Question

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

Answer 1

Before solving the problem, please note the problem has been changed as $(3x+y+z)p + (x+y+z)q = 2(z+y)$ instead of $(3x+y-z)p + (x+y+z)q = 2(z+y)$.

Consider, $(3x+y+z)p + (x+y+z)q = 2(z+y)$.

The auxiliary equation is,

$\dfrac{dx}{3x+y+z}=\dfrac{dy}{x+y+z}=\dfrac{dz}{2(z+y)}.$

Using the multipliers 1, -3, 1 each of the above ratio is equal to

$\dfrac{dx-3dy+dz}{3x+y+z-3x-3y-3z+2z+2y}=\dfrac{dx-3dy+dz}{0}$

Thus,

$dx-3dy+dz = 0$

Integrating we get, $x-3y+z=c_{1}.$

Each of the ratio in the auxiliary equation is equal to

$\dfrac{dx+dy+dz}{3x+y+z+x+y+z+2z+2y}=\dfrac{dx-dy-dz}{3x+y+z-x-y-z-2z-2y}\\~\\ \dfrac{dx+dy+dz}{4(x+y+z)}=\dfrac{dx-dy-dz}{2(x-y-z)}\\~\\ \dfrac{dx+dy+dz}{2(x+y+z)}=\dfrac{dx-dy-dz}{(x-y-z)}\\~\\ \text{Integrating,} \\ \dfrac{1}{2}\ln(x+y+z)=\ln(x-y-z)+\ln c_{2}\\~\\ \ln\bigg(\dfrac{\sqrt{x+y+z}}{x-y-z}\bigg)=\ln c_{2}\\~\\ \dfrac{\sqrt{x+y+z}}{x-y-z} = c_{2}.$

Hence the general solution is, $\phi(c_{1},c_{2})=0$

$\phi\bigg(x-3y+z, \dfrac{\sqrt{x+y+z}}{x-y-z}\bigg)=0$