Solving coupled second order ODEs via Laplace transforms & Function theory. I have used Laplace transforms to transform a system of 2 coupled second order ODEs into 2 simultaneous equations.
1st ode:
$$\frac{3d^2y}{dt^2}+\frac{dy}{dx}=0$$
2nd ode:
$$\frac{5d^2y}{dx^2}-\frac{dy}{dt}=0$$
y is a function of two variables $y(x,t)$
I define $L_t$ to be the Laplace transform wrt to $t$. 
I now have an equation for ${Y_t}({s_t})$ in terms of $s_x$ and $s_t$.
1st question (now solved) is how I work out the inverse Laplace transform wrt to t $L^{-1}_t$ of functions like $s_x$ ?
2nd question (unanswered): If y is a function of x and t, does this mean that x is a function of t?
3rd question (unanswered / not clear): Why are the above differential equations PDEs not ODEs?
Updates since comments:
I made a mistake on my working (I took transforms of x also when it should have been t) - I have now got $Y_t$ in terms of $s_t$ only, and can hence solve.
 A: Let $y(x)=z(x)$ (otherwise, after the Laplace transform they look like the same functions):
$$
\begin{cases}
3\cdot\frac{\text{d}^2y(t)}{\text{d}t^2}+\frac{\text{d}z(x)}{\text{d}x}=0\\
5\cdot\frac{\text{d}^2z(x)}{\text{d}x^2}-\frac{\text{d}y(t)}{\text{d}t}=0
\end{cases}
$$
Take the Laplace transform of both sides:
-First ODE:
$$3\cdot\left(s^2\text{Y}(s)-sy(0)-y'(0)\right)+\left(s\text{Z}(s)-z(0)\right)=0$$
 -Second ODE:
$$5\cdot\left(s^2\text{Z}(s)-sz(0)-z'(0)\right)-\left(s\text{Y}(s)-y(0)\right)=0$$
Now, simplify and factor out:
-First ODE:
$$\text{Y}(s)=\frac{3sy(0)+3y'(0)+z(0)-s\text{Z}(s)}{3s^2}$$
 -Second ODE:
$$\text{Z}(s)=\frac{5sz(0)+5z'(0)+s\text{Y}(s)-y(0)}{5s^2}$$
Now, using substitution:


*

*For $\text{Y}(s)$:
$$\text{Y}(s)=\frac{y(0)+15s^2y(0)+15sy'(0)-5z'(0)}{s+15s^3}$$

*For $\text{Z}(s)$:
$$\text{Z}(s)=\frac{z(0)+15s^2z(0)+3y'(0)+15sz'(0)}{s+15s^3}$$
Now, with Inverse Laplace transform:


*

*For $y(t)$:
$$y(t)=y(0)+5z'(0)\left(\cos\left(\frac{t}{\sqrt{15}}\right)-1\right)+y'(0)\sqrt{15}\sin\left(\frac{t}{\sqrt{15}}\right)$$

*For $z(x)$:
$$z(x)=z(0)-3y'(0)\left(\cos\left(\frac{x}{\sqrt{15}}\right)-1\right)+z'(0)\sqrt{15}\sin\left(\frac{x}{\sqrt{15}}\right)$$
