How to solve the system $t\frac{dx}{dt}=-x+yt$, $t\frac{dy}{dt}=-2x+yt$?

Question

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.

$$\frac{dx}{dt}t=-x+yt$$ $$\frac{dy}{dt}t=-2x+yt$$

Answer 1

$(tD+1)x=ty$ 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 $y=w+x$.

Answer 2

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ 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.