Question

solve the pde yzp+zxq-xy=0​

Answer 1

Lagrange's Solution of a Linear Partial Differential Equation Pp + Qq = R

To find: Solution of the partial differential equation $yzp+zxq-xy=0$.

Solution: The given partial differential equation is

$\quad\quad yzp+zxp=xy$

The Lagrange's subsidiary equations are

$\quad\quad \frac{dx}{yz}=\frac{dy}{zx}=\frac{dz}{xy}$

Taking the first two ratios, we get

$\quad \frac{dx}{yz}=\frac{dy}{zx}$

$\Rightarrow \frac{dx}{y}=\frac{dy}{x}$

$\Rightarrow x\:dx=y\:dy$

On integration, we get

$\quad\int x\:dx=\int y\:dy$

$\Rightarrow \frac{1}{2}\:x^{2}=\frac{1}{2}\:y^{2}+\frac{1}{2}\:c_{1}$ where $\frac{1}{2}\:c_{1}$ is integral constant

$\Rightarrow x^{2}-y^{2}=c_{1}$ .....(1)

Again taking the last two ratios, we get

$\quad \frac{dy}{zx}=\frac{dz}{xy}$

$\Rightarrow \frac{dy}{z}=\frac{dz}{y}$

$\Rightarrow y\:dy=z\:dz$

On integration, we get

$\quad \int y\:dy=\int z\:dz$

$\Rightarrow \frac{1}{2}\:y^{2}=\frac{1}{2}\:z^{2}+\frac{1}{2}\:c_{2}$ where $\frac{1}{2}\:c_{2}$ is integral constant

$\Rightarrow y^{2}-z^{2}=c_{2}$ .....(2)

Answer:

∴ the required general solution is

$\quad \phi(x^{2}-y^{2},\:y^{2}-z^{2})=0$ where $\phi$ is arbitrary.