How do I solve a certain characteristic system?

I am studying PDEs and have the following (seemingly simple) problem:

Find a surface that passes through the curve $$x^2+y^2=z=1$$ and is orthogonal to the family of surfaces $$z(x+y)=c(3z+1)\qquad(c\in\Bbb R)$$

After writing down the orthogonality condition (assuming my calculations are correct $(*)$), this yields the following equation: $$u(3u+1)(u_x+u_y)-x-y=0$$

We usually solve such equations by using the method of characteristics, which tells us (using assumption $(*)$ again) to solve the following characteristic system:

\begin{align}\dot x=&u(3u+1)\\\dot y=&u(3u+1)\\\dot u =&x+y\end{align}

Differentiating the last equation of this system with respect to $t$ gives us $\ddot u=\dot x+\dot y$, which using the first two equations gives us $$\ddot u = 2u(3u+1)$$

After staring at this equation for some time, I decided to ask Wolfram|Alpha. The result seems pretty ugly, so the following questions arise:

Did I make a mistake/am I missing something? Is my approach correct? How do I proceed?

Thanks.

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Dejan, since you're studying PDEs too, you may wanna be interested in this question I posted: math.stackexchange.com/questions/154515/… – Weltschmerz Jun 8 '12 at 7:00

The intersection of the surfaces would be the set of points $(x,y,z)$ such that \begin{cases}u(x,y)-z=0\\z(x+y)-3cz-c=0,\end{cases} that is, those that lie both on the solution surface and the given surface (determined by $c$, and this must hold for every $c$).
The surfaces are orthogonal means (I'm guessing) their respective gradients at the intersection points must be orthogonal. The gradient of the solution surface is $(u_x,u_y,-1)$, and the gradient of the given surface(s) is $(z,z,x+y-3c)$. So the condition is $zu_x+zu_y-x-y+3c=0$. Since this occurs when $z=u$ we can write it as $$uu_x+uu_y=x+y-3c.$$ We can clear $c$ from the intersection equations: $c=\frac{(x+y)u}{1+3u}$, and plugging it in the last equation gives us $$uu_x+uu_y=(x+y)(\frac{1}{1+3u}).$$