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Math Help - Simple one on first-order linear DEs

  1. #1
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    Simple one on first-order linear DEs

    Alright, I know this is a simple exponent issue, but I can't remember if it's x^2 or 2x in this case so bear with me.

    The problem states to find the particular solution that satisfies the initial condition.

    y'+y=6e^x is the DE and the conditions are y=3 when x=0

    P(x)=1 in this case so the IF=e^{\int1dx}=e^x

    Then I did e^x(y'+y)=(6e^x)e^x and that's where I'm forgetting my exponent rules...would (6e^x)e^x become 6e^{x^2} or 6e^{2x}?
    Sorry for the stupid question.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by emttim84 View Post
    Alright, I know this is a simple exponent issue, but I can't remember if it's x^2 or 2x in this case so bear with me.

    The problem states to find the particular solution that satisfies the initial condition.

    y'+y=6e^x is the DE and the conditions are y=3 when x=0

    P(x)=1 in this case so the IF=e^{\int1dx}=e^x

    Then I did e^x(y'+y)=(6e^x)e^x and that's where I'm forgetting my exponent rules...would (6e^x)e^x become 6e^{x^2} or 6e^{2x}?
    Sorry for the stupid question.
    6e^{x}(e^{x})=6(e^{x})^2=6e^{2x}
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