(a) It looks like the ODE is
y' = 4x √(1 - y ^2)
which is separable:
dy/dx = 4x √(1 - y ^2) => dy/√(1 - y ^2) = 4x dx
Integrate both sides. On the left, substitute y = sin(t ) and dy = cos(t ) dt :
∫ dy/√(1 - y ^2) = ∫ 4x dx
∫ cos(t ) / √(1 - sin^2(t )) dt = ∫ 4x dx
∫ cos(t ) / √(cos^2(t )) dt = ∫ 4x dx
∫ cos(t ) / |cos(t )| dt = ∫ 4x dx
Since we want the substitutiong to be reversible, we implicitly assume that -π/2 ≤ t ≤ π/2, for which cos(t ) > 0, and in turn |cos(t )| = cos(t ). So the left side reduces completely and we get
∫ dt = ∫ 4x dx
t = 2x ^2 + C
arcsin(y) = 2x ^2 + C
y = sin(2x ^2 + C )
(b) There is no solution for the initial value y (0) = 4 because sin is bounded between -1 and 1.