Question

3.

Answer 1

(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.