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Given two generic functions $x(t)$ and $y(t)$ I want to prove that $x(t) = y(t)$.

To do so, I take the derivative, which turn out to be: $$\dot x(t) = Ax(t) + B$$ $$\dot y(t) = Ay(t) + B$$ where $A$ and $B$ are the same in both derivatives.

Is this sufficient to say both functions are equal?

The reason I ask is that, in general, $x(t)$ and $y(t)$ could have vastly different forms (ie. $x(t)$ could be the resultant of a complicated integral $y(t)$ or something similar). Because of this, I am wondering if I have to go through the trouble of reducing $x(t)$ to $y(t)$ or vice-versa.

Thanks in advance!

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  • $\begingroup$ Are $A,B$ constants or functions of $x,y,t$ ? $\endgroup$ – Yves Daoust Mar 4 '20 at 13:52
  • $\begingroup$ @YvesDaoust they are constants $\endgroup$ – Clark Mar 4 '20 at 15:38
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They both satisfy the linear first order differential equation:

$\begin{equation*} f'(t) = A f(t) + B \end{equation*}$

You need at least one other condition, say prove that $x(t_0) = y(t_0)$ for some value $t_0$, or perhaps the same derivative at a point. Note that the solution is

$\begin{equation*} f(t) = c e^{A t} - B/A \end{equation*}$

here $c$ is an unknown constant, to be determined by other conditions.

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    $\begingroup$ do you mean $x(t_0) = y(t_0)$? So if I can prove both functions have the same value at one point, they should have the same value at all points? (ie. from existence and uniqueness?) $\endgroup$ – Clark Mar 4 '20 at 13:13
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    $\begingroup$ @Clark, yes; and fixed. $\endgroup$ – vonbrand Mar 4 '20 at 13:15
  • $\begingroup$ Awesome, thanks! It's all coming back $\endgroup$ – Clark Mar 4 '20 at 13:18
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If $x(t)$ and $y(t)$ are $C^1$ functions such that $x(t_0) = y(t_0)$ for some $t_0$ furthermore $x'(t) = y'(t) = f(t)$ for some function (Lipschitz-continuous) $f$, then according to the uniqueness of the solution of a differential equation you have $x(t)=y(t)$ for all $t\geq t_0$.

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By subtraction,

$$\dot x-\dot y=\dot{(x-y)}=A(x-y)$$ so that the two given functions can differ by a solution of this equation, given by

$$x-y=Ce^{At}.$$

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  • $\begingroup$ This doesn't contradict the other answers right? Let's say $x(t_0) = y(t_0) = 0$ where $t_0 = 0$ for simplicity. This would then give $C = 0$ and, therefore $x(t) = y(t)$ correct? $\endgroup$ – Clark Mar 4 '20 at 15:47
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    $\begingroup$ @Clark: absolutely. My goal was to show how to solve the question without integrating the original equations and focusing on the difference. (Though the benefit is tiny.) $\endgroup$ – Yves Daoust Mar 4 '20 at 15:50

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