Sum of square of function

Question

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

Answer 1

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, $(f(10))^{2}+(g(10))^{2}=8$.

Answer 2

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:

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

So, we get that:

1. $$\text{F}(s)=\frac{\text{G}(s)+f(0)}{s}$$
2. $$\text{G}(s)=\frac{g(0)-\text{F}(s)}{s}$$

Now use substitution, to get:

1. $$\text{F}(s)=\frac{sf(0)+g(0)}{1+s^2}$$
2. $$\text{G}(s)=\frac{sg(0)-f(0)}{1+s^2}$$

With inverse Laplace transform:

1. $$f(x)=f(0)\cos(x)+g(0)\sin(x)$$
2. $$g(x)=g(0)\cos(x)-f(0)\sin(x)$$

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:

$$f(10)^2+g(10)^2=8$$