I can't find anywhere how to solute PDE's.

[lukaszh]

I can't find anywhere how to solute PDE's. For exaple ODE:
$$\frac{du}{dt}=u\Rightarrow\ln u=t+\ln C\Rightarrow u=Ce^t$$
But this?
$$\frac{du}{dt}+\frac{du}{dx}=u$$

$$dudx+dudt=udtdx\Rightarrow udx+udt=udtdx$$

I haven't had pde's yet, but I'm interested in solving these equations :-(

 

[Ben Niehoff]

For many equations, there is a technique that works called "separation of variables". To separate the equation, we assume u(x,t) can be factored into two one-variable functions:

$$u(x,t) = X(x)T(t)$$

And then

$$\frac{\partial u}{\partial x} = T(t)\frac{dX(x)}{dx}$$

$$\frac{\partial u}{\partial t} = X(x)\frac{dT(t)}{dt}$$

Then, for your equation, we can write:

$$X(x)\frac{dT(t)}{dt} + T(t)\frac{dX(x)}{dx} = X(x)T(t)$$

Next, divide by X(x)T(t):

$$\frac{1}{T(t)}\frac{dT(t)}{dt} + \frac{1}{X(x)}\frac{dX(x)}{dx} = 1$$

Now, notice that we've separated the variables into the form

$$P(t) + Q(x) = 1$$

Since this must be true for all x and t, the functions P and Q must individually be constant. This gives

$$a + b = 1$$

$$\frac{1}{T(t)}\frac{dT(t)}{dt} = a$$

$$\frac{1}{X(x)}\frac{dX(x)}{dx} = b$$

These have the solution

$$X(x) = e^{bx}$$

$$T(t) = e^{at}$$

and so

$$u(x,t) = Ce^{at}e^{bx} = Ce^{at + (1-a)x}$$

for arbitrary constants C and a. Also note that any linear combination of u's (with different values for a) are also solutions, because the equation is linear. This can be used to construct a more general solution, given some boundary conditions.