1. ## Differential Equations

The solution to y" = 0 is a straight line y = C + Dt. Convert to a matrix equation:
d/dt[y y'] =[0 1 ; 0 0] has the solution [y y'] = e ^At[y(0) y'(0)].

This matrix A cannot be diagonalized. Find A^2 and compute e^At = I + At + .5A^2t^2... Multiply your e^At times y(0), y'(0) to check the line y(t) = y(0) +y'(0)t

I'm not even sure what A is in this question...

2. Originally Posted by veronicak5678
The solution to y" = 0 is a straight line y = C + Dt. Convert to a matrix equation:
d/dt[y y'] =[0 1 ; 0 0] has the solution [y y'] = e ^At[y(0) y'(0)].

This matrix A cannot be diagonalized. Find A^2 and compute e^At = I + At + .5A^2t^2... Multiply your e^At times y(0), y'(0) to check the line y(t) = y(0) +y'(0)t

I'm not even sure what A is in this question...
The solution to dy/dt= Ay is, of course, $y= Ce^{At}$. That is trivially true if y, A, and t are numbers and extends to the case, here, that A is a matrix and y a vector. That is, here, $A= \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$.

$A^2=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}= \begin{bmatrix}0(0)+ 1(0) & 0(1)+ 1(0) \\ 0(0)+ 0(0) & 0(1)+ 0(0)\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$

Do you see the point now?