- #1

lukaszh

- 32

- 0

[tex]\frac{du}{dt}=u\Rightarrow\ln u=t+\ln C\Rightarrow u=Ce^t[/tex]

But this?

[tex]\frac{du}{dt}+\frac{du}{dx}=u[/tex]

[tex]dudx+dudt=udtdx\Rightarrow udx+udt=udtdx[/tex]

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