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7.7.1. Case 1: Repeated Roots and Duplication

Interactive Audio Lesson

Session 1: Understanding Repeated Roots

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Sarah
SarahInstructor

Today, we will talk about repeated roots in differential equations. Can anyone tell me what we mean by 'repeated roots'?

Noah
Noah

I think it’s when the solutions to the characteristic equation are the same.

Sarah
SarahInstructor

Exactly! When the roots of the characteristic or auxiliary equation are repeated, we must adjust our approach in the undetermined coefficients method.

Isabella
Isabella

How do we know if there will be duplication in our solutions?

Sarah
SarahInstructor

Great question! Duplication occurs when the forcing function has terms that already appear in the complementary function. This is what leads us to modify our trial solutions.

Session 2: Trial Solutions and Duplication

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Robert
RobertInstructor

Let’s consider our approach to creating a trial solution. If our complementary function has a term e^(2x), and our forcing function is also e^(2x), how should we modify our trial solution?

Akash
Akash

I think we should just guess Ae^(2x) as the solution.

Robert
RobertInstructor

Not quite! Because there's duplication, we must multiply our trial solution by x raised to the power of m. So we guess Ax * e^(2x).

Ananya
Ananya

And if e^(2x) showed up twice?

Robert
RobertInstructor

In that case, we would try Ax^2 * e^(2x) to ensure we eliminate all duplication.

Session 3: Applying the Method

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Sarah
SarahInstructor

Let's apply what we have learned. Consider the equation y'' - 4y' + 4y = e^(2x). The characteristic equation would lead to repeated roots.

Noah
Noah

So, we start by finding the complementary function first?

Sarah
SarahInstructor

Correct! The complementary function will be y_c = C1e^(2x) + C2xe^(2x). Now our forcing function, being e^(2x), indicates we modify our trial solution to y_p = Ax^2e^(2x).

Isabella
Isabella

What happens next after setting up the trial solution?

Sarah
SarahInstructor

We substitute our trial solution back into the original equation, derive the necessary derivatives, and solve for the coefficients A. This ensures we account for all overlapping terms.