
biharmonic equation
verify that the following is a solution to the biharmonic equation
F(x,y)=(Acosbx+Bsinbc+Cxcosbx+Dxsinbx)(Ecoshby+Fsi nhby)
i have tried taking the derivates and have gotten up to the laplacian but this is a very long and tedious way,and the result up to second order is really long, like a page long,there must be an easier way. Can someone suggest an alternative method to prove this

The biharmonic eqiation is
$\displaystyle u_{xxxx} + 2u_{xxyy} + u_{yyyy} = 0$
Your solution is of the form
$\displaystyle u = A(x)B(y)$
Note that when you substitute into the biharmonic you get
$\displaystyle A^{(4)}(x)B (y)+ 2 A''(x)B''(y) + A(x) B^{(4)}(y) = 0$
With your $\displaystyle B(y) $ you have $\displaystyle B^{(4)}(y) = b^2B''(y) = b^4B(y)$ so what you really want to show is
$\displaystyle A^{(4)}(x)+ 2b^2 A''(x) + b^4A(x) = 0$.
Next, notice that this is
$\displaystyle \left(A'' + b^2A\right)'' + b^2\left(A'' + b^2A \right) = 0$
so compute $\displaystyle C = A'' + b^2A $ first and then show this satisfies $\displaystyle C'' + b^2C = 0$. It might make it a little more manageable.
