Answer to Question #141003 in Differential Equations for Rachit

Question

p − q = log( x + y)

Answer 1

$\displaystyle p - q = \log(x + y)\\ \textsf{The Langrange's auxiliary equation is}\\ \frac{\mathrm{d}x}{1} = \frac{\mathrm{d}y}{-1} = \frac{\mathrm{d}z}{\log(x + y)}\\ \textsf{Choosing}\, (1, 1, 0)\, \textsf{as multipliers}\\ \mathrm{d}x + \mathrm{d}y = 0\\ \textsf{Integrating both sides}\\ \int \mathrm{d}x + \int \mathrm{d}y = C_1\\ x + y = C_1\\ \textsf{Comparing the first and the last equation}\\ \frac{\mathrm{d}x}{1} = \frac{\mathrm{d}z}{\log(x + y)}\\ \int \mathrm{d}x = \int\frac{\mathrm{d}z}{\log(x + y)}\\ \int \mathrm{d}x = \int\frac{\mathrm{d}z}{\log(C_1)}\\ x = \frac{z + C_2}{\log(C_1)}\\ x\log(C_1) = z + C_2\\ x\log(C_1) - z = C_2\\ x\log(x + y)- z = C_2, \hspace{0.5cm}\{C_1 = x + y\}\\ \textsf{Thus, the solution to the partial}\\\textsf{differential equations is}\\ \phi(x + y, x\log(x + y) - z) = 0$