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Math Help - Dif EQ T/F

  1. #1
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    Dif EQ T/F

    State whether the following are true or false. Justify your choice.

    a.) Given that y = e^{-t}(\cos{t} + \sin{2t}) solves a 2nd order, linear, homogeneous ordinary Dif EQ (ODE), the ODE in equation would be of the form ay'' + by' + cy = 0 where a,b,c \in \mathbb{R} are constants.

    b.) Given that y_1 and y_2 are fundamental sol'ns to a 2nd order, linear, homogenous ordinary dif. eq (ODE) with each solution defined on an open interval I. On all I, the Wronskian W(y_1,y_2) is either strictly positive or it is strictly negative.
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  2. #2
    Senior Member tukeywilliams's Avatar
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    (a) False, general solution is  y = c_1e^{\lambda t} \cos(\mu t) + c_2e^{\lambda t} \sin(\mu t) . Here  t \neq 2t .

    (b)  W(y_1, y_2)(t) = W(y_1,y_2)(t_0)e^{-\int_{t_0}^{t} p(x) \ dx} = ce^{-\int p(t) \ dt}
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  3. #3
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    Quote Originally Posted by tukeywilliams View Post
    (a) False, general solution is  y = c_1e^{\lambda t} \cos(\mu t) + c_2e^{\lambda t} \sin(\mu t) . Here  t \neq 2t .

    (b)  W(y_1, y_2)(t) = W(y_1,y_2)(t_0)e^{-\int_{t_0}^{t} p(x) \ dx} = ce^{-\int p(t) \ dt}
    So for b, the wronskian has to be strictly positive, and so it's false? Since e^(anything) is always 0 or positive

    Thanks for the help.
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  4. #4
    Senior Member tukeywilliams's Avatar
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     c could be  0  . So the wronskian can be  0 .
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  5. #5
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    Quote Originally Posted by tukeywilliams View Post
     c could be  0 . So the wronskian can be  0 .
    Actually, isn't this true then? Because c could be negative, and hence the wronskian would be too
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  6. #6
    Senior Member tukeywilliams's Avatar
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    Quote Originally Posted by caeder012 View Post
    On all I, the Wronskian W(y_1,y_2) is either strictly positive or it is strictly negative.
    .

    You can just add: On all I, the Wronskian W(y_1,y_2) is either strictly positive,  0 , or strictly negative. So the above statement is false because it does not have  0 .
    Last edited by tukeywilliams; December 2nd 2007 at 08:15 PM.
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  7. #7
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    Got it. One more question, for the first one, you say t \neq 2t. What do you mean by this?
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  8. #8
    Senior Member tukeywilliams's Avatar
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    We need a  \cos \mu t and a  \sin \mu t .

    Instead we have  \cos t and  \sin 2t . We don't have a  \cos t, \ \sin t or  \cos 2t, \ \sin 2t .
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