A Little Bit of Diff Eq
The problem
$$\require{cancel}$$
$$y^\prime = 3x^2(y^2 + 1)$$
$$y(x) = tan(x^3 + C)$$
$$y(0) = 1$$
Find derivative
$$y^\prime(x) = {1 \over cos^2(x^3 + C)} \times 3x^2$$
Prove solution
$${1 \over cos^2(x^3 + C)} \times \cancel{3x^2} = \cancel{3x^2} \left[tan^2(x^3 + C) + 1 \right]$$
$${1 \over cos^2(x^3 + C)} = tan^2(x^3 + C) + 1$$
$${1 \over cos^2(x^3 + C)} - 1 = tan^2(x^3 + C)$$
Since:
$${1 \over cos^2(f(x))} - 1 = tan^2(f(x))$$
$$\cancel{tan^2(x^3 + C)} = \cancel{tan^2(x^3 + C)}$$
Find C
$$1 = tan(0 + C)$$
$$atan(1) = C$$
$$C = \pi / 4$$
Unique solution
$$y(x) = tan(x^3 + \pi / 4)$$