Solve this PDE $xu_x+yu_y+xyu_z=0,u(x,y,0)=x^2+y^2$ (problem with general solution)

Question

Find the particular solution of the partia differential equation $$ xu_x+yu_y+xyu_z=0 $$ that satisfies the Cauchy condition $$ u(x,y,0)=x^{2}+y^{2} $$ So: $$ \frac {dx}{x}=\frac {dy}{y} = \frac{dz}{xy} $$ which in turn implies $$ \begin{split} \int \frac{dx}{x} =\int \frac{dy}{y} &\implies \ln(x)=\ln(C_1y) \\ & \implies C_1=\frac{x}{y} \vee \int \frac{dx}{x}=\int \frac{dy}{y} \\ &\implies \ln(C_1x)=y \Rightarrow C_1=\frac{y}{x} \end{split} $$ Now I have a problem: I could proceed by using two different approaches

1. I don't know if this is a good approach: $$ \frac{dz}{xy}=\frac{dx+dy}{x+y} \implies \int dz=\int \frac{xy}{x+y} d(x+y)$$ and what is the solution of the integral on the right side $\to$ Is it possible to integrate over $d(x+y$) on Wolfram Alpha?
   Moreover is it true that $$\int \frac{xy}{x+y} d(x+y) \iff \int \frac{xy}{x+y} dx + \frac{xy}{x+y} dy$$ or not?

2. Maybe this second approach works better $\to$ but I don't know if it's possible: $$ \begin{split} \frac{dz}{xy}=\frac{dx\cdot dy}{x\cdot y} & \implies\int{dz}=\int{\left(\int{1\,dx}\right)dy} \\ & \implies z=xy+C_2y+C_3 \\ & \implies C_3=z-xy-C_2y \end{split} $$ So it's a problem with two constants and Wolfram Alpha gives this general solution for the query x*D[u[x,y,z],x]+y*D[u[x,y,z],y]+x*y*D[u[x,y,z],z]=0 $$ \implies u(x,y,z)=F\left(\frac{y}{x},z-\frac{xy}{2}\right) $$ so from Wolfram integration the result would be like $z=\frac{xy}{2}+C_{2}$

Maybe I should use approach 1, but how to calculate this integral?
I know that $$ \int \frac{d(x+y)}{x+y}=\ln(x+y)+\ln(C_2), $$ but I have $xy$ in numerator $\implies \int \frac{xy}{x+y} d(x+y)$.

I think there could be even a third approach:

3. I can write $$ \begin{split} \frac{dz}{xy}=\frac{dx}{x}+\frac{dy}{y} & \implies \int(dz)=\int(y dx)+\int(x dy) \\ & \implies z=2xy+C_2 \\ & \implies C_{2}=z-2xy \end{split} $$ But it doesn't agree with the Wolfram Alpha solution and I don't think it's mathematically correct to do so.

Please help me.

Answer 1

The Lagrange-Charpit equations for the PDE $xu_x+yu_y+xyu_z=0$ are given by $$ \frac {dx}{x}=\frac {dy}{y} = \frac{dz}{xy}=\frac{du}{0}, \tag{1} $$ the last expression meaning that $u=C_1$. The equation $\frac{dx}{x}=\frac{dy}{y}$ implies $y=C_2x$. Another independent equation is given by $\frac{dz}{xy}=\frac{ydx+xdy}{2xy}=\frac{d(xy)}{2xy}$, which has as solution $z=\frac{1}{2}xy+C_3$. Combining these three solutions we obtain $$ u(x,y,z)=F(C_2,C_3)=F\left(\frac{y}{x},z-\frac{1}{2}xy\right). \tag{2} $$ The function $F$ is determined by the condition $u(x,y,0)=x^2+y^2$, which implies $$ F\left(\frac{y}{x},-\frac{1}{2}xy\right)=x^2+y^2. \tag{3} $$ Let $\frac{y}{x}=t$ and $-\frac{1}{2}xy=s$; then $$ ts=\frac{y}{x}\left(-\frac{1}{2}xy\right)=-\frac{1}{2}y^2 \implies y^2=-2ts \tag{4} $$ and $$ \frac{t}{s}=\frac{y}{x}\left(-\frac{2}{xy}\right)=-\frac{2}{x^2} \implies x^2=-\frac{2s}{t}. \tag{5} $$ Equations $(3)$-$(5)$ imply $F(t,s)=-2s\left(t+\frac{1}{t}\right).$ Combining this result with Eq. $(2)$ we finally obtain $$ u(x,y,z)=(-2z+xy)\left(\frac{y}{x}+\frac{x}{y}\right). \tag{6} $$

Answer 2

The method of characteristics gives the following ODEs:

\begin{eqnarray} x'(t)&=x,\\ y'(t)&=y,\\ z'(t)&=xy,\\ u'(t)&=0 \end{eqnarray} The first two ODEs solve to $x(t)=x_0 e^t$, $y(t)=y_0 e^t$ where $x_0=x(0),y_0(0)$. Therefore the third solves as $$z'(t)=x_0 y_0 e^{2t}\implies \frac12 x_0 y_0(e^{2t}-1)$$ where I have taken $z(0)=0$ for future convenience. The last ODE then yields $$u(x,y,z)=u(x_0,y_0,0)=x_0^2+y_0^2=e^{-2t}(x^2+y^2)$$

It remains to eliminate the $t$-dependence in favor of $x,y,z$. But from the above work we have $$z=\frac12 x_0 y_0 (e^{2t}-1)=\frac12 xy (1-e^{-2t}),$$ so $e^{-2t}=-2z/(x y)$. Hence $$u(x,y,z)=\left(1-\frac{2z}{xy}\right)(x^2+y^2)=(x y-2z)\left(\frac xy+\frac yx\right).$$