Solve the initial value problem.
y' = 4x3, y(2) = 20
First we find the general solution. If y' = 4x3 then
y = x4 + C.
Since we're given the initial condition that y = 20 when x = 2, we have
20 = (2)4 + C = 16 + C.
This means C = 4 and so the particular solution is
y = x4 + 4.