solve the sytem of differential equations u' = 4v; v' = -u; where u and v are functions of x. solve the sytem of differential equations u' = 4v; v' = -u; where u and v are functions of x. @Mathematics

There are a few ways to skin this cat. The easiest way is to convert it back into a single ODE. Let's do that first.

Starting with u' = 4v, differentiate once and u'' = 4v' --- (*) but as v' = -u (*) implies that u'' = -4u i.e., u'' + 4u = 0 Now that equation you should be able to solve in your sleep. Once you solve for u, use v' = -u to find v.

Are good with that?

yes i can run with that one :)

i appreciate your help

Good. Now let me just comment on how to think about this as a linear system

u' = 4v and v' = -u is equivalent to \[\left(\begin{matrix}u \\ v\end{matrix}\right)' = \left(\begin{matrix}0 & 4 \\ -1 & 0\end{matrix}\right) \left(\begin{matrix}u \\ v\end{matrix}\right)\]

Call that 2x2 matrix A, and the column vector (u, v)^t, x. Then we have the matrix equation x' = Ax The solution of that equation together with an initial condition x(0) is x = exp(At) x(0). I'll stop here for now, because you may not have spoken about this yet in lectures, but we can make sense of what exp(At) means by using the power series definition exp(At) = I + At + A^2t^2/2! + A^3.t^3/3! + ... and calculate it by writing A as the conjugate of a diagonal matrix with eigenvalues down the diagonal. If you understand what eigenvalues are, calculate them for A and you'll see that they are strictly imaginary. You know that when that is the case with the characteristic equation it implies the solutions are period. It is exactly the same idea here.

yea...u just flew up over my head with that one

not the matrix...the last statement

You're getting to it.

ok. Going to move on. Have fun!

If you've time and appetite, you begin to get a look-ahead from your lectures by watching these: http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-24-introduction-to-first-order-systems-of-odes/

...you *can* begin to ...

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