Differential Equations – Day 3

Model: Mixing Problem – 1 Tank

Rum and Coke solution.


1. Maintain constant tank volume.

2. Inlet and outlets supply and drain at the same rate.

3. Instantaneous & thorough mixing

How much rum is present in the tank at any time t?

We want to track the amount of rum at a given time.

Rate in = clip_image002

Rate out = clip_image004



More on Separable Equations

Exact Equation of the form: clip_image010

Where M and N are expressed entirely in terms of their paired derivative we are able to find …



When f(x,y)= 0, it follows that clip_image010[1]

If clip_image010[2] With clip_image016 is a differential of clip_image012[1] then the equation is an exact equation which means clip_image018 is an implicit solution of the original equation.

This comes out of the mixed partial work that we did in Math 1D. If the mixed partials are equal, the equation is exact. Meaning – clip_image020


This is an exact solution

Super position principle & Variation of Parameter

clip_image028 we would like to solve for clip_image030. The solution to the second is called the complementary function, called clip_image032. The solution to the first is clip_image034where clip_image036 is a particular solution for equation and compensates for the clip_image038 input function.


So our original equation can be rewritten as clip_image042orclip_image044where clip_image046 for our homogeneous solution.

clip_image048 where M(x) is a separable equation. Solve the homogeneous solution first then allow the form of the homogeneous equation to help you get going on the right track for the particular/non-homogeneous. Multiply by M(x) and solve for M(x). Do remember to check things out after the fact though.



Since My and Nx are not the same, we can do some fancy work to make things fun.


We can use this to solve our equation for f and find out what our integrating factor is.


Assume that Fy or Fxis 0 and solve for f=F(x) or f = F(y), respectfully.