Beedgighref28n

2022-04-23

Show that the change of variables $x=2u+v,y=3u+v$ transforms the linear system

$\{\begin{array}{rl}{x}^{\prime}(t)& =\phantom{3}x-2y\\ {y}^{\prime}(t)& =3x-4y\end{array}$

into

$\{\begin{array}{rl}{u}^{\prime}(t)& =-2u\\ {v}^{\prime}(t)& =-v\end{array}$

Verify that the u-axis maps to the line$y=\frac{3}{2}x$ and the v-axis maps to the line $y=x$ .

I thought for this I would just have to plug the substitutions into x'(t) and y'(t) but when doing this I receive

$\{\begin{array}{rlr}{u}^{\prime}(t)& =2u+v-2(3u+v)\phantom{\rule{-15pt}{0ex}}& =-4u-v\\ {v}^{\prime}(t)& =3(2u+v)-4(3u+v)\phantom{\rule{-15pt}{0ex}}& =-6u-v\end{array}$

into

Verify that the u-axis maps to the line

I thought for this I would just have to plug the substitutions into x'(t) and y'(t) but when doing this I receive

Gianna Travis

Beginner2022-04-24Added 11 answers

Step 1

You're on the right track, but what you've shown is that

$\{\begin{array}{rl}{x}^{\text{'}}& =-4u-v\\ {y}^{\text{'}}& =-6u-v\end{array}$.

By subbing in expressions in u and v for x and y, you can convert the right sides of these equations to the new variables. But you also need to convert the left sides (the derivatives), and for that you need to reverse that change of variables.

Details: using your favorite technique from linear algebra, you can find

$\{\begin{array}{rl}x& =2u+v\\ y& =3u+v\end{array}{\textstyle \phantom{\rule{2em}{0ex}}}\u27fa{\textstyle \phantom{\rule{2em}{0ex}}}\{\begin{array}{rl}u& =-x+\phantom{2}y\\ v& =\phantom{\rule{2.5pt}{0ex}}3x-2y\end{array}$

Step 2

Therefore,

$\begin{array}{rl}{u}^{\prime}& =(-x+y{)}^{\prime}\\ & =-{x}^{\prime}+{y}^{\prime}\\ & =-(x-2y)+(3x-4y)\\ & =2x-2y\\ & =2(2u+v)-2(3u+v)\\ & =-2u\end{array}$

I leave the analogous calculation that ${v}^{\prime}=-v$ up to you.

If you squint really hard, you can see that what you are doing is diagonalizing a square matrix by converting to basis of eigenvalues: ${P}^{-1}AP=D$, i.e.

$\left[\begin{array}{cc}-1& 1\\ 3& -2\end{array}\right]\left[\begin{array}{cc}1& -2\\ 3& -4\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 1\end{array}\right]=\left[\begin{array}{cc}-2& 0\\ 0& -1\end{array}\right]$

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