using separation of variables to solve $u_{tt}=16u_{xx}, u(x,t)$

Question

Question:

Find a nontrivial family of solutions of the following PDEs by the method of separation of variables. You need not find the most general solution obtainable in this way.

4c. $u_{tt}=16u_{xx}, u(x,t)$

My Attempt: I am using separation of variables to find a nontrivial family of solution for this particular PDE.

First, I need my product solutions so I need to let $u=u(x,t)$ be $ u =f(x)g(t)$. Subsituting this condition on my PDE, I get

$f(x)g''(y)=16f''(x)$

So I am going to divide by $f(x)$ to bring my f's on one side

$g''(y) = \frac{16f''(x)}{f(x)}$

I split these two up and end up with the following equations

$g''(y) = \lambda$ and $ \frac{16f''(x)}{f(x)}=\lambda$

while my $g''(y)$ stays the same I will have $16f''(x)=f(x) \lambda$ which becomes $16f''(x)-f(x) \lambda = 0$

So my equations are $g''(y) = \lambda$ and $16f''(x)-f(x) \lambda = 0$. The main question is on how to solve them. I think if I apply the ODE technique of characteristic equations I would have $r^2 = \lambda$ and by taking the square root I will have $r = \lambda$ and $-\lambda$. Now what would happen to $16f''(x)-f(x) \lambda = 0$?

Would it become $16r^2 - \lambda =0$

$16r^2 = \lambda$ ? But wait what is going to happen to my 16?

The answer is $(c_1e^{\lambda x}+c_2e^{-\lambda x})(d_1e^{4 \lambda t}+d_2e^{-4 \lambda t})$. I can see the $ \lambda $ and $- \lambda$ and I know that there is going to be a 4 in play if I take the square root of 16, but I'm not sure if that is right

Edit: Let me try this again.

4c. $u_{tt}=16u_{xx}, u(x,t)$

for $ u(x,t) = f(x)g(t)$ for product solutions, so I have to substitute them into my equation

$g''(t)f(x)=16f''(x)g(t)$

dividing with f(x) I will get $g''(t)=16 \frac{f''(x)}{f(x)} g(t)$

and dividing with $g(t)$ I will have

$\frac{g''(t)}{g(t)} = 16 \frac{f''(x)}{f(x)}$

so splitting these equations up I should get

$\frac{g''(t)}{g(t)} = \lambda$

$ g''(t) = \lambda g(t)$

$g''(t) - \lambda g(t) = 0$

$r^2 - \lambda = 0$

$r^2 = \lambda$

$ r = + \lambda$ and $ r = - \lambda$

Now for $16 \frac{f''(x)}{f(x)} = \lambda$

$ 16 f''(x) = f(x) \lambda$

$16f''(x) -f(x) \lambda = 0$

$16r^2 - \lambda = 0$

$ 16r^2 = \lambda$

and I am still messed up unless I divide the 16

$ r^2 = \frac{\lambda}{16}$

but then I will have $r = \frac{\sqrt{\lambda}}{4}$ and $ r = -\frac{\sqrt{\lambda}}{4}$