Sum of square of function If $f'(x) = g(x)$ and $g'(x) = - f(x)$ for all real $x$ and $f(5) =2 =f'(5)$ then we  have to find $f^2$$(10) + g^2(10)$
I tried but got stuck 
 A: Use Laplace transform:
$$
\begin{cases}
f'(x)=g(x)\\
g'(x)=-f(x)\\
f(5)=f'(5)=2
\end{cases}
$$
Take the Laplace transform of both sides:


*

*$$\mathcal{L}_x\left[f'(x)\right]_{(s)}=\mathcal{L}_x\left[g(x)\right]_{(s)}\Longleftrightarrow s\text{F}(s)-f(0)=\text{G}(s)$$

*$$\mathcal{L}_x\left[g'(x)\right]_{(s)}=\mathcal{L}_x\left[-f(x)\right]_{(s)}\Longleftrightarrow s\text{G}(s)-g(0)=-\text{F}(s)$$


So, we get that:


*

*$$\text{F}(s)=\frac{\text{G}(s)+f(0)}{s}$$

*$$\text{G}(s)=\frac{g(0)-\text{F}(s)}{s}$$


Now use substitution, to get:


*

*$$\text{F}(s)=\frac{sf(0)+g(0)}{1+s^2}$$

*$$\text{G}(s)=\frac{sg(0)-f(0)}{1+s^2}$$


With inverse Laplace transform:


*

*$$f(x)=f(0)\cos(x)+g(0)\sin(x)$$

*$$g(x)=g(0)\cos(x)-f(0)\sin(x)$$


And notice that we can say that $f(0)$ and $g(0)$ are constants.
Using the initial conditions:
$$f(0)=2(\cos(5)-\sin(5)),g(0)=2(\sin(5)+\cos(5))$$
So, using this gives us:
$$f(10)^2+g(10)^2=8$$
A: Suppose 
\begin{eqnarray}
I=\int f'(x) f(x) dx.
\end{eqnarray}
Putting $f(x)=t$, we get $f'(x)~dx=dt$, and thus, we have
\begin{eqnarray}
I=\int t dt=\frac{t^{2}}{2}+c=\frac{(f(x))^{2}}{2}+c,
\end{eqnarray}
where $c$ is a constant of integration. Using this formula, we get
\begin{eqnarray}
0=\int f'(x) f(x) dx+\int g'(x) g(x) dx=\frac{(f(x))^{2}}{2}+\frac{(g(x))^{2}}{2}+c.
\end{eqnarray}
Use $f(5)=2=f'(5)=g(5)$ to solve for $c$. We get $c=-4$. Thus, 
\begin{eqnarray}
\frac{(f(x))^{2}}{2}+\frac{(g(x))^{2}}{2}=4~\text{for all}~x.
\end{eqnarray}
Thus, $(f(10))^{2}+(g(10))^{2}=8$.
