Contents of this Calculus 2 episode:

Constant coefficient, Homogeneous equation, Homogeneous solution, Particular solution, Method of Undetermined Coefficients, Trial Functions Method, Quadratic polynomial, Exponential expression, Expression with sine or cosine, General solution.

A First-order constant coefficient linear differential equations

This type is a special case of the first-order linear differential equations.

They are called “constant coefficient” because in these equations the function is some constant.

Let’s see a completely new solution method for this special type.

We could solve it the same way as we did in the previous slideshow, but this solution is much more amusing.

The first step is to solve this so called homogeneous equation:

This is a very simple equation.

The homogeneous equation is:

The homogeneous solution is:

We can get the general solution of the equation by adding the particular solution to the homogeneous solution.

We can obtain the particular solution based on the function on the right side, using a very funny procedure called “Method of Undetermined Coefficients”, or “Trial Functions Method”.

We find the particular solution using the Undetermined Coefficients Method:

quadratic polynomial:

exponential expression:

sine or cosine:

Here is this equation:

Now we start looking for the particular solution.

The exact nature of this particular solution always depends on the function on the right side.

It seems that now there will be sine and cosine in the particular solution.

We substitute this into the original equation.

And then we figure out what A and B is.

Here, the particular solution will be polynomial.

We plug this back into the original equation, and figure out the values of A, B and C.

And then we figure out what A and B is.

In cases where the particular solution includes exponential expressions, well, we could face some problems.

The next slideshow will discuss that.

If the particular solution includes an term, then we may face some problems when trying to solve it.

The first step is to solve this so called homogeneous equation:

Then we proceed to the particular solution.

We substitute this into the original equation:

Next, let's see the meaning of resonance.

Calculus 2 episode