How do I solve the partial differential equation $u_{xx}=4u_{tt}$? This is basically the first partial differential equation I am solving by myself (we just started PDE's) and I need some help.
Here is the question:

Determine the solution of $$u_{xx}=4u_{tt}$$ where $t>0$ and $x
> \in(0,1)$. The inital conditions are: 
$u(x,0)=2+\sin{2\pi x} \\ u_t(x,0)=-x^2-5x+3 \\ u(0,t)=2+3t \\
 u(1,t)=2-3t$

I basically followed the "guide" in this video by Dr. Chris Tisdell: LINK
Here is what I have done so far:
Assume a solution: $u(x,t)=X(x)T(t)$
$$ \implies\begin{aligned} \ &  u_x(x,t)=X'(x)T(t) \\ &  u_t(x,t)=X(x)T'(t) \\ & u_{xx}(x,t)=X''(x)T(t) \\ & u_{tt}(x,t)=X(x)T''(t)\end{aligned}$$
Substituting into the original PDE:
$$\begin{aligned} \ & \implies 4X(x)T''(t)=X''(x)T(t) \\ & \iff \frac{4T''(t)}{T(t)}=\frac{X''(x)}{X(x)}\end{aligned}$$
Because the LHS and RHS only depend on $t$ and $x$ respectively, these terms must be constant:
$$\implies \frac{4T''(t)}{T(t)}=\frac{X''(x)}{X(x)}=\gamma=const.$$
The PDE can now be converted to two second order ODE's:
$$\implies \begin{aligned}\ & 4T''-\gamma T=0 \\ & X''-\gamma X=0 \end{aligned}$$
Solving the second equation first: $X''-\gamma X=0$. The characteristic polynomial is $\lambda^2=\gamma$
$$\implies   X(x) =
\begin{cases}
Ax+b,  & \text{if $\gamma=0$, roots are real and equal} \\
Ae^{\sqrt{\gamma}x}+Be^{\sqrt{\gamma}x}, & \text{if $\gamma>0$, two real distinct roots} \\
A \cos{(\sqrt{-\gamma }x)}+B \sin{(\sqrt{-\gamma}x)} , & \text{if $\gamma<0$, complex roots}
\end{cases}$$
Now I want to plug in and examine some of the boundry conditions:
$$\begin{aligned} \ & u(0,t)=2+3t=X(0)T(t) \\ & u(1,t)=2-3t=X(1)T(t)\end{aligned}$$
In the video I linked above, the boundry conditions were equal to zero which made it possible to conclude that $X(0)=0$ and $X(\pi)=0$. But here my boundry conditions equal to a function. What do I do here?
I am interested in any hints, tips, full solutions or references that might help me solve this. Thanks!
This is a continuation of my answer:
As suggested by an answer I used the substitution: $$v(x,t)=u(x,t)-(2+3t)(1-x)-(2-3t)x$$
Plugged into the original PDE, I get:
$$v_{xx}=4v_{tt}$$
I continued with my separation of variables approach and tried to find a solution for the second order ordinary differential equation: $$X''(x)-\gamma X=0$$
Using my new boundry conditions and looking at the cases I get:
Case 1: $X(x)=Ax+B \implies A=0, B=0 \implies \text{ignore because trivial solution}$
Case 2: $X(x)=Ae^{\sqrt{\gamma}x}+Be^{\sqrt{\gamma}x} \implies A=0, B=0 \implies \text{ignore because trivial solution}$
Case 3: $X(x)=A \cos{(\sqrt{-\gamma }x)}+B \sin{(\sqrt{-\gamma}x)} \implies A=0, B\sin{\sqrt{-\gamma}}=0$
So this case will be a solution for all $\sqrt{-\gamma}=n\pi$
Is my calculation correct?
 A: Hint: Consider $$v(x,t) := u(x,t) -(2+3t)(1-x)-(2-3t)x$$
Note that 
$$v(0,t) = u(0,t) -(2+3t)(1-0)-(2-3t)(0)=(2+3t)-(2+3t) = 0$$
$$v(1,t) = u(1,t) -(2+3t)(1-1)-(2-3t)(1)=(2-3t)-(2-3t) = 0$$
So we have zero boundary conditions for $v(x,t)$. Coming up with this function $v$ includes some trial and error. This is not the only candidate for $v$, however it is probably the simplest. 
1) I guessed it by knowing that the end goal is to arrive at 0 boundary conditions. The $(1-x)$ and $(x)$ terms are strategically placed to eliminate an entire bracket when $x = 1$ and $0$ respectively. 
2) This is allowed as we are now going to solve a new PDE in $v$. To see what PDE $v$ satisfies, we need to substitute $u = v - r(x,t)$ into the original PDE, where $r(x,t) = -(2+3t)(1-x)-(2-3t)x$.
To continue:
We substitute $ u = v - r(x,t)$ into the original PDE where $r$ is described above. 
$$(v - r)_{tt}=4(v-r)_{xx}\Rightarrow v_{tt}-r_{tt} = 4v_{xx}-4r_{xx}$$
Note that 
$$r_{tt} = 0,r_{xx} = 0$$
So we have
$$v_{tt} = 4v_{xx}$$
And the new boundary conditions are
$$v(0,t) = 0, v(1,t) = 0$$
And the initial conditions are
$$v(x,0) = u(x,0)-r(x,0)$$
$$ v_t(x,0) = u_t(x,0)-r_t(x,0)$$
Now $u(x,0)-r(x,0)$ and $u_t(x,0)-r_t(x,0)$ are known and I will leave that to you to calculate. 
Now you can solve the PDE for $v$ like you would normally as you have the $0$ boundary conditions that you desired. When you have the solution for $v$ you can very easily get the solution for $u$ since
$$u(x,t) = v(x,t)-r(x,t)$$
Some remarks on your answer:
In case 3 you must replace $\sin$ and $\cos$ with $\sinh$ and $\cosh$ respectively. This will lead you to the conclusion that the only solution is $A,B = 0$
In case 2 you should have $$ X = A\cos(\sqrt{\gamma}x)+B\sin(\sqrt{\gamma}x)$$ 
We see that $A = 0$ but $B\neq 0$ when $\sqrt{\gamma}\cdot 1 = 2\pi n$ $$\Rightarrow \gamma = (2\pi n)^2, n = 1,2,3,...$$ 
So we have the eigenfunction is, $\sin({\sqrt{\gamma}x})$.
