y'' = 3 => y' = 3x + C1
y'(2) = 4 => 4 = 3 * 2 + C1 => C1 = -2
y' = 3x - 2 => y = 3/2 * x^2 - 2x + C2
y(2) = 3 => 3 = 3/2 * 2^2 - 2 * 2 + C2
3 = 6 - 4 + C2 => C2 = 1
Ответ: y = 3/2 * x^2 - 2x + 1