# Prove Two Functions are Equal?

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.

• Are $A,B$ constants or functions of $x,y,t$ ? – Yves Daoust Mar 4 '20 at 13:52
• @YvesDaoust they are constants – Clark Mar 4 '20 at 15:38

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.

• 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?) – Clark Mar 4 '20 at 13:13
• @Clark, yes; and fixed. – vonbrand Mar 4 '20 at 13:15
• Awesome, thanks! It's all coming back – Clark Mar 4 '20 at 13:18

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 $$t\geq t_0$$.

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

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

• 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? – Clark Mar 4 '20 at 15:47
• @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.) – Yves Daoust Mar 4 '20 at 15:50