# Thread: Equilibrium Solution of autonomous ODE Draw Solution Curves?

1. ## Equilibrium Solution of autonomous ODE Draw Solution Curves?

I understand in an autonomous ODE a root of the derivative is a root of the function. But I will type the question and explain where I don't understand.

1. Consider the differential equation dy/dx = (y+4)(y-5).

(a) What are the two equilibrium solutions y(x) = c, where c is a constant, that make both sides of the equation zero?
(Question I) Would it be y = -4 and y = 5? (Question II) How would that make both sides of the equation zero?

(b) Draw solution curves, as in class notes, for curves with y(0) = -5, and y(0) = 0, and y(0) = 8.
(Question III) I just don't understand this. The y(x) =c statement from the first question seems like its y= c where c is a root of the derivative with respect to x.
How then can I throw these other constants, -5, 0, and 8 into the mix?

2. ## Re: Equilibrium Solution of autonomous ODE Draw Solution Curves? Originally Posted by physics I understand in an autonomous ODE a root of the derivative is a root of the function. But I will type the question and explain where I don't understand.

1. Consider the differential equation dy/dx = (y+4)(y-5).

(a) What are the two equilibrium solutions y(x) = c, where c is a constant, that make both sides of the equation zero?
Yes, the derivative of a constant is 0 so if you have an equation of the form dy/dx= f(y) then a constant solution must satisfy 0= f(y).

(Question I) Would it be y = -4 and y = 5?
Yes, with y constant we have dy/dx= 0= (y+ 4)(y- 5) so either y= -4 or y= 5.

(Question II) How would that make both sides of the equation zero?
Just as I said, with y= -4, dy/dx= 0= (0)(-9) and with y= 5, dy/dx= 0= (-1)(0).

(b) Draw solution curves, as in class notes, for curves with y(0) = -5, and y(0) = 0, and y(0) = 8.
(Question III) I just don't understand this. The y(x) =c statement from the first question seems like its y= c where c is a root of the derivative with respect to x.
How then can I throw these other constants, -5, 0, and 8 into the mix?
First, although it does not ask you to do so, draw the horizontal lines y= -4 and y= 5. Those are themselves solution "curves" and will help you draw the others. (You can only draw rough sketches, not exact graphs of the others.)

Now, what about the solution curve that contains y(0)= -5? Recall that solution curves cannot cross! (That is the "uniqueness" part of the "existence and uniqueness theorem"- through every point there is one and only one solution curve.) Since -5< -4, the entire solution curve passing through (0, -5) must be below y= -4. When y= -5, dy/dx= (-5+ 4)(-5- 4)= -1(-10)= 10. In fact, for all y< -4, y+ 4 and y- 5 are both negative so dy/dx= (y+4)(y- 5) is positive and so the graph is increasing. Since it is increasing but cannot cross y= -4, the solution curve, for x> 0 must rise up toward and the approach y= -4 as horizontal asymptote. For x< 0 there is no lower bound so the graph must just keep going down. The solution curve here rises from negative infinity, passes through (0, -5) and as x increases approaches y= -4 as a horizontal asymptote.

What about the solution curve that contains y(0)= 0? That curve must always lie between y= -4 and y= 5. When y= 0, y+ 4= 4 and y- 5= -5. dy/dx= (0+ 4)(0- 5)= -20< 0. In fact, for all y between -4 and 5, (y+ 4)(y- 5) is negative so the solution curve is [I]decreasing[/i The solution curve must start, on the left at y= 5 as a horizontal asymptote, pass through (0, 0), then continue, on the right down to y= -4 as a horizontal asymptote.

Finally, what about a solution curve that contains y(0)= 8? That curve must always be above y= 5. When y= 8, y+ 4= 12 and y- 5= 3. dy/d= (8+ 4)(8- 5)= (12)(3)= 36> 0. In fact, for all y greater than 5, (y+ 4)(y- 5) is positive so the solution curve is increasing. The curve starts on the left with y= 5 as a horizontal asymptote, passes through (0, 8), then keeps increasing to the right toward infinity.

3. ## Re: Equilibrium Solution of autonomous ODE Draw Solution Curves?

Ah thank you. You have removed all doubt with reason! Bravo my friend.

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