Answer to Question #138536 in Differential Equations for Jagadeshreddu

Question

Solve y zp + x²zq = y2x

Answer 1

$\displaystyle\textsf{The given partial differential equation is}\\ yz\, p + x^2 z\, q = y^2 x \\ \mathrm{d}z =\frac{\partial z}{\partial x} \mathrm{d}x + \frac{\partial z}{\partial y} \mathrm{d}y \\ \mathrm{d}z = xy^2, \mathrm{d}x = yz, \mathrm{d}y = x^2 z \\ \textsf{The Lagrange’s auxiliary equations}\\\textsf{for the given PDE is}\\ \frac{\mathrm{d}x}{yz} = \frac{\mathrm{d}y}{x^2 z} = \frac{\mathrm{d}z}{xy^2}\\ \textsf{Choosing}\, (x^2, -y, 0)\, \textsf{as multipliers, we have}\\ x^2\mathrm{d}x - y\mathrm{d}y = 0 \\ \textsf{Integrating both sides, we have} \\ \int x^2\mathrm{d}x - \int y\mathrm{d}y = C_1 \\ \frac{x^3}{3} - \frac{y^2}{2} =c \\ \textsf{Comparing the first and the last differential equations}\\ \frac{\mathrm{d}x}{yz} = \frac{\mathrm{d}z}{y^2 x}\\ yx\mathrm{d}x - z\mathrm{d}z = 0\\ \textsf{But}\\ x^3/3 - y^2/2 = c \\ y = \sqrt{\frac{2x^3}{3} + C} \\ \therefore \int x\sqrt{\frac{2x^3}{3} + C}\, \mathrm{d}x - \frac{z^2}{2} = C_2\\ \textsf{The Integral cannot be expressed in simpler}\\ \textsf{terms and as such it can be expressed}\\\textsf{in terms of advanced functions like the}\\\textsf{hypergeometric function. We can hence}\\\textsf{write the Integral in terms of know}\\\textsf{functions if}\,C = 0.\\ \therefore \int x\sqrt{\frac{2x^3}{3}}\, \mathrm{d}x - \frac{z^2}{2} = C_2\\ \frac{2}{7}\sqrt{\frac{2}{3}} x^{\frac{7}{2}} - \frac{z^2}{2} = C_2\\ \textsf{So, a solution of the given PDE is}\, \\ \phi\left(\frac{x^3}{3} - \frac{y^2}{2}, \frac{2}{7}\sqrt{\frac{2}{3}} x^{\frac{7}{2}} - \frac{z^2}{2}\right) = 0$