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anonymous

  • 5 years ago

Find a particular solution of y"+y=2e^x

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  1. helpmeplease
    • 5 years ago
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    Use the annihilator method on the 2e^x. Think, what differential operator would result in the answer.

  2. anonymous
    • 5 years ago
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    is there a formula i can follow? this is supposed to be a non homogeneous second order differential equation.

  3. helpmeplease
    • 5 years ago
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    Well, you need to use the differential operator concept. What D would destroy 2e^x? Then use that differential operator and create the particular solution. Then plug that solution into the function, and see what the coefficient is.

  4. anonymous
    • 5 years ago
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    (e^x)"+(e^x)=2e^x

  5. helpmeplease
    • 5 years ago
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    D-1 = Ae^(1x) yp = Ae^(1x) e^x = Ae^x+A^ex = 2e^x

  6. helpmeplease
    • 5 years ago
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    sorry, function, not e^x

  7. helpmeplease
    • 5 years ago
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    also, use t for 1.

  8. helpmeplease
    • 5 years ago
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    I mangled that a little bit. Your y_p = Ae^x. You have to determine what A is. To do that, substitute the y_p back into the function, and solve for A.

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