Question

how to solve this Initial value Diff equation?
y"=6t
y(0)=3
y'(0)=-1

Answer 1

find the anti derivative twice
first time you get \(y'=3t^2+C\) and since \(y'(0)=-1\) you know \(C=-1\) and you have
\[y'(t)=3t^2-1\]

Answer 2

repeat the process to get \(y\)

Answer 3

thanks