When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

 $\displaystyle u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$\displaystyle u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$\displaystyle u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\displaystyle \bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$\displaystyle u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the wave equation is 

$\displaystyle u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$

When dealing with partial differential equations, there are phenomenons in the physical world that have specific equations related to them in the mathematical world.

Looking at the possible answer selections below, identify the physical phenomena each represents.

 $\displaystyle u_t-k(u_{xx}+u_{yy}+u_{zz})=0$ is known as the heat equation.

$\displaystyle u_{tt}-c^2(u_{xx}+u_{yy}+u_{zz})=0$ is known as the wave equation.

$\displaystyle u_{xx}+u_{yy}+u_{zz}=0$ is known as the Laplace equation.

$\displaystyle \bigtriangledown^2u=f(x,y,z)$ is known as the Poisson equation.

$\displaystyle u_{tt}+c^2\bigtriangledown^4u=0$ is known as the biharmonic wave equation.

Therefore, the correct answer for the heat equation is 

$\displaystyle u_t-k(u_{xx}+u_{yy}+u_{zz})=0$

Since the question states to use separation of variables the solution looks as follows.

$\displaystyle u(x,0)=cos(\pi x)$

Let 

$\displaystyle U(x,t)=T(t)X(x)$

therefore the partial differential equation becomes

$\displaystyle \\X(x)T'(t)=T(t)X''(x) \\\\\frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=k$

$\displaystyle k$ is some constant therefore making the ordinary differential equation,

$\displaystyle \\T'(t)-kT(t)=0 \\X''(x)-kX(x)=0$

In this particular case the constant must be negative.

$\displaystyle \\k=-\lambda^2 \\T'(t)+\lambda^2T(t)=0 \\X''(x)+\lambda^2X(x)=0$

Solving for $\displaystyle X$ and $\displaystyle T$ results in the following.

$\displaystyle \\T(t)=Ae^{-\lambda^2t} \\X(x)=B\cos(\lambda x)+C\sin(\lambda x)$

From here solve for the general transient solution

$\displaystyle \\U(x,t)=e^{-\lambda ^2t}(A\cos(\lambda x)+B\sin(\lambda x)) \\U_x(x,t)=e^{-\lambs ^2t}(-A\lambda \sin(\lambda x)+B\lambda \cos(\lambda x))$

Lastly apply the boundary conditions to solve for the constants and in turn solve the general solution.

$\displaystyle \\U_x(0,t)=B\lambda e^{-\lambda ^2t}=0 \\B=0 \\u(x,t)=e^{-\pi^2t}\cos(\pi x)$