# How to show $y(T) = x(1)$ for $y'=f(y)$ and $x'=Tf(x)$

Suppose I have two ODEs, $$\frac{dy}{dt}=f(y), \ \ \ \frac{dx}{dt}= Tf(x),$$ where $$t$$ is the time and $$T$$ is the terminal time. Also $$x(0)=y(0)=c$$.

How to show that $$y(T)=x(1)$$? This means the value of $$y(T)$$ for the LHS ODE is equal to $$x(1)$$ of the RHS ODE.

My effort: I only know that for any $$t_p\in [0,T]$$, LHS ODE $$\frac{dy}{dt}|_{t=t_p}= f(y(t_p))$$however, RHS ODE $$\frac{dx}{dt}|_{t=t_p}= Tf(x(t_p)),$$ i.e., at each point, the slope is $$T$$-times the previous one.

But I still have no idea about how to show $$x(1) = y(T)$$?

Consider an example:

Suppose $$x(0)=1$$, and consider the following MATLAB code

tf = 4;
x0 = 1;
tspan = [0 tf];
sspan = [0 1];
x0 = 1;
[t,x] = ode45(@(t,x) x, tspan, x0);
[s,y] = ode45(@(s,y) tf*y, sspan, x0);


In this case, $$x(4)=x(1)$$ (blue line is the second system.)

I just edited my question so I am sorry for the first answer from Kavi Rama Murthy.

• This is awesomely unclear. It would be good to use different letters for the two equations: $x'=f(x)$, $y'=Tf(y)$... Commented Jun 28, 2019 at 22:45
• Btw you can't possibly show $y(1) = x(T)$, because you didn't specify initial conditions - both equations have infinitely many solutions. Commented Jun 28, 2019 at 22:47
• @DavidC.Ullrich Sorry for that, I modify it a bit. Commented Jun 29, 2019 at 2:01
• @Denny You changed the question completely and then commented that my answer is wrong. You should mention in your question that you have edited the question. Commented Jun 29, 2019 at 4:45
• @KaviRamaMurthy I added that in my question. Sincerely sorry for my confusing statement of the previous edition of my questions! Commented Jun 29, 2019 at 4:52

I start with the claim that $$y(T \times t) = x(t)$$. Here, we can check that $$t = 0$$ we obtain $$y(0) = x(0) = c$$ for some constant $$c$$. Furthermore, at $$t = 1$$, we have $$y(T) = x(1)$$ and we are done.
From $$\frac{dy(\xi)}{d\xi} = f(y(\xi))$$, we pick $$\xi = Tt$$. Hence,
$$\frac{dy(\xi)}{d\xi} = f(y(\xi)) \iff \frac{dy(Tt)}{d(Tt)} = f(y(Tt)) \iff \frac{d(x(t))}{d(Tt)} = f(x(t)) \iff \frac{d(x(t))}{d(t)} = Tf(x(t))$$
Thus, $$y(T \times t) = x(t)$$ indeed satisfy both differential equations.
• You want to verify the claim that $y(Tt) = x(t)$. Why could you say $T(f(x(t))) = Tf(y(Tt))$? You have not done this, right? Commented Jun 29, 2019 at 2:40
• I feel that I just have to verify the claim, ie I want to say that all my solutions of $y$ and $x$ satisfy this relation of $y(Tt) = x(t)$. I can indeed check that this holds for both equations. Simply said, the two differential equations are related under a linear transformation. So if I want to probe the behavior in one of the differential equation, I can use the transformation to probe the behavior in the other equation. Commented Jun 29, 2019 at 2:55