Show that u(x,t)=cos(a*pi*x)e^-a^2*pi^2*t is a solution of the heat equation with k=1, on any interval [0,L].

Answer:Step-by-step explanation:We have the function[tex] u(x,t) = \cos(a\pi x)e^{-a^2\pi^2t}[/tex],while the heat equation in one spacial dimension is [tex]\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial^2 x} [/tex].So, to solve this exercise we only need to calculate the derivatives that appears in the equation. Let us start by the derivative with respect to t:[tex]\frac{\partial u}{\partial t} (x,t) = -a^2\pi^2\cos(a\pi x)e^{-a^2\pi^2t}[/tex].In the other hand[tex]\frac{\partial u}{\partial x} (x,t) =-a\pi\sin(a\pi x)e^{-a^2\pi^2t}[/tex],then[tex]\frac{\partial^2 u}{\partial^2 x} (x,t) =-a^2\pi^2\cos(a\pi x)e^{-a^2\pi^2t}[/tex] .Notice that,[tex]-a\pi\sin(a\pi x)e^{-a^2\pi^2t}=-a\pi\sin(a\pi x)e^{-a^2\pi^2t}[/tex]So, the function [tex]u[/tex] satisfies the heat equation with [tex]k=1[/tex] on any interval [0,L].