Auxiliary equations with repeated roots  Summary and Q&A
TL;DR
Learn how to solve secondorder linear differential equations with constant coefficients by finding the characteristic equation, solving for the roots, and using them to construct the building blocks of the solution.
Key Insights
 The roots of the characteristic equation determine the building blocks of the solution.
 If the roots are repeated, the solution contains both e^rt and t * e^rt.
 Reduction of order is used to find a second building block for the solution when the roots are repeated.
Transcript
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Questions & Answers
Q: How do you find the roots of the characteristic equation?
To find the roots, replace y' with r^2, y with r, and set the equation equal to zero. Solve for r using factoring or the quadratic formula.
Q: What is the purpose of the reduction of orders method?
The reduction of orders method is used to find a second building block for the solution when the roots of the characteristic equation are repeated. It involves finding a function v(t) that, when multiplied by e^rt, satisfies the original equation.
Q: Can the first and second building blocks of the solution be combined?
No, the first and second building blocks cannot be combined because they are linearly independent. However, they can be added together along with any constant multiples to obtain the general solution.
Q: Are the building blocks always constant multiple of e^rt or t * e^rt?
No, the building blocks are only constant multiples of e^rt or t * e^rt when the roots of the characteristic equation are repeated. Otherwise, the building blocks may be different functions of t.
Summary & Key Takeaways

This video explains how to solve secondorder linear differential equations with constant coefficients and the righthand side equal to zero.

The characteristic equation is obtained by replacing y' with r^2, y with r, and setting the equation equal to zero.

If the roots of the characteristic equation are repeated, the solution will contain both e^rt and t * e^rt.

The video introduces reduction of order as a method to find a second building block for the solution, involving finding a function v(t) that, when multiplied by e^rt, satisfies the original equation.