Hi, I'm unsure where the c has come from in your decoupled equations so starting with

and following with then substituting into the first equation

and as both sides are independant of each other can be written equal to a constant, say. This could be any constant, but the square allows for neatness in a minute, the negative allows for a solution in the cosine / sine form rather than hyperbolic solutions.

You would then get a solution of the form

As I said I don't quite follow where the c comes from but assuming it's important I'm guessingyour lecture notes lead to something similar to

?

Regardless the G function can be solved similarly but it's t dependant and t is unimportant so leave it as G.

Boundary condition at x=0 leads to B=0

Then differentiate u wrt x for the second BC

This equals zero for x=L, either G=0 which would be trivial so

k can't be equal to zero so it is set so the sine function is zero for all L,

i.e.

Then

and (I think) you have the answer assuming you have instead of my k?