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Math Help - "Constant Solutions" Question

  1. #1
    s3a
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    "Constant Solutions" Question

    I have no idea how to approach part (a) of this question. As for part (b), I thought I had a plan but it then failed.

    (The question and my work are both attached as PDF files).

    Any help in figuring out how to do this problem would be greatly appreciated!
    Thanks in advance!
    Attached Files Attached Files
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: "Constant Solutions" Question

    Quote Originally Posted by s3a View Post
    I have no idea how to approach part (a) of this question. As for part (b), I thought I had a plan but it then failed.

    (The question and my work are both attached as PDF files).

    Any help in figuring out how to do this problem would be greatly appreciated!
    Thanks in advance!
    Starting with...

    \frac{d y}{dt}= - y^{4} + 4\ y^{3} +21\ y^{2}= f(y) (1)

    ... it is evident that the 'constant solutions' are the zeroes of f(y), i.e. y(t)=0, y(t)=7 and y(t)=-3. Regarding the question b) is also evident that y(t) is increasing for -3<y<0 and 0<y<7 and is decreasing for y>7 and y<-3...

    A property of an ODE like (1) is that y^{'} is function of the y alone and that means that, if y(t) is solution of (1), then y(t+\tau)\ ,\ \tau \in \mathbb{R} is also solution of (1). A consequence of that is that for all the initial value y(0) different from 0, 7 and -3 the (1) has only one solution, but that isn't true if y(0)=0, y(0)=7 or y(0)=-3...

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 29th 2011 at 12:13 AM. Reason: eliminated errors according to s3a's post...
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  3. #3
    s3a
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    Re: "Constant Solutions" Question

    Thank for your answer but I still have some problems (Sorry if I did not understand some stuff). I now know (but not fully understand) that the constant solutions are -3,0, and 7 and that y is increasing when -3 < y < 7. (I agree with you that y decreases when -3 < y < 0 but that is apparently not the answer - unless I am just misinterpreting the answer).

    But, look at this plotted graph: http://www.wolframalpha.com/input/?i=constant+solutions+of+dy%2Fdt+%3D+-y^4+%2B+4y^3+%2B+21y^2

    Also, y'' = -4y^3 + 12y^2 + 42y, and if you plug in y = 6.9 as one example, you will get a negative answer which means it's decreasing so I don't see how -3 < y < 7 is true.

    Could you clear this up as well please?
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  4. #4
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    Re: "Constant Solutions" Question

    The differential equation is
    \frac{dy}{dt}= -y^4+ 4y^3+ 21y^2= -y^2(y^2- 4y- 21)= -y^2(y- 7)(y+ 3)
    The values y= 0, y= -3, and y= 7 make the right side of that 0, which means that dy/dt= 0 which means y is a constant.

    The crucial point is that y- a< 0 if y< a, y- a> 0 if y> a. If x< -3, all of the factors, -1, y= y- 0, y- 7, and y+ 3= y- (-3) are negative. There are a total of 5 such factors and the product of 5 negative numbers is negative. dy/dt is negative so y is decreasing. For -3< x< 0, y+ 3 is now positive but -1, y and y- 7 are still negative. There are four (don't forget to count y twice) negative factors so dy/dx is now positive and dy/dt is increasing. For 0< x< 7, -1 and y+ 7 are the only negative factors. Two two negative factors means that dy/dx is positive and y is still increasing (it is the fact that y is squared that causes that). If 7< x, only the "-1" factor is negative so dy/dt is negative and y is decreasing again.


    y'' tells you the convexity, NOT whether y is increasing or decreasing so the fact that y'' is positive or negative is not relevant.
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