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For what values of r does the function y=e^(rx) satisfy the equation y"+6y'+8y=0?

Mathematics
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just solve the quadratic formed by characteristic equation: r^2 +6r +8 = 0
the function \[y=e ^{rx}\] satisfy the equation \[y"+6y'+8y=0\]
can you elaborate a little more = (

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Other answers:

no but i will refer to a site that can :) http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx
this is a linear second order homogeneous differential equation that involves the first second and 0th derivatives of a function. Therefore the solution to this differential equation will have a solution of the form e^(rx) where r is uniquely determined by the coefficients of the 1st 2nd and 0th derivatives of y.
by using the quadratic formula like dumbcow advised, you can find the values of r which will serve as the coefficients of the linearly independent solutions of the form e^rx
Solving differential equations is all about classifying them by their order and type, and then solving accordingly by pre-prescribed methods
so you plug r in for y's? how come?
yes, sort of. you look at the equation of this form. ay"+by′+cy=0, and you take the coefficients a, b, and c and then plug them into a corresponding 2nd degree polynomial in terms of r. so for ay"+by′+cy=0, you have \[ar^2+br+c =0\] solving this polynomial will give you the values of r for which you will have solutions of the form \[e ^{rx}\]
so, lets say you obtained two real values of r 2 and 3. Therefore you will have a solution of the form \[Ke ^{3x}+Ce ^{2x}\] where C and K are arbitrary constants
each value of r gives you a linearly independent solution to the differential equation, by taking a linear combination of the two you obtain all possible solutions
ok so i understand the first part the solving this polynomial part is confusing and so we're using the quadratic formula with a=1 b=6 and c=8?
yes, however you should notice that it can be easily factored as well
I'm sorry you are helping me but your examples are a little fancy worded for me. I'm really good with actual problems but when math starts adding describing words I start getting confused lol = (
my b, lol
its ok lol
so if you factor it then what would you do with the values?
so yea, set up your 2nd degree polynomial using the coefficients from the initial equation: y"+6y'+8y=0 \[r^2 +6r+8=0\] like dumbcow said, the left side of this equation can be easily factored: \[(r+4)(r+2)=0\] the solutions to this equation are therefore r=-4 and r=-2 so therefore we have two linearly independent solutions \[y = e ^{-4x}\] and \[y = e ^{-2x}\] both of these equations are solutions to the differential equation y"+6y'+8y=0. Every possible linear combination of these equations is as well, ad since they are linearly independent, the sum of these equations is also a solution to y"+6y'+8y=0. Try all of these out use \[y = e ^{-4x}\], \[y = e ^{-2x}\], \[y = 4e ^{-4x}\] and \[y = e ^{-4x}+e ^{-2x}\]
you'll find that they will all satisfy y"+6y'+8y=0. So in order to provide a general equation that expresses all possible solutions we write \[y=Ke ^{-4x}+Ce ^{-2x}\] where K and C are arbitrary constants
and that's the answer
plug in any value for K and C and then plug that into y"+6y'+8y and you will get 0
ok i get it now = ) thank you for the help = )
np

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