I am being asked to solve a system of differential equations. We are provided with a hint to use substitution to simplify the process, but I don't follow how to use this process to move forward.
Define the vectors $\vec y$ = [$y_1$, $y_2$], $\vec y$' = $\left[ \frac{dy_1}{dt}, \frac{dy_2}{dt}\right]$ and the coefficient matrix A = $\begin{bmatrix} a & -b\\ b & a\end{bmatrix}$.
Then, $\vec y$' = Ay defines a system of linear homogeneous differential equations. Note that A has complex eigenvalues.
Hint: To solve the system of equations, write $z = y_1 + iy_2$ and calculate $z'$. For some $u$ defined in terms of $a$ and $b$ you should find that $z'=uz$, which is easily solved. Now, write the solution in terms of the original variables $y_1$ and $y_2$. You will need to use the identity $e^{itb}=cos tb+isin tb.
I've reached out to my instructor but it consistently takes over a week to get a response.
My attempt so far is below.
Finding $z'$:
$z' = \frac{dz}{dt} = \frac{d}{dt}\left(y_1 + iy_2\right) = \frac{dy_1}{dt} + i\frac{dy_2}{dt}$
Let us write all of this in a different form:
$\frac{dy_1}{dt} = ay_1 – by_2$
$\frac{dy_1}{dt} = by_1 + ay_2$
$\frac{dz}{dt} – (ay_1-by_2) + i(by_1 + ay_2)$
$z' = (a+ib)y_1 + (-b+ia)y_2$
Now we set $z' = uz$
$uz = u(y_1 + iy_2) = (a+ib)y_1 + (-b+ia)y_2$
$u = \frac{(a+ib)y_1+(-b+ia)y_2}{y_1+iy_2}$
My problem is I'm not really sure where I'm going at this point.
Any advice would be greatly appreciated!
Best Answer
$$\frac{dz}{dt} = (ay_1-by_2) + i(by_1 + ay_2)$$ $$\frac{dz}{dt} = a(y_1+iy_2) + ib(y_1 +i y_2)$$ This can be rewritten as: $$\frac{dz}{dt} = (a+ib)(y_1+iy_2) $$ $$\frac{dz}{dt} = (a+ib)z $$ $$\frac{dz}{dt} = uz $$ Where $u=a+ib$. Since $u$ is a function of $a,b$.