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 PF Gold P: 3,173 Ok thank you vela! This means that $\frac{\partial u }{\partial y} \big | _{y=0}=0$. Thus the PDE is equivalent to $\frac{\partial ^2 U_c (p,x)}{\partial x^2}-p^2 U_c (p,x)=0$. Since $p>0$, $U_c(p,x)=Ae^{px}+Be^{-px}$. Now I think it's time to take the inverse cosine transform.