Answer to Question #218786 in Differential Equations for Pratham Gawali

Question

x²(y-z)p + y²(z − x)q = z²(x-y)

Answer 1

Given the equation:

$$x^2(y-z)p + y^2(z − x)q = z^2(x-y)$$

The equation is a Lagrange Linear PDE and it's of the form:

$$Pp+Qq=R$$

with:

$$P=x^2(y-z); \quad Q=y^2(z − x); \quad R=z^2(x-y)$$

The auxiliary equation is in the form:

$$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$$

Thus:

$$\frac{dx}{x^2(y-z)}=\frac{dy}{y^2(z − x)}=\frac{dz}{z^2(x-y)}$$

We proceed to solve the equation.

Using the multipliers $\frac{1}{x},\frac{1}{y},\frac{1}{z}$

$$\frac{\frac{dx}{x}}{x(y-z)}=\frac{\frac{dy}{y}}{y(z − x)}=\frac{\frac{dz}{z}}{z(x-y)}\\$$

Adding up we have:

$$\frac{dx}{x}+\frac{dy}{y}+\frac{dz}{z}=0$$

Integrating through:

$$\int\frac{dx}{x}+\int\frac{dy}{y}+\int\frac{dz}{z}=\int0\\ \ln x + \ln y+ \ln z=\ln c\\ \ln(xyz)=\ln c\\ \bold{c_1 = xyz}$$

Using the multipliers $\frac{1}{x^2}, \frac{1}{y^2}, \frac{1}{z^2}$ as another triplet such that the denominator vanishes:

$$\frac{\frac{dx}{x^2}}{y-z}=\frac{\frac{dy}{y^2}}{z − x}=\frac{\frac{dz}{z^2}}{x-y}\\$$

Adding up, we get:

$$\frac{dx}{x^2}+\frac{dy}{y^2}+\frac{dz}{z^2}=0$$

Integrating through:

$$\int\frac{dx}{x^2}+\int\frac{dy}{y^2}+\int\frac{dz}{z^2}=\int 0\\ -\frac{1}{x}-\frac{1}{y}-\frac{1}{z}=c_2\\ \bold{c_2=-\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Big)}$$

Therefore the solution of the PDE is:

$$\phi(c_1,c_2) = \Big(xyz, -\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Big)\Big)$$