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Math Help - behavior of solutions

  1. #1
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    behavior of solutions

    If a>0, b>0, c=0, show that all solutions of ay''+by'+cy=0 approach a constant that depends on the initial conditions as t approaches to infinity. Determine this constant for the initial conditons y(0)=y_0, y'(0)=y_0'
    -------------------------------------------------------
    my approach:
     y=c_1e^{r_1 t}+c_2e^{r_2 t}
    b^2-4ac=b^2
    r_1=0, r_2= \frac{-b-b}{2a}
    y=c_1e^0+c_2e^{\frac{-bt}{a}}
    y(0)=c_1=y_0
    y'(t)=c_2(-{\frac{b}{a})e^{\frac{-bt}{a}}}
    t=0, c_2=y_0'(-{\frac{a}{b}})
    y=c_1e^0+c_2e^{\frac{-bt}{a}}
    =c_1+c_2e^{\frac{-bt}{a}}
    =y_0+{y_0'}(-{\frac{a}{b}})e^{\frac{-bt}{a}}
    which approaches to y_0 as t approaches to infinity.

    (according to the solution manual, the constant should be y_0+{\frac{a}{b}}y_0' )

    please help me out. thanks in advance.
    Last edited by elmo; October 13th 2009 at 02:45 PM.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello elmo
    Quote Originally Posted by elmo View Post
    If a>0, b>0, c=0, show that all solutions of ay''+by'+cy=0 approach a constant that depends on the initial conditions as t approaches to infinity. Determine this constant for the initial conditons y(0)=y_0, y'(0)=y_0'
    -------------------------------------------------------
    my approach:
     y=c_1e^{r_1 t}+c_2e^{r_2 t}
    b^2-4ac=b^2
    r_1=0, r_2= \frac{-b-b}{2a}
    y=c_1e^0+c_2e^{\frac{-bt}{a}}
    \color{red}y(0)=c_1=y_0
    y'(t)=c_2(-{\frac{b}{a})e^{\frac{-bt}{a}}}
    t=0, c_2=y_0'(-{\frac{a}{b}})
    y=c_1e^0+c_2e^{\frac{-bt}{a}}
    =c_1+c_2e^{\frac{-bt}{a}}
    =y_0+{y_0'}(-{\frac{a}{b}})e^{\frac{-bt}{a}}
    which approaches to y_0 as t approaches to infinity.

    (according to the solution manual, the constant should be y_0+{\frac{a}{b}}y_0' )

    please help me out. thanks in advance.
    See the line I've highlighted in red. Can you see your mistake? e^0=1, not 0.

    Grandad
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