rewriting the system in one variable ODE

Question

I want to write the following coupled system in one variable (not solving it!).

$$ x'=ax+by \\ y'=cx+dy $$

The aim is to substitute with a $u=\alpha x+\beta y$ in a way that the system can be written as $u^{'}=K u$, where $K$ is some constant, some combination of $a,b,c,d$.

In the case $a=d=0$ the substitution can be $u=\sqrt{b}y+\sqrt{c}x$, but I cannot find the right coefficients when $a,d$ are non-zero.

Note: I am not trying to solve it, rather find a suitable substitution.

Answer 1

Let $X = \left(\begin{array}{c}x \\ y\end{array}\right)$ and $A = \left(\begin{array}{cc}a&b \\ c&d\end{array}\right)$, then $X'=AX.$ Now, diagonalize $A$ (i.e. write it as $P\Lambda P^{-1}$, where $\Lambda= \left(\begin{array}{cc}\lambda_1&0 \\ 0&\lambda_2\end{array}\right)$ is the matrix of eigenvalues of $A$, and $P$ is a matrix whose columns are the eigenvectors of $A$).

Define $U:=P^{-1}X$, then $U' = P^{-1}X' = PAX = P^{-1}P\Lambda P^{-1} X=\Lambda U$. In particular both $u_1' = \lambda_1 u_1$ and $u_2' = \lambda_2 u_2$ are answers to your question.

More explicitly: $$ P = \left(\begin{array}{cc}\frac{a-d-\sqrt{(a+d)^2-4 (a d- b c)}}{2 c}&\frac{a-d+\sqrt{(a+d)^2-4 (a d- b c)}}{2 c} \\ 1&1\end{array}\right). $$ $$U = P^{-1}X = \frac1{\sqrt{(a+d)^2-4(ad-bc)}}\left(\begin{array}{c}-cx+\frac{a-d+\sqrt{(a+d)^2-4 (a d- b c)}}2 y \\ cx-\frac{a-d-\sqrt{(a+d)^2-4 (a d- b c)}}2 y\end{array}\right)$$ $$ \lambda_1 = \frac12\left(d+a-\sqrt{(a+d)^2-4 (a d- b c)}\right),\quad \lambda_2 = \frac12\left(d+a+\sqrt{(a+d)^2-4 (a d- b c)}\right).$$

Answer 2

Looks like a straight forward computation. If $u= \alpha x+ \beta y$ then $u'= \alpha x'+ \beta y'$$= \alpha(ax+ by)+ \beta(cx+ dy)= (\alpha a+ \beta c)x+ (\alpha b+ \beta d)y$. If that is to be written $Ku= K(\alpha x+ \beta y)= K\alpha x+ K\beta y$ we must have $K\alpha= \alpha a+ \beta c$ and $K\beta= \alpha b+ \beta d$ . That gives two equations to be solved for $\alpha$ and $\beta$.

We can write the first equation as $(K- a)\alpha= c\beta$. From that $\alpha= \frac{c}{K- a}\beta$. The second equation can be written as $(K- d)\beta= b\alpha$ so that $\alpha= \frac{K- d}{b}\beta$. Setting those equal to each other, $\frac{c}{K- a}\beta= \frac{K- d}{b}\beta$. From that $bc= (K-a)(k- d)$ which we can write as the quadratic equation $K^2- (a+ d)K+ ad- bc= 0$. There will exist a real value of K satisfying that as long as $(a+ d)^2+ 4bc\ge 0$.

Answer 3

Hint:

From the first equation we have: $$ x''=ax'+by' $$

now use $y'=cx+dy$ and $by=x'-ax$

Answer 4

Note that both $x$ and $y$ satisfy the same equation $\begin{cases}x''=(a+d)x'+(bc-ad)x\\y''=(a+d)y'+(bc-ad)y\end{cases}$

So any combination $u=\alpha x+\beta y$ will also satisfies this same equation $$u''=(a+d)u'+(bc-ad)u$$

The characteristic equation gives roots $\dfrac{(a+d)\pm\sqrt{(a+d)^2+4(bc-ad)}}2$ and you can calculate these are also the eigenvalues of the matrix $\begin{pmatrix}a & b\\c & d\end{pmatrix}$ as suggested in stochastic's answer as well as the $K$ found in user247327's one.