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#### MathsCasts: First Order Differential Equations: Separation of Variables 2

(Swinburne Commons, 2011)

This screencast illustrates the method of separation of variables for a medium-level first order differential equation: dy/dx=(2-y)/(1+x). Different ways of displaying the solution are discussed.

#### MathsCasts: First Order Differential Equations: Separation of Variables 3

(Swinburne Commons, 2011)

This screencast illustrates the method of separation of variables for a medium-level first order differential equation: dv/dpsi=v(tan(psi)-mu sec(psi)).

#### MathsCasts: Second Order Constant Coefficient Differential Equations: Part 1

(Swinburne Commons, 2011)

This screencast first derives the auxilary equation corresponding to a second order d.e. with constant coefficients. It then shows how to find the solution of this d.e. The auxiliary equation has complex roots, and the ...

#### MathsCasts: First Order Differential Equations: Separation of Variables

(Swinburne Commons, 2011)

This screencast illustrates the method of separation of variables for solving a first order ordinary differential equation.

#### MathsCasts: First Order Differential Equations: Solution of a Bernoulli Equation

(Swinburne Commons, 2011)

This screencast explains what is meant by a first order differential equation of Bernoulli type and solves a specific example.

#### MathsCasts: First Order Linear DEs: Integrating Factors: Part 2

(Swinburne Commons, 2011)

This screencast shows a medium-level example of the solution of a first order differential equation using the integration factors method.

#### MathsCasts: First Order Linear DEs: Integrating Factors: Part 1

(Swinburne Commons, 2011)

This screencast shows a simple example of the solution of a first order differential equation using the integration factors method.

#### MathsCasts: First Order Differential Equations: Separation of Variables 4

(Swinburne Commons, 2011)

This screencast illustrates the method of separation of variables for a more advanced (and applied) example with a boundary condition: v dv/dz=-g+kv^2, v(h)=0.