I need to solve this question:
Consider the IVP $y' = f(x, y)$, for $f(x, y) = x \sin(y)$ and $y(0) = π/2$ for $x ∈ [0, 3] =: I$.
Verify that $y(x) = π − \arctan ( 2 e^{x^ 2/ 2}/e^{x^2}-1)$ solves the IVP
Here is what I have so far:
Integrating on both sides, I get:
$\ln{\lvert\tan(y/2)\rvert} = {x^2 /2} + C$
$\tan(y/2) = e^{x^2} + C$
inserting $x=\pi/2$ and $y=0$ gives us, $C=0$
$y = (1/2)\arctan(e^{x^2} )$
I don't know how to proceed further to get