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!

Solutions Collecting From Web of "Solve the initial value problem $u_{xx}+2u_{xy}-3u_{yy}=0,\ u(x,0)=\sin{x},\ u_{y}(x,0)=x$"

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 $\psi(x)$.