Calculus of a Single Variable, Chapter 4, 4.1, Section 4.1, Problem 37

You need to use direct integration to evaluate the general solution to the differential equation:
int (8t^3 + 5)dt = int 8t^3 dt + int 5dt
int (8t^3 + 5)dt = 8t^4/4 + 5t + c
int (8t^3 + 5)dt = 2t^4 + 5t + c
You need to find the particular solution using the information provided by the problem, that h(1) = -4, such that:
h(1) =2*1^4 + 5*1 + c => -4 = 7 + c => c = -4 - 7 => c = -11
Hence, evaluating the particular solution to the given differential equation yields h(t) = 2t^4 + 5t - 11.