Non-linear Cauchy problem for $xu_x+uyu_y=-xy$ 
Consider the partial differential equation $$x\frac{\partial u}{\partial x}+uy\frac{\partial u}{\partial y}=-xy$$ for $x>0$ subject
  to $u=5$ on $xy=1$. Then
$(a)$ $u(x,y)$ exists when $xy\leq 19$ and $u(x,y)=u(y,x)$ for
  $x>0,y>0$.
$(b)$ $u(x,y)$ exists when $xy\geq 19$ and $u(x,y)=u(y,x)$ for
  $x>0,y>0$.
$(c)$ $u(1,11)=3$ and $u(13,-1)=7$.
$(d)$ $u(1,-1)=5$ and $u(11,1)=-5$.

I attempted the method of characteristics and obtained the equations as $$\frac{dx}{x}=\frac{dy}{uy}=\frac{du}{-xy}$$ Afterwards I took first, third and second, third ratio respectively and then adding them together to find one characteristic as $\displaystyle{(u+1)^2+2xy=c_1}$, $c_1$ being some arbitrary constant. Now if I use that equation to first and second ratio, it gives $$\frac{dx}{x}=\frac{dy}{y(\sqrt{c_1-2xy}-1)}$$ which is obviously too laborious to solve. I don't think actual solution to this problem would involve that much calculation. What is the best possible approach to this problem? If solution for $u(x,y)$ is the only way, what is the fault at my attempt above? Thanks in advance.
 A: I agree with your result : 
$$(u+1)^2+2xy=c_1$$
is a first characteristic equation.
Then one is tempted to look for a second characteristic equation in order to get the general solution.
But in the present case we don't need the general solution since the first characteristic equation alone can be fitted to the specified boundary condition :
$u=5$ on the curve $xy=1$, with $(5+1)^2+2(1)=c_1=38$.
Thus the particular solution fitting to the boundary condition is a curve lying on the particular characteristic curve whose implicit equation is :
$$(u+1)^2+2xy=38$$
The solution on explicit form is :
$$\boxed{u(x,y)=-1+\sqrt{38-2xy}}$$
Note that the part $u=-1-\sqrt{38-2xy}$ of the characteristic curve is rejected because the boundary condition wouldn't be satisfied on it.
With this explicit solution it is easy to find which proposed options are exact or not.
$$ $$
In addition, COMPLETELY SOLVING the PDE :
A second characteristic equation comes from 
$\frac{dx}{x}=\frac{du}{-xy}= \frac{2du}{(u-1)^2-c_1}$
$\ln|x|+\int\frac{2du}{c_1-(u-1)^2}=c_2$
$\ln|x|-\ln|c_1-(u+1)^2|-\frac{2}{\sqrt{c1}}\tanh^{-1}\left(\frac{u+1}{\sqrt{c_1}} \right)=c_2$
$\ln|x|-\ln|2xy|-\frac{2}{\sqrt{(u+1)^2+2xy}}\tanh^{-1}\left(\frac{u+1}{\sqrt{(u+1)^2+2xy}} \right)=c_2$
$-\ln|y|-\frac{2}{\sqrt{(u+1)^2+2xy}}\tanh^{-1}\left(\frac{u+1}{\sqrt{(u+1)^2+2xy}} \right)=c_2+\ln(2)=c'_2$
Genertal solution on the form of implicit equation $c_1=F(c'_2)$ :
$$\boxed{(u+1)^2+2xy=F\left(-\ln|y|-\frac{2}{\sqrt{(u+1)^2+2xy}}\tanh^{-1}\left(\frac{u+1}{\sqrt{(u+1)^2+2xy}} \right) \right)}$$
$F$ is an arbitrary fonction to be determined according to the boundary condition.
Condition : $u=5$ on the curve $xy=1$ :
$38=F\left(-\ln|y|-\frac{2}{\sqrt{38}}\tanh^{-1}\left(\frac{6}{\sqrt{38}} \right) \right)$
Due to $\ln|y|$ it is impossible that the function $(\ln|y|+C)$ be constant, except if the function itself is a constant function $F=38$
Now the function $F$ is determined and we see that the particular solution of the PDE which satisfies the boundary condition is the function already found above. 
A: The Lagrange's auxiliary equations for the given PDE are $$\frac{dx}{x}=\frac{dy}{uy}=\frac{du}{-xy}\tag1$$
Choosing $~y,~x,~(1+u)~$ as multipliers , each fraction of $(1)$
$$=\dfrac{ydx+xdy+(1+u)du}{xy+uxy-xy(1+u)}=\dfrac{ydx+xdy+(1+u)du}{0}$$
So $$ydx+xdy+(1+u)du=0\implies (1+u)^2+2xy=c$$where $~c~$ is constant.
Given that $~u=5\quad \text{on}\quad xy=1~$, so $~c=36+2=38$.
Therefore $$(1+u)^2+2xy=38\implies u=\sqrt{38-2xy~}-1$$



*

*$u(x,y)~$exists if $$~38-2xy\ge 0\implies x\le 19$$
Also here $$u(x,y)=\sqrt{38-2xy~}-1=\sqrt{38-2yx~}-1=u(y,x)\qquad\text{for$~~x>0,y>0$}$$So we can conclude that option $(a)$ is correct while option $(b)$ is not.



*

*$$u(1,11)=\sqrt{38-22~}-1=3$$
$$u(13,-1)=\sqrt{38-26~}-1=7$$
so option $(c)$ is correct.



*

*$$u(1,-1)=\sqrt{38+2~}-1=\sqrt{40}-1\ne 5$$
$$u(10,1)=\sqrt{38-22~}-1=3\ne 5$$
so option $(d)$ is not correct.

