Thread: Seperation of variables View Single Post

## Seperation of variables

This is the the first time I've encountered seperation with partial differential equations. There are no worked examples, so I need some help to work through this problem. The question seems to be somewhat hand holding, since it seems to be THE introduction.

Q: Apply seperation of variables $u_t = u_x$ by substituting $u=A(x)B(t)$ and then dividing by AB. If one side depends only on $t$ and the other only on $x$, they must equal a constant $k$; what are $A$ and $B$?

$$\frac{\partial u}{\partial t}-\frac{\partial u}{\partial x} = 0$$

$$u = A(x)B(t)$$

$$\frac{\partial}{\partial t} \left[ A(x)B(t) \right] - \frac{\partial}{\partial x} \left[ A(x)B(t) \right] = 0$$

$$A(x)B'(t)-A'(x)B(t)=0$$

$$\frac{A(x)B'(t)-A'(x)B(t)}{A(x)B(t)}$$

$$\frac{B'(t)}{B(t)}-\frac{A'(x)}{A(x)}=0$$

Now I was reading on various websites, that I can set each independent term equal to seperation constants to make two coupled (is this the proper word to use?) differential equations. I don't understand where this step comes from.

but...

$$\frac{B'(t)}{B(t)}=k$$

$$\frac{A'(x)}{A(x)}=k$$

Now solving for $A(x)$ and $B(t)$. I'm a little rusty here, so I don't know if this part is correct.

Rewriting the two equations above in Leibniz notation

$$\frac{dB(t)}{dt} \cdot \frac{1}{B(t)} = k$$

Seperating:

$$\frac{dB(t)}{B(t)} = k dt$$

$$\int \frac{dB(t)}{B(t)} = \int k\,\,dt$$

$$\ln B(t) = kt +c$$

$$B(t) = e^{kt+c}$$

And subsequently:

$$A(x) = e^{kx+c}$$

Does this make sense? :)