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.
 A: 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.
A: 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:


*

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

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