)+logc_2$

$\dfrac{x-y}{y-z}=c_2$

Hence General equation of the differential equation is-

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