Question

Find the integral surface of the partial differential equation
(x y) y p (y x) x q (x y ) z 2 2 2 2 − + − = +
through the curve , 0 2 xz = a y =

Answer 1

For equation

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

write an auxiliary equation

$$\displaystyle\frac{dx}{y^2(x-y)}=\displaystyle\frac{dy}{x^2(y-x)}=\displaystyle\frac{dz}{z(x^2+y^2)}\\[0.3cm] \displaystyle\frac{dx}{y^2(x-y)}=\displaystyle\frac{dy}{x^2(y-x)}\rightarrow\displaystyle\frac{dx}{y^2}=\displaystyle\frac{dy}{x^2(-1)}\\[0.3cm] y^2dy+x^2dx=0\rightarrow d\left(\frac{x^3}{3}+\frac{y^3}{3}\right)=0\\[0.3cm] \frac{x^3}{3}+\frac{y^3}{3}=C_1\rightarrow x^3+y^3=3C_1\\[0.3cm] \boxed{x^3+y^3=c_1}\\[0.3cm] \displaystyle\frac{dx}{y^2(x-y)}=\displaystyle\frac{dz}{z(x^2+y^2)}\rightarrow \displaystyle\frac{(x^2+y^2)dx}{(x-y)}=\displaystyle\frac{y^2dz}{z}\\[0.3cm] \int\left(\frac{(x^2-y^2)+2y^2}{(x-y)}\right)dx=\int\frac{y^2dz}{z}\\[0.3cm] \int\left(\frac{(x-y)(x+y)}{(x-y)}+\frac{2y^2}{(x-y)}\right)dx=y^2\ln|z|\\[0.3cm] \int\left(x+y+\frac{2y^2}{(x-y)}\right)dx=y^2\ln|z|\\[0.3cm] \boxed{\frac{x^2}{2}+xy+2y^2\cdot\ln|x-y|+c_2=y^2\ln|z|}$$

Then,

$$y=0\rightarrow 0^3+x^3=c_1\rightarrow\boxed{x=c_1^{1/3}}\\[0.3cm] y=0\rightarrow\frac{x^2}{2}+c_2=0\rightarrow\boxed{x=(-2x_2)^{1/2}}\\[0.3cm] c_1^2=(-2c_2)^3\\[0.3cm] \boxed{x^6=\left(-2\cdot\left(y^2\ln|z|-\frac{1}{2}\cdot(x^2-3y^2+2xy+4y^2\ln|x-y|)\right)\right)^3}$$

ANSWER

$$x^6=\left(-2\cdot\left(y^2\ln|z|-\frac{1}{2}\cdot(x^2-3y^2+2xy+4y^2\ln|x-y|)\right)\right)^3$$