Hey Guys, I am stuck on the following problem:

Show that U(x,t) = e^(lambda^2*alpha^2*t)*[A*sin(lambda*x) + B*cos(lambda*x) satisfied the PDE Ut = alpha^2 * Uxx for arbitrary A,B and lambda.

Here's what I have so far, but I don't know if it's right:

1.) General eqn to soln: U(x,t) = Summation * Bn * Rn * Tn
2.) (R*T)t = alpha^2 *(RT)xx
(R*T')/RT = (alpha^2 * Rxx * T)/RT
T/T' = (alpha^2*R'')/R = -lambda^2 (just some constant)
T' = - lambda^2*T = T(t) = e^(-lambda^2*t)
R'' = (-lambda^2 * R) / (alpha^2)
R(x) = A*cos((lambda*x)/alpha) + B*sin((lambda*x)/alpha)

3.) Plug in R(x) and T(x) into general solution
U(x,t) = e^-lambda^2*t * (A*cos((lambda*x)/alpha) + B*sin
((lambda*x)/alpha))

I know that this is not the final solution, and maybe it's even the wrong approach, but I don't know what to do next. Any and all help would be greatly appreciated. Thanks for the help!

Your are going over and above what is needed.

They GAVE you the solution. Just check that it is a solution.

Take one derivative with respect to t and two with respect to x and verify.

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Awesome! Thanks a lot!