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I have:

$\dfrac {dx} {dt}$=$-x+y$

$\dfrac {dy}{dt}$=$-x-y$

and I am trying to find $x(t)$ and $y(t)$ given that $x(0)=0$ and $y(0)=1$

I know to do this I need to decouple the equations so that I only have to deal with one variable but the decoupling is what I am having trouble with

Do I set them equal to each other and then just move like terms to separate sides getting two different equations and then integrate?

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    $\begingroup$ You can write the as a matrix equation $x' = Ax$, find the eigenvalues, vectors and then sub in the ICs. You can also go backward to original DEQ and solve that too. You should get $x(t) = e^{-t} \sin t, y(t) = e^{-t} \cos t$. $\endgroup$
    – Amzoti
    Jul 17, 2013 at 19:42

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Differentiate both w.r.t. t again, we end up with $$ d^2x/dt^2 = -dx/dt+dy/dt $$ $$ d^2y/dt^2 = -dx/dt-dy/dt $$

Then add both of them. Remember that $y=dx/dt + x$

We get
$$ d^2(x+y)/dt^2 = -2dx/dt $$ Substitute the value of $y = dx/dt +x $ You get, $$ d^3x/dt^3 + 2d^2x/dt^2 + 2dx/dt = 0 $$

Mathematica(or setting $dx/dt = u$) will give the answer as $$ x(t) = 1/2 exp(-t) ((c_2-c_1) sin(t)-(c_1+c_2) cos(t))+c_3 $$

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  • $\begingroup$ I am looking for $x(t)$ and $y(t)$ separately $\endgroup$
    – asdfghjkl
    Jul 17, 2013 at 19:50
  • $\begingroup$ sorry, I have edited. Please check. $\endgroup$ Jul 17, 2013 at 19:52
  • $\begingroup$ I think that there is probably a way to decouple these equations into separate equations for $x$ and $y$ without using any derivatives of order higher than $2$, since it is a two-dimensional system to begin with. I'll try to get to it . . . in a while. $\endgroup$ Jul 18, 2013 at 4:21
  • $\begingroup$ @RobertLewis Cher ami, but look, the ultimate differential equation looks like a third order one, when in actuality, by the substitution of ($dx/dt = u$) should make it a second order one. It is infact a well known form of a second order d.e, the damped oscillator! $\endgroup$ Jul 18, 2013 at 6:05
  • $\begingroup$ @Anurag Pallaprolu, achiy, et al: Whoops, I fear I didn't carefully read your equation much past the $\frac{d^3x}{dt^3}$ term! But I still suspect there is a line of attack involving only derivatives or order two or less. Also, the presence of three arbitrary constants in your Mathematica solution is puzzling. Shalom. $\endgroup$ Jul 18, 2013 at 6:53
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If I understand your question correctly, you are seeking for a possiblity to separate the equations as common with some of the systems, by shifting the system parallel to the existing $x$ and $y$ coordinates. In simple words you want to find a way by addition or subtraction or multiplication or division. A fair answer is that this is not possible with this system, you can not just scale it. You need to rotate it.

You would see this from the solution of the equation with

$$x(t) = e^{-t} \sin t$$ $$y(t) = e^{-t} \cos t$$

you see there the trigonometric term and $\sin$ and $\cos$ depicting that $y$ and $x$ are shifted at a rotation of $\pi/2$ against each other.

But a priori you dont have of course the solution and the only way to know about the demand for a rotation is due to the fact that you feel by yourself intuitively the system has a resistence and you can not just stretch it. The positive and negative signs just make it impossible to manage a separation by only scaling - following your words.

There is no other way to solve the system by separation except via transformation in the eigenspace accepting a rotation.

You will see a way to caculation of the eigenvalues and solution of the system here>>>

It wont make sense to copy/paste all.

The solution path for such equations via transformation is standard straight forward and cares about all stretching/scaling and rotation required. You can find it also in other books/literature.

Hope this helps out of all confusion.

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Suppose you change your variables from $(x,y)$ to $(u,v)$ according to the rule $$\begin{cases}x=a u+bv \\y=cu+dv \end{cases}. $$ This can be written in a matrix form $\mathbf x=P \mathbf u$ where $P=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

The ODE transforms accordingly, from $$\frac{ \mathrm d \mathbf x}{\mathrm d t}=A \mathbf x $$ to $$P\frac{ \mathrm d \mathbf u}{\mathrm d t}=AP \mathbf u, $$ or $$\frac{ \mathrm d \mathbf u}{\mathrm d t}=P^{-1}AP \mathbf u. $$ If you choose the "right" matrix $P$ the equations for $u,v$ become uncoupled and you can solve them. Then it takes a simple multiplication by $P^{-1}$ to get back to $x,y$.

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    $\begingroup$ isnt this practically identical to solving the system through transformation into eigenspace via evalues and evectors? $\endgroup$ Jul 18, 2013 at 5:38
  • $\begingroup$ @al-Hwarizmi It is. But I wasn't sure that OP was familiar with the method, so I elaborated. $\endgroup$
    – user1337
    Jul 18, 2013 at 8:40

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