Question

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

Answer 1

The given partial differential equation is

(y+ x)p + (y− x)q = z

The Lagrange’s auxiliary equations are

$\frac{dx}{y+x}=\frac{dy}{y-x}=dz\\ \text{integrating the first two ratio, we have}\\ \frac{dx}{y+x}=\frac{dy}{y-x}\\ \frac{dy}{dx}=\frac{y-x}{y+x}\\ \text{solving by homogeneous first order differential equation}\\ y=vx\\ \frac{dy}{dx}=v+x\frac{dv}{dx}\\ v+x\frac{dv}{dx}=\frac{vx-x}{vx+x}\\ x\frac{dv}{dx}=\frac{v-1}{v+1}-v\\ x\frac{dv}{dx}=\frac{-(v^2+1)}{v+1}\\ \frac{dx}{x}=-\frac{v+1}{v^2+1}dv\\ \frac{dx}{x}+\frac{v+1}{v^2+1}dv\\=o\\ \int \frac{dx}{x}+\int \frac{v+1}{v^2+1}dv\\=c_1\\ c_1=log_ex+\frac{1}{2}log_e(v^2+1)+\frac{1}{v}tan^{-1}v\\ C_1=\frac{1}{2}log_e(\frac{x^3}{y}+1)+\frac{y}{x}tan^{-1}\frac{x}{y}\\ \text{integrating the last two ratio}\\ \frac{dy}{y-x}=\frac{dz}{z}\\ y=vz\\ \frac{dy}{dz}=v+z\frac{dv}{dz}\\ v+z\frac{dv}{dz}=\frac{vz-x}{z}\\ z\frac{dv}{dz}=\frac{vz-x}{z}-v\\ z\frac{dv}{dz}=\frac{-x}{z}\\ {z^2}dv=-xdz\\ \int {z^2}dv=\int -xdz\\ z^2v+xz=c_2\\ \frac{z^2y}{x}+xz=c_2\\ \phi(\frac{1}{2}log_e(\frac{x^3}{y}+1)+\frac{y}{x}tan^{-1}\frac{x}{y}, \frac{z^2y}{x}+xz)=0\\ \phi \\ \text{is an arbitrary constant}$