How to solve this differential equation, involving leibniz notation? I thought I was pretty good at calculus, but this one has stumped me. I can do many almost identical examples, but I can't seem to extrapolate the skills needed to this one tricky problem.
$${dy \over dx} = y+2$$
The solution is listed in the book as $y=Ce^x-2$
I can't seem to wrestle the equation into any kind of form that I know how to work with, which is frustrating as I realize this is quite simple.
I want to multiply by dx, but it just seems to complicate things.
 A: This is a first-order linear ordinary differential equation, and it is separable. You may write it as: $$\frac{dy}{y+2}=dx$$ Then, integrate both sides. Thus, you have that: $$\int \frac{dy}{y+2}=\int dx$$ $$\ln\left|y+2\right|=x+C$$ $$\left|y+2\right|=e^{x+C}=e^Ce^x \iff y+2=\pm e^Ce^x=c_1e^x \iff y=c_1e^x-2$$
A: This is actually a fairly easy ordinary differential equation-it is what we call a separable equation.i.e. the variables can be "separated" as follows: 
                   $$dx = \frac {1}{y+2}dy$$  
Integrating both sides gives x = In (y+2) + C . Taking the natural exponential of both sides gives:
             $e^{x}$ = $e^{c}(y+2)$  
Multiplying through by $e^{-c}$ gives:
             $e^{-c}e^{x}$ = y+2
Now we simply rearrange the equation and collect terms to give the answer in your book. Do it yourself, it's just algebra. 
Ain't calculus grand? 
A: These equations are almost always solved by collecting the same variables on either side of the equation so that it results in a normal expression. In this case you would want to write it as: 
$$\frac{dy}{y+2}=dx$$
Then you can integrate both sides with respect to each variable to get:
$$\int\frac{dy}{y+2}=\int dx$$
$$ln(y+2)+lnC=x$$
$$y=Ce^x-2$$
