Here's the question you clicked on:
bettyboop8904
For what values of r does the function y=e^(rx) satisfy the equation y"+6y'+8y=0?
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 = (
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 = (
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 = )