Question

Find the equation of integral surface to the differential equation

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

Answer 1

$2y(z-3)p+(2x-z)q=y(2x-3)\\ \displaystyle p = \frac{\partial z}{\partial x}, q = \frac{\partial z}{\partial y}\\ \textsf{The equation for the total}\\\textsf{differential is thus given as}\\ \mathrm{d}z = \frac{\partial z}{\partial x} \mathrm{d}x + \frac{\partial z}{\partial y}\mathrm{d}y \\ \textsf{The differentials are;}\\ \begin{aligned} \mathrm{d}z &= 2xy - 6y\\ \mathrm{d}x &= 2yz - 6y\\ \mathrm{d}y &= 2x - z \end{aligned} \\ \textsf{The Lagrange’s auxiliary equations}\\\textsf{for given PDE is}\\ \frac{\mathrm{d}z}{y(2x - 3)}= \frac{\mathrm{d}y}{(2x - z)} = \frac{\mathrm{d}x}{2y(z-3)} \\ \textsf{Choosing}\hspace{0.1cm} (x,3y,−z) \hspace{0.1cm}\\ \textsf{as multipliers, we have}\\ x\mathrm{d}x + 3y\mathrm{d}y - z\mathrm{d}z = 0\\ \textsf{Integrating both sides, we have thus,}\\ x² + 3y² - z² = C_1 \\ \textbf{\textsf{Or}}\\ \textsf{Choosing}\hspace{0.1cm} (l,3m,−n) \hspace{0.1cm} \\\textsf{as multipliers, we have}\\ l\mathrm{d}x + 3m\mathrm{d}y - n\mathrm{d}z = 0\\ \Rightarrow lx + 3my - nz = C_2.\\ \textsf{Thus, the solution of the given PDE is}\\ \phi(lx + 3my - nz, x² + 3y² - z²) = 0 \hspace{0.2cm}\forall l,m,n \in \mathbb{R}\hspace{0.1cm}\&\\ C_1 \hspace{0.1cm}\textsf{and}\hspace{0.1cm}C_2\hspace{0.1cm}\textsf{are arbitrary constants}$