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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just solve the quadratic formed by characteristic equation: r^2 +6r +8 = 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the function \[y=e ^{rx}\] satisfy the equation \[y"+6y'+8y=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you elaborate a little more = (

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no but i will refer to a site that can :) http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Solving differential equations is all about classifying them by their order and type, and then solving accordingly by preprescribed methods

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you plug r in for y's? how come?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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 \[e ^{rx}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0each 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, however you should notice that it can be easily factored as well

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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 = (

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if you factor it then what would you do with the values?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 \[y = e ^{4x}+e ^{2x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you'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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and that's the answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0plug in any value for K and C and then plug that into y"+6y'+8y and you will get 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i get it now = ) thank you for the help = )
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