Annihilator Method 1: Real Linear Factors

ODEs: We use the annihilator method to solve y" - 2y' - 3y = b(x), where b(x) = (a) e^{2x}, (b) e^{3x}, and (c) e^{3x} + e^{-x}. We explain the method as an application of characteristic polynomials.

[aroyy]

great video. explains way better than diff eq textbook

[WilliamBlake-yjyu]

To get the y particular are you subtracting the exponents of e^(3x) +^(-x) to get e^(2x)

[MathDoctorBob]

It's in there before the e^2x in the box. In the check for yp, the -3 and -1/3 cancel out.

[MathDoctorBob]

Yes. For these equations, we need to find an equation for which b(x) is the solution. e^rx is always a solution of y'-ry=0.

[rpsingh]

Great video! Increased my understanding of this method.

[kelvinfeliciano]

Did you work like backwards. e^2x = (r-2) = (yprime - 2y) thats the only thing that I was able to come up with.

[sayruhh]

Your videos are fantastic. Thanks for your help!

[kelvinfeliciano]

When you are explaining (a) On the right side of the board I cant see where the equation
yprime-2y=0 is coming from.

[kelvinfeliciano]

why at the end of problem (a) you did not include the -1/3 for the value of c

[MrLuke]

Are there are any differences for higher level derivatives, like Y triple prime and beyond?

[sgtcojonez]

I like his explanation, but he looks really intimidating.

[kasafarow]

Thanks for your help, but I think it would be better if you write while explaining because this is math