Answer to Question #266938 in Differential Equations for Devil

Question

Solve the partial differential equation

Px+q=p^2

Answer 1

Its subsidiary equations are given by

$$\dfrac{dx}{-f_p}=\dfrac{dy}{-f_q}=\dfrac{dz}{-pf_p-qf_q}=\dfrac{dp}{f_q+pf_z}$$

$$=\dfrac{dq}{f_y+qf_z}=\dfrac{d\phi}{0}$$

$$\dfrac{dx}{2p-x}=\dfrac{dy}{-1}=\dfrac{dz}{2p^2-xp-q}=\dfrac{dp}{p+0}$$

$$=\dfrac{dq}{0}=\dfrac{d\phi}{0}$$

Taking $dq = 0 ⇒ q = c (constant)$

$p^2-xp-q=0$ becomes $p^2-xp-c=0$

$$p=\dfrac{x\pm\sqrt{x^2+4c}}{2}$$

Thus

$$dz=pdx+qdy$$

$$dz=\dfrac{x\pm\sqrt{x^2+4c}}{2}dx+cdy$$

Integrate

$$z=\int(\dfrac{x\pm\sqrt{x^2+4c}}{2})dx+c\int dy+C_1$$

$$\int \dfrac{\sqrt{x^2+4c}}{2}dx=c\ln(|\sqrt{x^2+4c}+x|)+\dfrac{x}{4}\sqrt{x^2+4c}$$

$$z=\dfrac{x^2}{4}\pm(c\ln(|\sqrt{x^2+4c}+x|)+\dfrac{x}{4}\sqrt{x^2+4c})$$

$$+cy+C_2$$