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

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 :-(
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 Recognitions: Science Advisor 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.