PDE Lagrange method - $xuU_x + yuU_y = -xy$

Question

I'm not sure where I went wrong with this one:

the PDE: $$xuU_x + yuU_y = -xy$$

my try: I Wrote the characteristic lines equation: $$ \frac{dx}{xu} = \frac{dy}{yu} = \frac{du}{-xy}$$ eq (1): $$ \frac{dx}{xu} = \frac{du}{-xy} \Longrightarrow \ \frac{dx(-xy)}{x} = udu$$ solving this I got the first surface $\phi_1 = u^2 +2xy$

eq (2): $$ \frac{dx}{xu} = \frac{dy}{yu} \Longrightarrow \ln|x| + C = \ln|y| $$ which gave me the second surface: $$ \phi_2 = \frac{y}{x} $$

according to the solution: $$\phi_1 = (u^2 +xy)\cdot \frac{x}{y} \\ \phi_2 = \frac{x}{y} $$

I don't understand how I got $\phi_{1,2}$ wrong, I'm worried that there is something wrong with my understanding of the theory,

help \ hints would be appreciated, thank you!

Answer 1

$$xuu_x + yuu_y = -xy$$

$$ \frac{dx}{xu} = \frac{dy}{yu} = \frac{du}{-xy}\quad\text{ is OK.}$$ eq (1): $$ \frac{dx}{xu} = \frac{du}{-xy} \Longrightarrow \ \frac{dx(-xy)}{x} = udu\quad\text{ is OK.}$$ First surface $$\phi_1 = u^2 +2xy\quad \text{is FALSE.}$$ Why ?

$$-y\:dx=u\:du \quad\implies\quad -\int y\:dx=\frac12 u^2$$ $$\int y\:dx\neq y\:dx\quad\text{ because } y \text{ is not constant.}$$ So you cannot integrate since $y(x)$ is unknown at this stage.

One have to proceed differently.

$$ y\frac{dx}{yx} = x\frac{dy}{xy}=\frac{ydx+xdy}{yx+xy}=\frac{d(xy)}{2xy} \tag 3$$

$$\frac{d(xy)}{2xy}= \frac{udu}{-xy}$$ $$d(xy)=-2udu$$ $$xy=-u^2+\text{constant}$$ $$\phi_1 = u^2+xy$$

eq (2): $$ \frac{dx}{xu} = \frac{dy}{yu} \Longrightarrow \ln|x| + C = \ln|y| $$ $$ \phi_2 = \frac{y}{x} \quad\text{ is OK.}$$

NOTE:

If you don't understand the above calculus from Eq.$(3)$ there is an alternative method : $$-y\:dx=u\:du $$ $$\phi_2 = \frac{y}{x} \quad\implies\quad y=\phi_2x$$ $$-(\phi_2x)\:dx=u\:du $$ Now you can integrate because $\phi_2=$constant. $$-\frac12\phi_2x^2=\frac12 u^2+\text{constant}$$ $$\phi_1=u^2+\phi_2x^2$$ $$\phi_1=u^2+(\frac{y}{x})x^2$$ $$\phi_1=u^2+xy$$