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haganmc

  • 3 years ago

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

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  1. anonymous
    • 3 years ago
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    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 \[y'(t)=3t^2-1\]

  2. anonymous
    • 3 years ago
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    repeat the process to get \(y\)

  3. haganmc
    • 3 years ago
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    thanks

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