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Math Help - Find a solution of the second-order IVP

  1. #1
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    Find a solution of the second-order IVP question (First chapter in my Diff Eq's book)

    Hey all. I have this problem:

    x=c_{1}e^x+c_2e^{-x}

    I am supposed to use this to find out what c_1 and c_2 are:

    y(0)=1
    y'(0)=2

    What happens when I use the first equation is I get
    1=c_{1}e^0+c_2e^{-0}
    1=c_{1}(1)+c_2(1)
    1=c_{1}+c_2

    leaving me know way to solve for them. The answer in the back of the book says:

    y=\frac{3}{2}e^x-\frac{1}{2}e^{-x}

    Can anyone show me what I should do? If this one is answered, I will be able to complete all the other problems. Because I'm getting stuck on the same thing, having 1 equation and 2 variables c_1 and c_2

    Thanks,
    DHS1
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by DHS1 View Post
    Hey all. I have this problem:

    x=c_{1}e^x+c_2e^{-x}

    I am supposed to use this to find out what c_1 and c_2 are:

    y(0)=1
    y'(0)=2

    What happens when I use the first equation is I get
    1=c_{1}e^0+c_2e^{-0}
    1=c_{1}(1)+c_2(1)
    1=c_{1}+c_2

    leaving me know way to solve for them. The answer in the back of the book says:

    y=\frac{3}{2}e^x-\frac{1}{2}e^{-x}

    Can anyone show me what I should do? If this one is answered, I will be able to complete all the other problems. Because I'm getting stuck on the same thing, having 1 equation and 2 variables c_1 and c_2

    Thanks,
    DHS1
    I assume you meant y=C_1e^x+C_2e^{-x}. Just plug and calculate y(0)=C_1e^0+C_2e^{-0}=C_1+C_2=1 and y'(0)=C_1e^{0}-C_2e^{-0}=C_1-C_2=2
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  3. #3
    MHF Contributor Calculus26's Avatar
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