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Agent_A

  • one year ago

Differential Equations problem. Solve the initial Value Problem: (see the given, below) I have the solution, and I know that we have to use integration by parts, but I'd like a better (clearer) solution, please. Just send me the whole thing. I want to see the way you solve it, in one run. Thanks!

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  1. Agent_A
    • one year ago
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    \[\frac{ dy }{ dt } = y + 2t\] \[y(0) = -2\]

  2. ganeshie8
    • one year ago
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    As a start, multiply \(e^{-t}\) through out

  3. ganeshie8
    • one year ago
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    \(\dfrac{ dy }{ dt } = y + 2t\) \(\dfrac{ dy }{ dt } -y = 2t\) \(\color{red}{e^{-t}}\dfrac{ dy }{ dt } - \color{red}{e^{-t}}y = \color{red}{e^{-t}}2t\) Now, do you notice anythign special about the left hand side ?

  4. Jhannybean
    • one year ago
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    Can this form be defined as \(\dfrac{dy}{dt} - P\cdot y = Q\) ?

  5. Jhannybean
    • one year ago
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    Oh I kind of see it now.

  6. ganeshie8
    • one year ago
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    Yes, it's indeed a linear equation..

  7. Astrophysics
    • one year ago
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    Note the integrating factor is \[e^{-t}\] now what happens when you take the derivative of this \[(e^{-t}y)'\]

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