Using the method of characteristics to find a general solution to PDE I want to find the general solution to $3U_x-4U_y = x^2$ using the method of characteristics. I'm given the answer which is $U(x,y)=\frac{x^3}{9}+F(3y+4x)$ but I'm having trouble getting to this solution.
Here is my attempt so far:
$$
x=x(t,s)
$$
$$
y=y(t,s)
$$
Using the chain rule:
$$
\frac{dU}{dt} = U_x\frac{d x}{d t}+U_y\frac{d y}{d t}
$$
So $\frac{dU}{dt}=x^2$ if: 
$$
\frac{d x}{d t}=3\implies x(t,s)=3t+c_x(s)
$$
and
$$
\frac{d y}{d t}=-4\implies y(t,s)=-4t+c_y(s)
$$
Now force $(x,y)$ to match $(t,s)$ at $x = t=0$:
$$
x(t,s) \vert _{t=0}=0 \implies c_x(s)=0
$$
$$
y(t,s) \vert _{t=0}=s \implies c_y(s)=s
$$
so
$$
x(t,s)=3t\implies t=\frac{x}{3}
$$
$$
y(t,s)=-4t+s\implies s=y+\frac{4x}{3}
$$
Now $\frac{dU}{dt}=x^2=9t^2$ and solving for $U$ reveals:
$$
U=3t^3+F(s)
$$
Since $U(t,s) \vert _{t=0}=U(0,s)=F(s)$. Substitute for $t$ and $s$:
$$
U(x,y)=\frac{x^3}{9}+F(y+\frac{4x}{3})
$$
This is close to the correct answer but the function $F(y+\frac{4x}{3})$ should be $F(3y+4x)$.
Thanks for any help you can give!
Edit: Fixed the solution a little but it still seems I'm missing a factor of 3 in the function $F$.
Can I simply multiply everything in the function $F$ by 3 without changing the solution, and if so, is it because this is just a general solution (no initial conditions provided) or is there another reason?
 A: I used the method of characteristics to solve this problem, but in a slightly different way.  Your pde is invariant to a Lie stretching group: 
$$
G(x,y,u)=(\lambda x, \lambda^\beta y, \lambda^\alpha u), \lambda_o=1
$$In other words,
$$
\lambda^\alpha u(x,y)=u(\lambda x, \lambda^\beta y)
$$If we take partial derivatives of both sides of this equation with respect to $\lambda$ and set $\lambda = \lambda_o =1$, we get
$$
\alpha u=xu_x+\beta yu_y
$$NOW use the method of characteristics to find the characteristic equation.
$$
\frac{du}{\alpha u}=\frac{dx}{x}=\frac{dy}{\beta y}
$$
This equation has two independent integrals which are constants of integration:  $\frac{u}{x^\alpha}$ and $\frac{y}{x^\beta}$  The most general solution for your pde is to take one invariant and set it equal to a function of the other.  (Sophus Lie referred to these as differential invariants, but by modern standards we know them as group stabilizers for the Lie group.  This becomes evident when you put the invariants through the group transformation:  they remain the same.)
$$
\frac{u}{x^\alpha}=f\bigg(\frac{y}{x^\beta}\bigg)
$$or
$$
u=x^\alpha f\bigg(\frac{y}{x^\beta}\bigg)
$$
Now we look at the specific pde, and note that
$$
u'_{x'}=\lambda^{\alpha-1}u_x
$$and
$$
u'_{y'}=\lambda^{\alpha -\beta}u_y
$$and
$$
(x')^2=\lambda^2 x^2
$$
For invariance, the $\lambda$ coefficients must cancel out, which means that
$$
\alpha -1=\alpha - \beta =2
$$which implies $\beta =1$ and $\alpha =3$.
This means that the most general solution to our pde is
$$
u=x^3f(\mu),  \mu=\frac{y}{x}
$$Taking partial derivatives,
$$
u_x=3x^2f+x^3f_\mu \mu_x=3x^2-x^2\mu f_\mu
$$
$$
u_y=x^3f_\mu \mu_y=x^2f_\mu
$$Substituting these back into the pde and canceling out the $x^2$ terms, we have
$$
9f-3\mu f_\mu -4 f_\mu=1
$$which rearranges to form a separable equation:
$$
\frac{d\mu}{3\mu+4}=\frac{df}{9f-1}
$$Letting $m=3\mu+4$ and $n=9f-1$,
$$
\frac{1}{3}\frac{dm}{m}=\frac{1}{9}\frac{dn}{n}\rightarrow 3\frac{dm}{m}=\frac{dn}{n}
$$This integrates to
$$
ln(m^3)=ln(n)+ln(A)\rightarrow m^3=An \rightarrow (3\mu+4)^3=A(9f-1)
$$Therefore,
$$
f=\frac{A(3\mu+4)^3+1}{9}=C(3\mu+4)^3+\frac{1}{9}
$$where $C=\frac{A}{9}$.  Since $u=x^3f$ and $\mu=\frac{y}{x}$, the most general solution to your pde is
$$
u=C(3y+4x)^3+\frac{x^3}{9}
$$So your function $F(3y+4x)$ is $C(3y+4x)^3$.  
