anonymous
  • anonymous
how to solve this Initial value Diff equation? y"=6t y(0)=3 y'(0)=-1
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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 \[y'(t)=3t^2-1\]
anonymous
  • anonymous
repeat the process to get \(y\)
anonymous
  • anonymous
thanks

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