• Support PF! Buy your school textbooks, materials and every day products Here!

A function y(t)

  • Thread starter Jbreezy
  • Start date
  • #1
582
0

Homework Statement



A function y(t) satisfies the differential equation dy/dt = y^4 - 6y^3 +5y^2

Homework Equations


Part a. What are the constant solutions of the equation5?
I just set it equal to 0 and solved. I got y = 0, y = 5, y = 1


The Attempt at a Solution



b. For what values of y is y increasing?
c. For what values of y is y decreasing?

So for this one I was thinking Since I know where the eq is not changing. I could make up intervals and just do the first derivative test?


So I did. I had 0, 5,1 and I tested -1 (1/2), 2, and 6 for the intervals.

I just plugged in my test values into my equation the very first one I wrote on here.
So am I doing this right? I'm confused on how to report my answer.
It is greater than 0 from up until my test number of 2. But I don't know..ah
It is negative before that also I just checked. It is negative at 1.5
 

Answers and Replies

  • #2
582
0
Second derivative would be better
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement



A function y(t) satisfies the differential equation dy/dt = y^4 - 6y^3 +5y^2

Homework Equations


Part a. What are the constant solutions of the equation5?
I just set it equal to 0 and solved. I got y = 0, y = 5, y = 1


The Attempt at a Solution



b. For what values of y is y increasing?
c. For what values of y is y decreasing?

So for this one I was thinking Since I know where the eq is not changing. I could make up intervals and just do the first derivative test?


So I did. I had 0, 5,1 and I tested -1 (1/2), 2, and 6 for the intervals.

I just plugged in my test values into my equation the very first one I wrote on here.
So am I doing this right? I'm confused on how to report my answer.
It is greater than 0 from up until my test number of 2. But I don't know..ah
It is negative before that also I just checked. It is negative at 1.5
There are four y-intervals in which the right-hand-side is nonzero; these are ##(-\infty,0), (0,1), (1,5)## and ##(5,\infty)##. What is the sign of the right-hand-side in each interval, and how can you tell?
 
  • #4
582
0
Test the original equation to see what the sign is? For the intervals you wrote I have + , +,- , + This is from testing values in the equation.
 
  • #5
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751
There are four y-intervals in which the right-hand-side is nonzero; these are ##(-\infty,0), (0,1), (1,5)## and ##(5,\infty)##. What is the sign of the right-hand-side in each interval, and how can you tell?
Probably more than the problem intends, but there is more than meets the eye to this. Try to draw a graph of y = f(x) crossing y = 5 satisfying the required slope signs and continuous derivative.
 
  • #6
582
0
I can't do that.
 
  • #7
582
0
So what my stuff is wrong?
 
  • #8
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751
I can't do that.
So what my stuff is wrong?
No, no, your derivative signs are correct, and that's all you are asked to do. I'm guessing you are soon going to be studying direction fields which will give you more insight to what the solutions to this problem must look like.
 
  • #9
582
0
LCKurtz you are good. That is section 9.2 the next one for tomorrow. I have a question where I'm given a graph and I'm asked which differential equation out of 3 is a solution to this graph. So I was wondering I have to go backwards in a sense and think...well if I have y' then the graph must change ...ah nvm I can't think of how to ask this.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
955
There are two ways to determine where y' is positive and where it is negative. Since it is equal to a polynomial in y, it can change sign only where it is 0. So it is sufficient to check the sign at one point in each interval.

Or, you probably determined where y' is 0 by factoring: y'= y^2(y^2- 6y+ 5)= y^2(y- 1)(y- 5).
Now, just use "y- a is negative if y< a, positive if y> a".

If y< 0, all factors are negative but there are four of them- their product is positive. y' is positive, y is increasing. If 0< y< 1. "y" is negative while both y-1 and y- 5 are still negtive. The product of two negative numbers is still positive. If 1< y< 5, y and y- 1 are positive. Now we have only one negative so y' is negative and y is decreasing. Finally, if y> 5, all factors are positive, y' is positive, y is increasing.
 
  • #11
582
0
OK thanks
 

Related Threads on A function y(t)

Replies
1
Views
2K
  • Last Post
Replies
4
Views
6K
Replies
5
Views
641
Replies
1
Views
461
Replies
2
Views
738
Replies
1
Views
3K
Replies
3
Views
523
Replies
3
Views
1K
Replies
43
Views
4K
Replies
0
Views
1K
Top