Neumann BVP steady state solution

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Homework Statement


U_t = u_{xx} - 4U
u_x (0, t) = 0, u_x (\pi, t) = 1
u(x, 0) = 4cos(4x)

Find a steady state solution to the boundary value problem.


Homework Equations


n/a


The Attempt at a Solution


Well I'm quite comfortable solving dirichlet/ mixed boundary value problems of this form. I was under the impression that you must search for a solution of the form
u(x,t) = w(x,t) + v(x,t) where, w(x, t) = ax^2 + bx + ct Since there cannot be a steady state solution because the boundary value rates of change are not equal, so there cannot be a solution that doesn't change with time.

I have the solution, and it sets U_t = 0 and goes about finding the 'steady state' solution and finding the eigenfunctions, but that seems wrong to me because of the reasons mentioned above. So I'm just wondering where my thinking is going wrong and why you can find a steady state solution to a neumann boundary value problem of this type.
 
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What about separation of variables, transform methods (Laplace transform in t perhaps) or even Greens functions.
 
Separation of variables along with eigenfunction expansion are the only methods i have learned for solving PDEs as of yet, so maybe I just haven't learned the proper method for this question.

The solution I have uses separation of variables, but I'm just questioning why you can use separation of variables to find a steady state solution, when it seems to me that one shouldn't exist due to the boundary conditions.
 
I am pretty sure that taking the Laplace transform in t will solve your problem nicely
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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