## Solve this Kitty

Images on March 5th, 2011 by nyoki | Report This Post | Click to Add to favorites | Tags: LOLcats, math

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March 5, 2011 at 12:08 am

so kitties equal cats?

March 5, 2011 at 12:32 pm

Can anyone tell me why they are saying y”’=m^3 ? Or explain the rest of the 2nd line.

March 5, 2011 at 3:32 pm

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.

March 5, 2011 at 4:00 pm

Ah, differential equations. My mortal enemy.

March 5, 2011 at 4:13 pm

We’re working on them in Linear Algebra right now.

March 5, 2011 at 9:42 pm

The only course I ever got a B in was the 2nd term of differential equations and series.