Reminder

2.1 Pretty direction fields

f[x_,y_]:= x Sin[x y]
StreamPlot[{1,f[x,y]}, {x,-3,3},{y,-3,3}]

2.2 Separable Equations

Just like in Calc an ODE of the form
	dy/dx = g[x]h[y]
is separable.  You get the general solution by separating variables and integrating
	∫1/h[y]dy = ∫g[x]dx+C.

2.3 Linear 1st Order ODE

Standard form
	y'[x]+P[x]y[x]=f[x]
Compute μ[x]=(e^(∫P[x]dx)) and multiply eq by μ and collapse product rule terms and integrate to get general solution
	({{μ[x]y'[x], +, μ[x]P[x]y[x], =, μ[x]f[x]}, {μ[x]y'[x], +, μ'[x]y[x], =, μ[x]f[x]}, { ,  , (μ[x] y[x])', =, μ[x] f[x]}, { ,  , μ[x]y[x], =, ∫μ[x]f[x]dx+C}})
The red entries in the first two lines work because of the formula for μ.  We got it by solving the separable ODE 
	({{μ'[x], =, μ[x]P[x]}})
As always if I have an initial condition I plug it into the general solution.

2.4: Exact

If I have a function f[x,y] then f[x,y]=c defines a one parameter familly of curves. If

f[x_,y_]:= x Sin[x +(y^2)]
ContourPlot[ f[x,y],{x,0,3},{y,0,3},
Contours->10,
ContourShading->False,
ContourLabels->All]

The first question is what ODE do they satisfy?  Implicit differentiation of
	f[x,y[x]]=c	 
answers this.  These curves satisfy
	({{((∂f)/(∂x)), +, ((∂f)/(∂y)), (dy/dx), =, 0}})
or equivalently and symmetrically 
	({{((∂f)/(∂x))dx, +, ((∂f)/(∂y)), dy, =, 0}})

An ODE in this form is called exact and you solve it by finding f[x,y].

The Standard form here is

M dx +N dy=0

We need to know how to tell if and ODE in std form is exact

Remember (((∂^2)f)/(∂x∂y))=(((∂^2)f)/(∂y∂x)) sooooo an exact ODE has ((∂M)/(∂y))=((∂n)/(∂x))

We need to know how to solve an exact ODE.

All we need to do is find f from M=((∂f)/(∂x)) and N=((∂f)/(∂y))

Integrate wrt x to get

f=∫((∂f)/(∂x))dx+g[y]=∫M[x,y]dx+g[y]

The differentiate wrt y to get

((∂f)/(∂y))=(∂/(∂y))((∂/∂)∫M[x,y]dx )+({{g'[y], =, N}})

Solve for g'[y] and integrate to get g.

If you have an x in g’[y] you have messed up somewhere.

The general solution is f[x,y]=c

As always if you have an IC you plug into the general solution.