Proof of Euler identity using ODE

I am writing an article to prove Euler identity :$e^{i\pi}+1=0$
Here the main part:
Consider the function :$ \mathbb{R} \rightarrow \mathbb{C}, f(x)=e^{ix}$
Differentiating twice,we get : $f''(x)=-f(x)$
So the function $f$ satisfies the linear 2nd order ODE : $y''+y=0$,which has general solution $y= a\cos x + b\sin x$,where $a$ and $b$ are complex constants
Is my reasoning correct ?
Edit:I will write the complete reasoning ( as recommended by a comment):
Now as we know that $f(x)=e^{ix} = a\cos x+ b\sin x$
Letting $x = 0$,we get $1 =a$,So $f(x) = \cos x + b\sin x$
Letting $x = \pi$,we get $f(\pi)=e^{i\pi}=-1+0$,and so $e^{i\pi}+1=0$,as required .