# Find $f, g$ such that $f'=g, g'=f, f(0)=1, g(0)=0.$

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. 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.

• What? Please make your post more legible. – user532449 Feb 19 '18 at 1:49
• 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. – fleablood Feb 19 '18 at 2:22
• 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. – fleablood Feb 19 '18 at 2:24
• Oh... wait ... f=g=e^x obvious don't satisfies f (0)!=g (0). – fleablood Feb 19 '18 at 2:25

## 2 Answers

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.

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

• Wait a sec.. Is it differential equation, isn't it? . But I met this PROBLEM in high school math... – user143993 Feb 19 '18 at 2:30
• Yup... system of coupled ode's. Check out some refs. If you know a little linear algebra, you can grok it! – EDZ Feb 19 '18 at 2:31
• 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! – EDZ Feb 19 '18 at 4:29