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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.

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

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I'm curious, where did this problem arise. Are you sure about the formulation? After playing with it for a bit, I wonder, is that $t$ really there? –  James S. Cook Oct 25 '13 at 4:47
I had the same question, so I asked my friend, who asked me about this question, whether or not this system was accurately incited from an introductory ODE book. I haven't gotten any response, but I have the same feeling as yours. –  user12345 Oct 28 '13 at 14:19
I'm actually quite happy it wasn't a simple Cauchy Euler in the end. The answer by Felix Marin has given me something to think about. @Felix Marin –  James S. Cook Oct 28 '13 at 16:16

2 Answers 2

up vote 3 down vote accepted

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

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Thanks for telling me not only the solution but also the type of system which my equations constitute. –  user12345 Oct 24 '13 at 3:10
Did you transform $t\frac{dx}{dy}=-2x+yt$ to $(tD-1)y=-2x$? I think it should be $(tD-t)y=-2x$. After substituting $y=t^r$, I obtained $\frac{r^2+r}{r-1}=t$ from the equation $(tD+1)(tD−t)y=ty$. This system is not Cauchy Euler system, I guess. –  user12345 Oct 24 '13 at 20:10
@User you're right. I'll adjust my answer and see if I can work this (not quite Cauchy Euler) problem out. Sorry, that $t$ just slipped pass me. –  James S. Cook Oct 25 '13 at 3:30
Interestingly, the series method stops at exponent 2 and can only indicate the particular solution $w(t)=t-2$. But then, the substitution $w(t)=(t-2)u(t)$ yields a linear differential equation in $(u',u'')$. Solving it, one sees that $u'(t)=cv(t)$ with $v(t)=\mathrm e^t /(t(t-2))^2$, which yields an explicit formula for $w$ in terms of primitives of $v$ on every interval avoiding the points $t=0$ and $t=2$. Note that the singularity of $v(t)$ at $t=2$ might be cancelled by the prefactor $t-2$... –  Did Oct 25 '13 at 8:13
Thanks for helpfulness of your modification. I also appreciate other answers. I finally could reach the solution. –  user12345 Oct 28 '13 at 17:12

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

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