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Math Help - finding solution by initial conditions

  1. #1
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    finding solution by initial conditions

    The solution of a certain differential equation is of the form

    y(t)=aexp(4t)+bexp(6t), where a and b are constants.
    The solution has initial conditions y(0)=1 and y'(0)=3.
    Find the solution by using the initial conditions to get linear equations for a and b.

    can anyone help me with this one.
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  2. #2
    Super Member Random Variable's Avatar
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     y(t) = Ae^{4t} + Be^{6t}

     y(0) = 1= Ae^{0} + Be^{0} = A+B


     y'(t) = 4Ae^{4t} + 6Be^{6t}

     y'(0) =3 = 4Ae^{0} + 6Be^{0} = 4A+6B

    so solve the system

    A+B=1
    4A+6B=3
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  3. #3
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    i dont know really what to do.
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  4. #4
    Super Member Random Variable's Avatar
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    multipy the first equation by -4 and add it to the second equation to get 2B=-1 or  B= -\frac{1}{2}

    then  A - \frac{1}{2} = 1 or  A = \frac{3}{2}

    so  y(t) = \frac{3}{2}e^{4t} - \frac{1}{2}e^{6t}
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