Answer to Question #138214 in Differential Equations for Nikhil

Question

Using LaGrange's method, solve the differential equations

p - qyIn(y) = zIn(y)

Answer 1

$\displaystyle\textsf{The given partial differential equation is}\\ p - q\, y\ln(y) = z\ln(y)\\ \mathrm{d}z =\frac{\partial z}{\partial x} \mathrm{d}x + \frac{\partial z}{\partial y} \mathrm{d}y \\ \mathrm{d}x = 1, \mathrm{d}y = -y\ln(y), \mathrm{d}z =z\ln(y) \\ \textsf{The Lagrange’s auxiliary equations}\\\textsf{for the given PDE is}\\ \frac{\mathrm{d}x}{1} = \frac{\mathrm{d}y}{-y\ln(y)} = \frac{\mathrm{d}z}{z\ln(y)}\\ \textsf{Choosing}\, \left(0, \frac{1}{y}, \frac{1}{z}\right)\, \textsf{as multipliers, we have}\\ \frac{1}{y}\mathrm{d}y + \frac{1}{z}\mathrm{d}z = 0 \\ \textsf{Integrating both sides, we have} \\ \int \frac{1}{y}\,\mathrm{d}y - \int \frac{1}{z}\,\mathrm{d}z= C_1 \\ \ln(y) - \ln(z) = C_1 \\ \ln\left(\frac{y}{z}\right) = C_1 \\ \therefore \frac{y}{z} = C_2, \hspace{1cm} \{\textsf{where}\, C_2 = e^{C_1}\}\\ \textsf{Choosing}\, \left(1, \frac{1}{y\ln(y)}, 0\right)\, \textsf{as multipliers, we have}\\ \mathrm{d}x + \frac{\mathrm{d}y}{y\ln(y)} = 0 \\ \textsf{Integrating both sides, we have} \\ \int \mathrm{d}x + \int \frac{\mathrm{d}y}{y\ln(y)} = C_3\\ x + \int \frac{\mathrm{d}(\ln(y))}{\ln(y)} = = C_3\\ x + \ln(\ln(y)) = C_3\\ \ln(e^x\ln(y)) = C_3\\ e^x \ln(y) = C_4 \hspace{1cm} \{\textsf{where}\, C_4= e^{C_3}\}\\ \textsf{So, the solution of the given PDE is}\, \\ \phi\left(e^x \ln(y),\,\frac{y}{z}\right) = 0$