Answer to Question #297053 in Differential Equations for Tokyo

Question

<e> Solve using the method of separation of variables the pde partial du/dt+du/dx+2e^tu=0

Answer 1

By the method of separation of variables we suppose that $u$ is of the form $u(t,x)=\phi(t)\psi(x)$ which gives us :

$\phi'(t) \psi(x)+\phi(t)\psi'(x)+2e^t \phi(t) \psi(x)=0$

dividing both parts by $u(t,x)=\phi(t)\psi(x)$ gives us

$\phi'(t)/\phi(t) + 2e^t+\psi'(x)/\psi(x)=0$

$\phi'(t)/\phi(t)+2e^t=-\psi'(x)/\psi(x)$

As the left part is a function of $t$ and the right is the function of $x$ the equality gives us

$\begin{cases} \phi'/\phi +2e^t=\lambda \\ -\psi'/\psi = \lambda \end{cases}$ with $\lambda\in \mathbb{R}$ a constant.

The second equation easily gives us $\psi_\lambda (x)= Ae^{-\lambda x}$, $A$ a constant

The first one gives us

$\ln \phi(t) - \ln\phi(0)=\lambda t -2e^t+2$

$\phi_\lambda(t)=Be^{\lambda t - 2e^t}$ , $B$ a constant

Combining two solutions gives us

$u_\lambda (t,x)= Ce^{\lambda(t-x)} e^{-2e^t}$