Solve the initial value problem $u_{xx}+2u_{xy}-3u_{yy}=0,\ u(x,0)=\sin{x},\ u_{y}(x,0)=x$ 
Solve the partial differential equation $$u_{xx}+2u_{xy}-3u_{yy}=0$$ subjet to the initial conditions $u(x,0)=\sin{x}$, $u_{y}(x,0)=x$.

What I have done
$$
3\left(\frac{dx}{dy}\right)^2+2\frac{dx}{dy}-1=0
$$
implies 
$$\frac{dx}{dy}=-1,\frac{dx}{dy}=\frac{1}{3}
$$
and so
$$
x+y=c_{1},\ 3x-y=c_{2}.
$$
Let $ \xi=x+y$, $\eta=3x-y$. Then
\begin{align}
u_{xx}&=u_{\xi\xi}+6u_{\xi\eta}+9u_{\eta\eta} \\ u_{yy}&=u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta} \\
u_{xy}&=u_{\xi\xi}-2u_{\xi\eta}-3u_{\eta\eta}.
\end{align}
Applying substitutions,
$$
u_{\xi\eta}=0.
$$ 
Thus,
\begin{align}
u(\xi,\eta)&=\varphi(\xi)+\psi(\eta) \\
u(x,y)&=\varphi(x+y)+\psi(3x-y).
\end{align}
Applying the initial value condition,
\begin{align}
u(x,0)&=\varphi(x)+\psi(3x)=\sin{x} \\
u_{y}(x,0)&=\varphi'(x)-\psi'(3x)=x
\end{align}
Therefore,
\begin{align}
\varphi(x)&= \frac{1}{2} \left(\sin{x}+\int_{x_{0}}^{x} \tau \, d\tau \right)+\frac k2 \\
ψ(3x)&=\frac{1}{2} \left(\sin{x}-\int_{x_{0}}^{x}\tau \, d\tau \right)-\frac k2.
\end{align}
I have no idea how to get $ψ(x)$. Does anyone could help me to continue doing this question? Thanks very much!
 A: After getting your general solution of
$$
u(x, y) = \phi(x + y) + \psi(3x - y)
$$
We can notice the following:
$$
\phi(x) + \psi(3x) = \sin x \\
\phi'(x) - \psi'(3x) = x
$$
Now we can differentiate our first equation to get
$$
\phi'(x) + 3\psi'(3x) = \cos x
$$
Do you see where to go from here? (try adding three of the second equation to the first equation)

I misunderstood what the issue was, leaving the above for the sake of the community. As for having $\psi(3x)$ and wanting $\psi(x)$ try putting $3x = y$ and then you'll get an expression for $\psi(y)$ which you can then rewrite as $\psi(x)$.
A: We use Laplace transform method and free CAS Maxima
http://maxima.sourceforge.net/

Answer:
$$u=\frac{\sin(x+y)}{4}+\frac{y^2}{3}+xy+\frac34\sin\left(x-\frac{y}{3}\right)$$
2 method


*

*$D_x^2+2D_xD_y-3D_x^2=(D_x-D_y)(D_x+3D_y)$

*General solution of $u_x-u_y=0$ is $u_1=f(x+y)$

*General solution of $u_x+3u_y=0$ is $u_2=g\left(x-\frac{y}{3}\right)$

*$u=u_1+u_2=f(x+y)+g\left(x-\frac{y}{3}\right)$

*From initial conditions we get
$$f(x)+g(x)=\sin(x),\\f'(x)-\frac13g'(x)=x$$

*After the integration of the second equation we get
$$f(x)+g(x)=\sin(x),\\f(x)-\frac13g(x)=\frac{x^2}{2}+c$$

*Then
$$f(x)=\frac{\sin{(x)}}{4}+\frac{3 {{x}^{2}}}{8}+\frac{3 c}{4},\\
g(x)=\frac{3 \sin{(x)}}{4}-\frac{3 {{x}^{2}}}{8}-\frac{3 c}{4}
$$

*$$u=f(x+y)+g\left(x-\frac{y}{3}\right)\\=\frac{\sin{\left( x+y\right) }}{4}+\frac{3 {{\left( x+y\right) }^{2}}}{8}-\frac{3 {{\left( x-\frac{y}{3}\right) }^{2}}}{8}-\frac{3 \sin{\left( \frac{y}{3}-x\right) }}{4}\\=
\frac{\sin(x+y)}{4}+\frac{y^2}{3}+xy+\frac34\sin\left(x-\frac{y}{3}\right)
$$

