How to solve the system $t\frac{dx}{dt}=-x+yt$, $t\frac{dy}{dt}=-2x+yt$? Could you show me how to solve the following simultaneous differential equations? I tried substitution such that $u=xt$, yet I couldn't find the solution.
$$\frac{dx}{dt}t=-x+yt$$
$$\frac{dy}{dt}t=-2x+yt$$
 A: $(tD+1)x=ty$ and $(tD-t)y=-2x$ thus $\frac{1}{-2}(tD+1)(tD-t)y=ty$. Now substitute $y=t^r$ and find a condition for $r$ then you can derive $x$. This is not a Cauchy Euler system. 
Added After Original Answer overlooked an unfortunate $t$: If we write $t\frac{dx}{dt}+x=yt$ and $t\frac{dy}{dt}+2x=yt$ then subtracting yields:
$$ t\frac{d}{dt}\left[ y-x \right]+x=0 $$
Let $w=y-x$ hence $y=w+x$ and we find:
$$ t\frac{dw}{dt}+x=0 \qquad \& \qquad t\frac{dx}{dt}+x=t(w+x) $$
Eliminating $x$ via $x=-t\frac{dw}{dt}$ yields:
$$ t\frac{d^2w}{dt^2}+(2-t)\frac{dw}{dt}+w = 0. $$
I think we can solve this by the series method, then $x = -t \frac{dw}{dt}$ hence we can calculate that and find $y$ from $y=w+x$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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Let's define
$$
\vec{r}\pars{t} \equiv {\,x\pars{t} \choose \,y\pars{t}\,}\,,
\quad
A \equiv
\pars{%
\begin{array}{cc}
0 & 1
\\
0 & 1
\end{array}}\,,
\quad
B \equiv
\pars{%
\begin{array}{cc}
-1 & 0
\\
-2 & 0
\end{array}}\,,
\quad
M\pars{t} \equiv A + {B \over t}
$$
The pair of coupled equations for $x\pars{t}$ and $y\pars{t}$ can be written as
$$
\totald{\vec{r}\pars{t}}{t} = M\pars{t}\vec{r}\pars{t}
$$
Since $\bracks{A,B} \not= 0$, it follows that
$\bracks{M\pars{t},M\pars{t'}} \not= 0$. In Quantum Mechanics the solution is written as
$$
\vec{r}\pars{t}
=
{\rm T}\exp\pars{-\int_{t_{0}}^{t}M\pars{t'}\,\dd t'}\,\vec{r}\pars{t_{0}}
\tag{1}
$$
where $T$ is the $\it\mbox{Dyson Chronological Operator}$. It's well known that it is a 'formal solution'. Its real meaning is, in principle, an infinite serie. Fortunately, in Quantum Mechanics, $M\pars{t}$ usually has 'nice properties' that yield nice theorems to manipulate the Dyson order. The ugly task with Eq. $\pars{1}$ is that $\bracks{M\pars{t}, M\pars{t'}} \not= 0$. See any many-body physics textbook. For example,
this one.
