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Math Help - First Order Linear Equations: Am I doing it right?

  1. #1
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    First Order Linear Equations: Am I doing it right?

    Just seeing if I understand this correctly.

    Here is the problem.

    e^x(dy/dx) + (2e^x)y = 1

    I rewrote the problem to this form:
    dy/dx + ((2e^x)/e^x)y = 1/e^x

    then I found e^(int((2e^x)/e^x) and got e^(2x)

    Then I found the  int(e^(2x) * 1/e^x) and got  e^x

    So I wrote the general formula as (e^x + c)/e^(2x)

    At this point I wasn't sure if I was finished. I read a tutorial online that said I take it further to find C but they used an equation that was given at the start. y(0) = 1

    in which case I would input 0 to get C+1 = 1 so C = 0

    But that y(0) = 1 was never given.

    So am I finished then? Is the answer just the general formula of  (e^x + c)/e^(2x) or is there more yet to do. Or did I just flat out do it wrong? Can someone help me!
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  2. #2
    MHF Contributor
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    Hello TheTaoOfBill

    Welcome to Math Help Forum!
    Quote Originally Posted by TheTaoOfBill View Post
    Just seeing if I understand this correctly.

    Here is the problem.

    e^x(dy/dx) + (2e^x)y = 1

    I rewrote the problem to this form:
    dy/dx + ((2e^x)/e^x)y = 1/e^x

    then I found e^(int((2e^x)/e^x) and got e^(2x)

    Then I found the  int(e^(2x) * 1/e^x) and got  e^x

    So I wrote the general formula as (e^x + c)/e^(2x)

    At this point I wasn't sure if I was finished. I read a tutorial online that said I take it further to find C but they used an equation that was given at the start. y(0) = 1

    in which case I would input 0 to get C+1 = 1 so C = 0

    But that y(0) = 1 was never given.

    So am I finished then? Is the answer just the general formula of  (e^x + c)/e^(2x) or is there more yet to do. Or did I just flat out do it wrong? Can someone help me!
    Your answer is correct, but you could have simplified the equation to get:

    \frac{dy}{dx}+2y=e^{-x}

    before using the integrating factor e^{\int2dx}=e^{2x}

    You can easily check that the solution
    y = \frac{e^x+c}{e^{2x}}=e^{-x}+ce^{-2x}
    is correct, by differentiating:
    \frac{dy}{dx}=-e^{-x}-2ce^{-2x}
    and then substituting into the original differential equation.

    Grandad
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