Contents of this Calculus 3 episode:

Second-order equations, Constant coefficient, Homogeneous equation, Characteristic equation, Solution of characteristic equation, Complex solution, Homogeneous solution, Particular solution, Method of Undetermined Coefficients, Trial Functions Method, Quadratic polynomial, Exponential expression, Expression with sine or cosine, General solution, Resonance.

The homogeneous equation and its solution:

If it has two real solutions:

If it has one real solution:

If it has two complex solutions:

Particular solution (Undetermined Coefficients Method)

Here is this equation that is nonhomogeneous:

In such cases we solve the homogeneous equation first,

and then find the particular solution using the Undetermined Coefficients Method.

To solve the homogeneous equation, we solve the usual

characteristic equation.

And now we are ready for the particular solution.

We can obtain the particular solution based on the function on the right side.

This time the function on the right side happens to be a polynomial, so we try to find the particular solution in that form, too.

But the right-side function could be exponential,

or even trigonometric.

The particular solution is:

Let's see what we get if we substitute this into the original equation.

And the general solution is:

Then, we have this other nonhomogeneous equation.

But there is a snag.

Just like with first-order equations, there could be resonance here, too.

Resonance occurs if a term of the homogeneous solution is equal to a term of the particular solution.

Now there is no resonance,

but there will be some in the next slideshow.

Compared to first-order equations, this resonance business gets a bit more complicated at second-order equations.

Here is this equation:

The solution of the homogeneous equation is:

And now we are ready for the particular solution.

We always figure out the particular solution based on the function on the right side.

One term of the homogeneous solution is equal to a term of the particular solution. That means that sadly, there is resonance.

In this case the constant multiplier doesn’t matter.

Due to the resonance, an x will come in here.

Now we compute the first and second derivative of the particular solution.

Next, we substitute these into the original equation.

When the characteristic equation has only one real solution, there could be dual resonance.

The resonance appeared.

Therefore, a multiplier of x will be needed in the particular solution.

But then there will be resonance with the second term...

so we need another x multiplier.

This is what we call dual resonance.

From here, the solution proceeds as usual.

Boring, as usual.

So let’s not solve it now. Instead, let’s see what kind of resonance could appear when the characteristic equation has two complex roots.

Here are these two equations:

The characteristic equations are:

For the complex solution all we need to know is this:

In such cases resonance occurs if:

And then the trial function is done.

Calculus 3 episode