Non-Linear Techniques This Question is a practice exercise assigned to us. I have the answer but I don't find it very intuitive. The answer I have is on this pdf : http://math.stanford.edu/~ralph/math53h/solution5.pdf
Find a global change of coordinates that linearizes the system
$x' = x + y^2$
$y' = −y$
$z' = −z + y^2$
Please do not give negative rating before giving me a chance to make corrections if any errors are found in the question.
 A: We are asked to find a global change of coordinates that linearizes the nonlinear system:
$\tag 1 x' = x + y^2$
$\tag 2 y' = −y$
$z\tag 3' = −z + y^2$
From $(2)$, we get:
$\tag 4 y = y_0 e^{-t}$ (You know how to solve this, correct?)
Lets now substitute this into $(1)$ and solve that DEQ:
We have $x' - x = y^2 = (y_0 e^{-t})^2 = y_0^2 e^{-2t}$. Lets work the details of this one as a sample manually.
Homogeneous: $m - 1 = 0 \rightarrow m = 1 \rightarrow x_h = x_0e^t$
Particular: Assume $x_p = Ae^{-2t}$, so $x' - x = -2Ae^{-2t} - A e^{-2t} = y_0^2 e^{-2t}$, so $-3A = y_0^2 \rightarrow A = -\frac{1}{3} y_0^2$.
Thus, we get $x = x_h+x_p, so$:
$\tag 5 \displaystyle x(t) =  x_0 e^t - \frac{1}{3} y_0^2e^{-2t}.$
Lets now repeat this process for $(3)$ and solve that DEQ:
We have $z' + z = y^2 = (y_0 e^{-t})^2 = y_0^2 e^{-2t}$, and solving for $z$ yields:
$\tag 6 z(t) =  z_0 e^{-t} - y_0^2e^{-2t}.$
Next, our goal is to get rid of the nonlinear terms in $x, y$ and $z$ by finding a 'global change of coordinates'.
What if we choose a change a variables that guarantees those nonlinear terms disappear? How can we do that?
Let: $\displaystyle u = x + \frac{y^2}{3} = x_0 e^t - \frac{1}{3} y_0^2e^{-2t} + \frac{1}{3} y_0^2e^{-2t} = x_0e^t.$
So, we have: $u = x_0e^t$.
If you take the derivative of $u$, what do you get?
$u' = x_0e^t$. In other words $u' = u.$ (That nicely linearized.)
Next, if we let: $v = y = y_0e^{-t}$, what is $v'$? Well, $v' = -y_0e^{-t} = -v$. (That nicely linearized.)
If we let (note - there is a slight typo in the solution key here, they meant $y^2$):
$w = z + y^2 = z_0e^{-t} -y_0^2e^{-2t} + y_0^2e^{-2t} = z_0e^{-t}.$
If we take the derivative of this, we get $w' = -z_0e^{-t} = -w$. (That nicely linearized.)
We are now left with a linearized system:
$\tag 7 u' = u$
$\tag 8 v' = -v$
$\tag 9 w' = -w$
They wrote the solution in a very compact/condensed form and maybe that was throwing you off. I would recommend comparing the above with what they wrote and make sure it is clear! 
Hope that makes sense!
