Why was this change of variables made in this system of differential equations?

Question

I'm trying to understand an example from my notes. I'm given a system of linear differential equations as follows $$x'=2x-y$$ $$y'=2x-2$$ The notes solve these by making the change of variables $$u=x-1$$$$v=y-2$$ to give $$u'=2u-v$$ $$v'=2u$$ and this then gives a system of the form $y'=Ay$ which is solveable with the methods given. My question is: how do you know to make this particular choice of variables to change to? I don't see the logical jump in making $u=x-1$ and $v=y-2$. What am I missing?

Answer 1

If you look at the second equation, you see that there is an opportunity to get rid of the constant.

That can always be helpful to simplify a DEQ.

If we made $x=u+1$, we have for the second equation:

$$2(u+1) - 2 = 2u.$$

Hence $u = x - 1$. However, now that we did that, it creates a new constant that we would like to remove in the first equation as:

$$2(u + 1) - y = 2u -y + 2$$

What can be done to get rid of the $2$ term?

What if we wrote:

$$ 2u -(v+2) + 2 \implies y = v+2 \implies v = y - 2$$

Also, from the substitutions themselves, we have $u' = x', v' = y'$ and our new system becomes:

$$u'=2u-v \\ v'=2u$$

By the way, it is interesting to note that if you look at the phase portraits, this was shifting the critical point of the original system $(x , y) = (1,2)$ to $(u, v) = (0,0)$. Notice that the phase portraits are identical in both cases (see the following pre and post phase portraits):

It is now easier to solve the shifted system and then to shift those answers using the original transformations. We have:

$$u(t) = c_1 e^t (\sin t+\cos t)-c_2 e^t \sin t \\ v(t) = 2 c_1 e^t \sin t + c_2 e^t (\cos t- \sin t)$$

Using the shift transformations ($u = x - 1, v = y - 2$) for the original system, we have:

$$x(t) = c_1 e^t (\sin t+\cos t) -c_2 e^t \sin t+1 \\ y(t) = 2 c_1 e^t \sin t+ c_2 e^t (\cos t-\sin t) + 2$$