I am struggling with

If $f, g$ satisfy $f'=g, g'=f, f(0)=1, g(0)=0$, find $[f(10)+g(10)]\cdot [f(10)-g(10)]$.

And the following flash across my mind.enter image description here

So, I guess the answer of $[f(10)+g(10)]\cdot [f(10)-g(10)]$ is $1$.

Could you please solve or prove this. Really thank you.

  • $\begingroup$ What? Please make your post more legible. $\endgroup$ – user532449 Feb 19 '18 at 1:49
  • $\begingroup$ You have a good future ahead of you. That was a good thought. But clearly $f (x)=g (x)=e^x $ is a solution. can you combine those reasults. $\endgroup$ – fleablood Feb 19 '18 at 2:22
  • $\begingroup$ You don't actually have to solve fo f and g I think. Just express those in terms of derivatives and... something will probably happen. $\endgroup$ – fleablood Feb 19 '18 at 2:24
  • $\begingroup$ Oh... wait ... f=g=e^x obvious don't satisfies f (0)!=g (0). $\endgroup$ – fleablood Feb 19 '18 at 2:25

You want $f^2-g^2$. Differentiating that, gives $2ff’-2gg’=0$ by the given conditions, so $f^2-g^2$ is a constant. Plug in $x=0$ to get the constant.

  • $\begingroup$ Shortest possible answer. Does exactly what is needed and nothing more. +1 $\endgroup$ – Paramanand Singh Feb 19 '18 at 3:38
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    $\begingroup$ MARVELOUS... wow... $\endgroup$ – user143993 Feb 19 '18 at 4:16

Congrats, you've just solved a simple dynamical system:

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x'\\ y'\\ \end{bmatrix} = \begin{bmatrix} y\\ x\\ \end{bmatrix}$$

with initial condition:

$$ \begin{bmatrix} x(0)\\ y(0)\\ \end{bmatrix} = \begin{bmatrix} 1\\ 0\\ \end{bmatrix}$$

The above has general solution:

$$\begin{bmatrix} x(t)\\ y(t)\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2}c_1e^{-t}(e^{2t}+1) + \frac{1}{2}c_2e^{-t}(e^{2t}-1)\\ \frac{1}{2}c_1e^{-t}(e^{2t}-1) + \frac{1}{2}c_2e^{-t}(e^{2t}+1)\\ \end{bmatrix}$$

using the initial conditions, we get the following system of equations: $$ \begin{bmatrix} c_1\\ c_2\\ \end{bmatrix} = \begin{bmatrix} 1\\ 0\\ \end{bmatrix}$$

Therefore the particular solution is:

$$x(t) = \frac{e^t + e^{-t}}{2},$$ $$y(t) = \frac{e^t - e^{-t}}{2}$$

This solution is unique since the matrix mapping is continuous.

Note that you can also look at the system as a standard second order ODE with:

$$x(t) = x''(t)$$

and initial conditions: $$x(0) = 1$$ $$x'(0) = 0$$

particular solution is:

$$x(t) = \frac{e^t + e^{-t}}{2}$$

$x'(t)$ follows via differentiation.

For more info, check out:

  1. https://www.ru.ac.za/media/rhodesuniversity/content/mathematics/documents/thirdyear/linearcontrol/AM32LC2%20Linear%20Dynamic%20Sys.pdf

  2. https://en.wikipedia.org/wiki/Linear_dynamical_system

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    $\begingroup$ Wait a sec.. Is it differential equation, isn't it? . But I met this PROBLEM in high school math... $\endgroup$ – user143993 Feb 19 '18 at 2:30
  • $\begingroup$ Yup... system of coupled ode's. Check out some refs. If you know a little linear algebra, you can grok it! $\endgroup$ – EDZ Feb 19 '18 at 2:31
  • $\begingroup$ Also, just because something arose in high school math, doesn't mean it doesn't have deeper implications — watch out for that and keep asking why! $\endgroup$ – EDZ Feb 19 '18 at 4:29

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