comment on degree of f(x) and g(x) let f be a twice differentiable function such that f"(x)=-f(x) and f'(x)=g(x). if h'(x)= [{f(x)}^2 +{g(x)}^2] , h(1)=8 and h(0)=2 , then h(2)=?. i approached this question by twice differentiating h(x) and as it comes out to be 0 so i can say that h(x) is a linear function. now h'(x)= some constant and in question it is given that h'(x)=[{f(x)}^2+{g(x)}^2] so from here can i comment on degree of f(x) and g(x) or it's not possible to say anything 
 A: If $h'(x)=a$, then $h(x)=ax+b$. From your conditions $h(0)=2$ means $b=2$ and $h(1)=8$ means $a=6$, then $h(2)=14$.
In case you are wondering, the solution for the equation $f''(x)=-f(x)$ has the form $A\sin(x+\phi)$
A: Yes, you are right: $h(x)=ax+b$. $h(0)=2$ gives $b=2$. Then $h(1)=8$ shows:
$8=a+b=a+2$, hence $a=6$.
Conclusion: $h(x)=6x+2$, therefore $h(2)=14$, and you are done !
A: Regarding your question to Andrei about the form of $f$.
Plan: $f$ behaves similarly to $\cos$ in that $f''=f$, so use it to define some sort of "exponential function" $E$ and use the standard proof of the uniqueness (up to scalar multiple) of the differential equation $\frac{d}{dx} \exp(x) = \exp(x)$.
You can define the smooth function $E : \mathbb{R} \to \mathbb{C}$ like so
$$E(x) := f(x)+if'(x)$$
Then $E'(x) = -iE(x)$ and we can compare it to the complex exponential function:
$$(e^{ix}E(x))' = ie^{ix}E(x)-ie^{ix}E(x) = 0$$
Thus $E(x) = \lambda e^{-ix}$ for some $\lambda \in \mathbb{C}$.
We can write in polar coordinates $\lambda = Ae^{i\phi}$.
By Euler's formula
$$ f(x) = \Re(\lambda e^{-ix}) = A\Re(e^{i(\phi-x)}) = A\cos(\phi-x)$$
