# 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.

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

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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. –  Aran Komatsuzaki 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

$(tD+1)x=ty$ 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. –  Aran Komatsuzaki 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. –  Aran Komatsuzaki 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. –  Aran Komatsuzaki Oct 28 '13 at 17:12


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