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:
$ln|tan(y/2)| = {x^2 /2} + C$
$tan(y/2) = e^{x^2} + C$
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 How can I get to $y(x) = π − arctan ( 2 e^{( x^ 2/ 2 )}/e^{(x^2)}-1)$