Find the general solutions of the PDEs below... 
a)$3u_x-4u_y = x^2$


b)$u_x - 4u_y + u = x + y + 1$

I'm starting in PDEs, I saw two resolution methods, but I think I didn't understand them right.
a) $\frac{dx}{3}= \frac{dy}{-4}=\frac{dz}{x^2}$
$\frac{dx}{3}= \frac{dy}{-4} \to -4x + c = 3y \to c_1= 4x+3y$
$\frac{dy}{-4}=\frac{dz}{x^2} \to x^2y = -4z + c_2 \to c_2 = x^2y + 4z $
$u= \frac{x^2y - f(4x+3)}{-4} $
$\dfrac{dx}{dt}=3 \to x(0)=0 \to x=3t$
$\dfrac{dy}{dt}=-4 \to y(0)=y_0 \to y= -4t +y_0 \to y= \frac{-4x}{3} + y_0$
$\dfrac{du}{dt}=x^2 = (3t)^2 \to u(0)=f(y_0) \to u(x,y)=3t^3 + f(y_0) = x^3+f(\frac{4x} {3}+y)$
b) $\frac{dx}{1}= \frac{dy}{-4}=\frac{dz}{x + y + 1}$
$\frac{dx}{1}= \frac{dy}{-4} \to -4x + c = y \to c_1= 4x+y$
$\frac{dy}{-4}=\frac{dz}{x + y + 1} \to x^2 + y + x = -4z + c_2 \to c_2 = x^2 + y + x  + 4z $
$u= \frac{x^2 + y + x - f(4x+y)}{-4} $
$\dfrac{dx}{dt}=1 \to x(0)=0 \to x=t$
$\dfrac{dy}{dt}=-4 \to y(0)=y_0 \to y= -4t +y_0 \to y= -4x + y_0$
$\dfrac{du}{dt}=x + y + 1 = -3t + 1 \to u(0)=f(y_0) \to u(x,y)=\frac{-3t^2}{2} + t f(y_0) = \frac{-3t^2}{2} + t + f(4x + y)$
Shouldn't the answers be the same? Are any of them correct?
Thanks.
 A: a)
$$3u_x-4u_y=x^2$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{3}=\frac{dy}{-4}=\frac{du}{x^2}$$
A first characteristic equation comes from solving $\frac{dx}{3}=\frac{dy}{-4}$ :
$$4x+3y=c_1$$
A second characteristic equation comes from solving $\frac{dx}{3}=\frac{du}{x^2}$ :
$$u-x^3=c_2$$
The solution of the PDE expressed on implicit form $c_2=F(c_1)$ is :
$$u-x^3=F(4x+3y)$$
$F$ if an arbitrary function.
$$\boxed{u(x,y)=x^3+F(4x+3y)}$$
This is equivalent to
$$u(x,y)=x^3+f\left(\frac{4x}{3}+y\right)$$
$f$ is an arbitrary function. Both arbitrary functions $F$ and $f$ are related : $F(3X)=f(X)$.
Thus you correctly found the solution of problem (a).
b)
$$u_x-4u_y+u=x+y+1$$
$$u_x-4u_y=-u+x+y+1$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{1}=\frac{dy}{-4}=\frac{du}{-u+x+y+1}$$
You forgot $u$ in the denominator. That is a cause of failure of your calculus.
A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{-4}$ :
$$4x+y=c_1$$
A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{-u+x+y+1}$. This cannot be donne rightaway because $x$ and $y$ are related on the caracteristic curves. In the present case $y=c_1-4x$.
$\frac{dx}{1}=\frac{du}{-u+x+(c_1-4x)+1}\quad\implies\quad \frac{dx}{1}=\frac{du}{-u-3x+c_1+1}\quad\implies\quad \frac{du}{dx}=-u-3x+c_1+1.\quad$ Solving it leads to :
$u=c_2e^{-x}-3x+c_1+4$
The solution of the PDE expressed on implicit form $c_2=F(c_1)$ is :
$u=-3x+c_1+4+e^{-x}F(c_1)$
$F$ is an arbitrary function.
$u=-3x+(4x+y)+4+e^{-x}F(4x+y)$
$$\boxed{u(x,y)=x+y+4+e^{-x}F(4x+y)}$$
