In order to solve this, you need to assume the function y(x) = e^(m*x) so that the function y and all of it’s derivatives are the same, just multiplied by a different constant.
If y(x) = e^(m*x), then y’ = m*e^(m*x), and y” = m^2*e^(m*x), and thusly y”’ = m^3*e^(m*x)
Once you have all those differential y functions replaced with the respective C*e^(m*x) functions you can factor out the e^(m*x) and you’re left with the cubic equation in the second line.
The three answers for m come from factoring the cubic into:
(m + 1)*(m – 2)*(m + 3) = 0
Since all of these values for m are valid, we have to put all of them back into the original differential, they’re just multiplied by different constants.
You begin with y(x) = C1*e^(-1*x) + C2*e^(2*x) + C3*e^(-3*x)
Since the three constants are arbitrary and we’re going to solve for them, we can replace them with anything; in this case, pictures of kittens.
The matrix below are the three equations you achieve by taking the first and second derivatives of your assumed y(x) function and set them equal to the initial values given at the beginning of the problem.
You then solve the matrix and get the values for each of the kittens, and plug those values back into the assumed y(x) function and you have your answer.
so kitties equal cats?
Can anyone tell me why they are saying y”’=m^3 ? Or explain the rest of the 2nd line.
It’s a third-order differential equation.
In order to solve this, you need to assume the function y(x) = e^(m*x) so that the function y and all of it’s derivatives are the same, just multiplied by a different constant.
If y(x) = e^(m*x), then y’ = m*e^(m*x), and y” = m^2*e^(m*x), and thusly y”’ = m^3*e^(m*x)
Once you have all those differential y functions replaced with the respective C*e^(m*x) functions you can factor out the e^(m*x) and you’re left with the cubic equation in the second line.
The three answers for m come from factoring the cubic into:
(m + 1)*(m – 2)*(m + 3) = 0
Since all of these values for m are valid, we have to put all of them back into the original differential, they’re just multiplied by different constants.
You begin with y(x) = C1*e^(-1*x) + C2*e^(2*x) + C3*e^(-3*x)
Since the three constants are arbitrary and we’re going to solve for them, we can replace them with anything; in this case, pictures of kittens.
The matrix below are the three equations you achieve by taking the first and second derivatives of your assumed y(x) function and set them equal to the initial values given at the beginning of the problem.
You then solve the matrix and get the values for each of the kittens, and plug those values back into the assumed y(x) function and you have your answer.
Ah, differential equations. My mortal enemy.
We’re working on them in Linear Algebra right now.
The only course I ever got a B in was the 2nd term of differential equations and series.