Getting into differential equations I'm just getting into differential equations now and I've got to show that the given $y(x)$ is a solution to the differential equation: $$u'+u = 0 \ , \ y(x) = Ce^{-x}$$

How do I tackle this? I know nothing about differential equations and my book has the strangest explanations.
The point is that you plug $y(x) = Ce^{-x}$ into the differential equation at the place of $u$ , and of course the derivative of $y(x)$ at the place of $u'$ (do you know how to differentiate $y(x)$?). Then you must show that the equality in the differential equation holds, that is $y'(x) + y(x) = 0$. So
\begin{align}
y'(x) + y(x) = -Ce^{-x} + Ce^{-x} = (-C+C)e^{-x} = 0\cdot e^{-x} = 0.
\end{align}
So it is, in fact, a solution since it equals zero as it should.
This is a first order differential equation so called because the highest derivative is the first derivative (that is u'). To solve in this instance you simply separate the variables.  Let dy/dx = u' and y = u ;  dy/dx  + y = 0 ====> dy/dx = -y. From here you see that multiplying both sides by dx/y yields dy/y = - dx. Apply the integral operator and you get ln|y| = -x + ln C. Taking the antilogarithm of both sides and remembering you're rules of logarithms gives you the answer. Y = Ce^(-x) the general solution