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

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dumbcowBest ResponseYou've already chosen the best response.1
just solve the quadratic formed by characteristic equation: r^2 +6r +8 = 0
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
the function \[y=e ^{rx}\] satisfy the equation \[y"+6y'+8y=0\]
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
can you elaborate a little more = (
 one year ago

dumbcowBest ResponseYou've already chosen the best response.1
no but i will refer to a site that can :) http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
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.
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
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
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
Solving differential equations is all about classifying them by their order and type, and then solving accordingly by preprescribed methods
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
so you plug r in for y's? how come?
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
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 \[e ^{rx}\]
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
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
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
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
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
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?
 one year ago

dumbcowBest ResponseYou've already chosen the best response.1
yes, however you should notice that it can be easily factored as well
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
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 = (
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
so if you factor it then what would you do with the values?
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
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 \[y = e ^{4x}+e ^{2x}\]
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
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
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
and that's the answer
 one year ago

sweet1137Best ResponseYou've already chosen the best response.0
plug in any value for K and C and then plug that into y"+6y'+8y and you will get 0
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
ok i get it now = ) thank you for the help = )
 one year ago
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