I've been trying to prove the following,
$\sin^2(x) + \cos^2(x) = 1$ from Abel's theorem, and from the fact that $y_1 = \cos(x)$ and $y_2 = \sin(x)$ form a set of fundamental solutions for the standard harmonic equation $y'' + y = 0$, that satisfy the initial conditions at x = 0; $y_1(0)=1, y'_1(0)=0, y_2(0)=0, y'_2(0)=1$
(hint: furthermore you may use the definition of the Wronskian, of course) which kinda confused me, to be honest, but I've decided to use it. Using the definition of the Wronskian I've found
$W = c \cdot exp \left( -\int p(x)dx \right) = ce^{-x}$
$ce^{-x} = y_1y'_2 - y_2y'_1 = -sin^2(x) - cos^2(x)$
Then by using $y_1(0)=1$ and $y_2(0)=0$ we can find $c=1$ therefore $W = -e^{-t}$
Now I know that by simply deriving $sin^2(x)+cos^2(x)=1$ we find $0=0$, which proves it. However, I cannot find a way to prove it with Abel's Theorem, which states:
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if $y_1$ and $y_2$ are solutions of the second order lineaer differential equation $L[y] = y''+p(t)y'+q(t)y=0$
where p and q are continuous on an open interval I, then the Wronskian $W[y_1,y_2](t)$ is given by
$W[y_1, y_2](t) = c \cdot \exp\left(\int p(t)dt\right)$
where c is a certain constant that depends on $y_1$ and $y_2$, but not on $t$. Further, $W[y_1, y_2](t)$ either is zero for all t in I or else is never zero in I
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Now I'm pretty much stuck on what to do, so any help is appreciated