Forming differential equation I'm trying to get from:
$$e^{\lambda t} (\frac{dN}{dt} + \lambda N) = re^{\lambda t} $$
To:
$$ \frac {d}{dt}(Ne^{\lambda t}) = re^{\lambda t} $$
However I'm not sure what procedure to use to go about doing it. Thanks for the help in advance.
 A: Instead of thinking about going from the first equation to the second, let's go the other way. That means we have to take the derivative in your second equation:
$$\frac{d}{dt}\left(Ne^{\lambda t}\right) = \frac{dN}{dt}e^{\lambda t} + N\frac{d}{dt}e^{\lambda t} = e^{\lambda t}\frac{dN}{dt} + \lambda N e^{\lambda t} = e^{\lambda t}\left(\frac{dN}{dt}+\lambda N\right).$$
Going from the first equation to the second is kind of like factoring a quadratic equation. It takes a little experience and comfort with the operations to see how to do it. However it's really easy to expand a product of two binomial terms like $(x-a)(x-b)$ which is what going from the second equation to the first equation is like.
A: Proceed as follows:
Note that, by the distributive law,
$e^{\lambda t}(\dfrac{dN}{dt} + \lambda N) = e^{\lambda t} \dfrac{dN}{dt} + \lambda e^{\lambda t} N; \tag{1}$
also note that, by the Leibniz rule for differentiation of a product,
$\dfrac{d}{dt}(Ne^{\lambda t}) = e^{\lambda t} \dfrac{dN}{dt} + \lambda e^{\lambda t} N, \tag{2}$
since
$\dfrac{de^{\lambda t}}{dt} = \lambda e^{\lambda t}; \tag{3}$
comparing (1) and (2) we see that
$\dfrac{d}{dt}(e^{\lambda t} N) = e^{\lambda t}(\dfrac{dN}{dt} + \lambda N); \tag{4}$
inserting (4) into the original equation yields
$\dfrac{d}{dt}(e^{\lambda t} N) = re^{\lambda t}, \tag{5}$
and the process terminates!
A: This is technique applies the product rule for differentiation in reverse.  Essentially you are utilizing the fact that the exponential function is proportional to its derivative to create an expression that is the derivative of a product. This technique is known as the  method of integrating factors.  
The product rule states $$\frac {d (uv)}{dx}=u\frac {dv}{dx}+v\frac{du}{dx}$$
To illustrate let $u=N $, $x=t $ hence $\frac {du}{dx}=\frac {dN}{dt}$
So far we have: $$\frac{d(Nv)}{dt}=N\frac {dv}{dt}+v\frac {dN}{dt}$$
So all we need is a nice friendly function  $v $ that is proportional to  its own derivative. This is where the $e^{\lambda t}$ comes in.
