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anonymous

  • 5 years ago

Determine the solution to the initial value differential equation: 3y'-5y=e^(2x), y(0)=13

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  1. anonymous
    • 5 years ago
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    Use undetermined coefficients or variation of paramaters

  2. anonymous
    • 5 years ago
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    Get the equation in the form y' -5/3*y=e^(2x)/3, then get the multiply both sides by the integrating factor e^(-5x/3) to get \[(e ^{-5x/3}*y)'=1/3*e^{2x}*e^{-5x/3} \rightarrow (e ^{-5x/3}*y)' = 1/3*e^{x/3}\] All you have to do next is take the antiderivative of both sides and then divide both sides by \[e^{-5x/3}\]

  3. anonymous
    • 5 years ago
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    oh oops, nevermind what i said thought it was 3y'' lol jprahman is right

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