# "Constant Solutions" Question

• Sep 27th 2011, 02:57 PM
s3a
"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!
• Sep 27th 2011, 11:41 PM
chisigma
Re: "Constant Solutions" Question
Quote:

Originally Posted by s3a
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!

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 and $0 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

$\chi$ $\sigma$
• Sep 28th 2011, 03:18 PM
s3a
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?
• Nov 21st 2011, 09:04 AM
HallsofIvy
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.