How to solve this initial-value problem (differential eqns)

(y^3)*(dy/dx)=(8y^4+14)*cos(x); y(0)=C

Then you need to put the equation in the form of y^4 = ...

when I initially tried to solve this, I got 7(8*sin(x)-((1/7)*(y^4)))

(y^3) (dy/dx) = (8y^4 + 14)*cos(x); y(0) = C

(y^3)/(8y^4 + 14) dy = cos(x) dx

Integratring both sides:

1/32*ln(8y^4 + 14) = sin(x) + D

ln(8y^4 + 14) = 32*sin(x) + D

8y^4 + 14 = D*e^(32*sin(x))

8C^4 + 14 = D

y^4 = [(8C^4 + 14)*e^(32*sin(x)) - 14]/8

It is rather unusual to have an initial condition involving the constant of integration as you have here, and it gives complex numbers when solving for the constant.

Find the general solution by separating the variables then integrating:
y³(dy / dx) = (8y⁴ + 14)cosx
y³ / (8y⁴ + 14) dy = cosx dx
∫ 32y³ / (8y⁴ + 14) dy / 32 = ∫ cosx dx
ln(8y⁴ + 14) / 32 = sinx + C
ln(8y⁴ + 14) / 32 = C + sinx
ln(8y⁴ + 14) = C + 32sinx
8y⁴ + 14 = ℮^(C + 32sinx)
8y⁴ + 14 = ℮ᶜ℮^(32sinx)
8y⁴ + 14 = C℮^(32sinx)
8y⁴ = C℮^(32sinx) - 14
y⁴ = C℮^(32sinx) - 7 / 4

Find the particular solution by solving for the constant:
When x = 0, y = C
C - 7 / 4 = C⁴
4C - 7 = 4C⁴
4C⁴ - 4C + 7 = 0
C ≈ 0.824 ± 0.613i, -0.824 ± 0.991i
y⁴ = (0.824 + 0.613i)℮^(32sinx) - 7 / 4, y⁴ = (0.824 - 0.613i)℮^(32sinx) - 7 / 4, y⁴ = (-0.824 + 0.991i)℮^(32sinx) - 7 / 4, y⁴ = (-0.824 - 0.991i)℮^(32sinx) - 7 / 4

separetion variable method and integrate both sides --->(1/16)ln(4y^4+7) = 2sinx + K , find K in terms of C then y^4 = ...