Answer to Question #286892 in Differential Equations for Mary

Question

Solve, using the method of separation of variables, the PDE

∂u

∂t +

∂u

∂x + 2e

tu = 0

Answer 1

Since, given equation is unclear, we assume the given equation to solve is $\dfrac{∂u}{∂x} =2\dfrac{∂u}{∂t}+u=0 ; u(x,0)=6e^{-3x}$

Solution:

Given, $\dfrac{∂u}{∂x} =2\dfrac{∂u}{∂t}+u=0$ ...(i)

Let the solution be $u(x,t)=X(x).T(t)$ ...(ii)

Using (ii) in (i).

$X'T=2XT'+XT \\ \Rightarrow (X'-X)T=2XT' \\ \Rightarrow \dfrac{X'-X}X=\dfrac{2T'}T=k\ \ (say)$

Now, solving $\dfrac{X'-X}X=k$

$\Rightarrow \dfrac{X'}X=k+1$

On integrating,

$\log X=(k+1)x+\log C_1 \\ \Rightarrow \log X=(k+1)x\log e+\log C_1 \\ \Rightarrow \log X=\log e^{(k+1)x}+\log C_1 \\ \Rightarrow X=C_1e^{(k+1)x}$

On solving $\dfrac{2T'}T=k$

$\Rightarrow \dfrac{T'}T=\dfrac k2$

On integrating,

$\log T=\dfrac k2t+\log C_2 \\\Rightarrow T=C_2e^{kt/2} \\ \therefore u(x,t)=X.T=C_1e^{(k+1)x}.C_2e^{kt/2} \\=C_1C_2e^{(k+1)x+kt/2}$

When t=0,

$u(x,0)=C_1C_2e^{(k+1)x}=6e^{-3x}$

On comparing

$C_1C_2=6,k+1=-3\Rightarrow k=-4 \\ \therefore u(x,t)=6e^{-3x-2t}=6e^{-(3x+2t)}$