The excerpts in this post are from Gilbert Strang's Introduction To Linear Algebra Book, Page 315 and 316.

What we're solving for

I did the math, and i saw that he was just writing y'' in terms of the position y. But i still don't understand how he came up with the formulas for Yn+1 and Zn+1 and what he was trying to explain. Any help is much appreciated, thanks in advance.

Steps he took to solve it

A visual representation of a Forward Euler Spiral


1 Answer 1


You start with

$y'' = -y$

define $z = y'$ or $y'=z$

Then $z' = y'' = -y$

If we let some small amount of time go by.

$y(t + \Delta t) = y(t) + y'(t) \delta t\\ z(t + \Delta t) = z(t) + z'(t) \delta t$

But we know that $y',z'$ can be described in terms of $y,z$

$y(t + \Delta t) = y(t) + z(t) \Delta t\\ z(t + \Delta t) = z(t) - y(t) \Delta t$

$\begin{bmatrix} y(t+\Delta t)\\z(t + \Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}\begin{bmatrix} y(t)\\z(t)\end{bmatrix}$

If we start at $t = 0$

$\begin{bmatrix} y(\Delta t)\\z(\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}\begin{bmatrix} y(0)\\z(0)\end{bmatrix}$

$\begin{bmatrix} y(2\Delta t)\\z(2\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}\begin{bmatrix} y(\Delta t)\\z(\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}^2\begin{bmatrix} y(0)\\z(0)\end{bmatrix}$

$\begin{bmatrix} y(n\Delta t)\\z(n\Delta t)\end{bmatrix} = \begin{bmatrix} 1&\Delta t\\-\Delta t&1\end{bmatrix}^n\begin{bmatrix} y(0)\\z(0)\end{bmatrix}$

It might be simpler to say:

$\begin{bmatrix} y\\z\end{bmatrix}' = \begin{bmatrix} 0&1\\-1&0\end{bmatrix}\begin{bmatrix} y\\z\end{bmatrix}$

Turning a second degree diff eq. into a system of 1st degree equations

If it wasn't a system of equations, you would say:

$y' = ay\\ y = y(0) e^{at}$

Just because you have vectors and matrices, no reason to change that...

$\mathbf x' = A\mathbf x\\ \mathbf x = e^{At}\mathbf x_0$


$e^{at} = \sum \frac {a^nt^n}{n!}$

If $A$ can be diagonalized, then

$A = P^{-1}\Lambda P\\ A^n = P^{-1}\Lambda^nP\\ e^{At} = P^{-1}(\sum \frac {\Lambda^nt^n}{n!})P\\ \mathbb x' = P^{-1} e^{\Lambda t} P \mathbb x_0$


$P\mathbb x' = e^{\Lambda t} P \mathbb x_0$

The eigenvalues of $A$ have most of the information that describes your differential equation.

  • $\begingroup$ omg, thanks so much, I've been sitting on this question for days. I wish i could +1 you but the least I could do is reply for how grateful I am $\endgroup$ Nov 16, 2018 at 4:49

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