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Consider the system of linear differential equations

\begin{align} x'(t) = \frac{x(t)}{t} - y(t) \tag1\\ y'(t) = \frac{y(t)}{t} + x(t) \tag2 \end{align}

for $t>0$.

An elegant method to solve this system is to multiply $(2)$ by the imaginary unit $i$ and adding that to $(1)$ to obtain

\begin{align} x'(t) + iy'(t) = \frac{x(t) + iy(t)}{t}+ix(t)-y(t) = \frac{x(t) + iy(t)}{t}+i(x(t)+iy(t)) \tag3 \end{align}

Letting $z(t):=x(t)+iy(t)$, $(3)$ becomes

\begin{align} z'(t) = \left(\frac{1}{t}+i\right)z(t) \tag4 \end{align}

which is a seperable equation that can easily be solved. This gives us

\begin{align} z(t)=Cte^{it} \tag5 \end{align}

where $C$ is an arbitrary constant.

From there, $x(t)$ and $y(t)$ arise as the real and imaginary part of $z(t)$.

What I'm having trouble with are the constants of integration. As this is a system of two first-order differential equations, shouldn't there be two constants of integration, one for each equation? What happened to the second one?

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    $C$ is a complex number $C_1+iC_2$, which contains two constants. – 2017-01-01
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    Of course! Thanks a lot. In hindsight, it's obvious. Then $x(t)=t(C_1\cos t - C_2\sin t)$ and $y(t) = t(C_1\sin t + C_2\cos t)$, right? – 2017-01-01
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    Yes, exactly. :) – 2017-01-01

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