General solution to PDE 
Consider the equation $$xu_x+(1+y)u_y=x(1+y)+xu.$$
Find the general solution.
Now assume an initial condition for the form $u(x,6x-1)=\phi(x).$ Find a necessary and sufficient condition for $\phi$ that guranantees the existence of a solution to the problem. Solve the problem for the appropriate $\phi$ that you found.

I am not sure how to solve this problem? In particular, I am having difficulty understanding and applying the concept of finding independent integrable identities that, when integrated, do not involve the parameter $t$.
 A: Your PDE is linear and hence can be solved applying the method of characteristics, which for your particular case reads:
$$ \frac{\mathrm{d} x }{x} = \frac{\mathrm{d} y}{1+y} = \frac{\mathrm{d} u}{x(1+y) + xu} = -\frac{\mathrm{d} p}{blabla} = -\frac{\mathrm{d} q }{blabla}, \tag{1} $$
where $blabla$ are things I'm not interested in. Note that from $1$st and $2$nd fractions we can write:
$$\log{x} = \log{(1+y)} + k \implies x = C \, (1+y), \quad C \in \mathbb{R}, \tag{2}$$
which gives us a specific relation between the independent variables called characteristic curves or simply characteristics. Note now that we can write $1$st and $3$rd or $2$nd and $3$rd as a single equation involving $x$ and $u$ or $y$ and $u$, respectively, by using the fixed relation given by $(2)$. Indeed, we end up with:
$$ \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}(C x^2  + xu) , \tag{3}  $$ which is a 1st order linear ODE for $u= u(x)$ which can be easily solved up to a constant of integration. Put this constant of integration, say $B$, as a function of $C = x/(1+y)$ to end up with the general integral of the solution of the form (if I made no mistakes):

$$ \color{blue}{ u = C(x,y) e^x \int e^{-x}x\, \mathrm{d}x + B(C) e^x } \tag{4}$$

I'm sure you can take it from here.
Cheers!
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1+y$ , letting $y(0)=0$ , we have $y=e^t-1$
$\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t=x_0(y+1)$
$\dfrac{du}{dt}=x(1+y)+xu=x_0e^{2t}+x_0e^tu$ , we have $u(x,y)=f(x_0)e^{x_0e^t}-e^t-\dfrac{1}{x_0}=f\left(\dfrac{x}{y+1}\right)e^x-y-1-\dfrac{y+1}{x}$
$u(x,6x-1)=\phi(x)$ :
$f\left(\dfrac{1}{6}\right)e^x-6x-6=\phi(x)$
$\therefore$ The PDE have solutions only when $\phi(x)=ke^x-6x-6$
A: Using Maple I am obtaining
$$u \left( x,y \right) =-1-y+{{\rm e}^{x}}{\it F} \left( {\frac {1+y}
{x}} \right) +{\frac {-1-y}{x}}
$$
Applying the mentioned condition we obtain
$$-6-6\,x+{{\rm e}^{x}}{\it F} \left( 6 \right) =\phi \left( x
 \right)$$
Do you agree?
