# Determining equilibrium solutions to differential equations

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• chwala

#### chwala

Gold Member
TL;DR Summary
In determining the solutions; would i be correct to to say that one has to first solve the differential equation to get the function then only then can one determine the solutions by checking on the family of curves- substituting for ##c## values.
I am going through this,

I noted that, i shall have a separation of variables, that leads to

$$\left[\int \dfrac{1}{y(y-1)} dy\right]= \int \dfrac {1}{6} dt$$

and using partial fraction, i then have,

$$\left[\int -\dfrac{1}{y} dy - \int \dfrac{1}{y-1} dy\right] = \int \dfrac {1}{6} dt$$

$$-(\ln y + \ln (y-1) = \dfrac{1}{6} t + c$$

$$-\ln y(y-1) = \dfrac{1}{6} t + c$$

$$\ln y(y-1) = -\left[\dfrac{1}{6} t + c\right]$$

$$e^{{-\frac{1}{6} t - c}} = y^2-y$$

which is an implicit solution to get an explicit solution one may solve the quadratic equation,

$$y^2-y-e^{-\frac{1}{6} t - c}=0$$

...

Edit;
I just amended ##x## to ##t## so as to have an autonomous ode.

... from this point ( that is, at the implicit solution) i was able to get the shape of the function. Using different assigned values of ##c## should give me the directional field that is shown... unless there is another approach then i stand to be informed hence reason of this post.

My specific question is this,
Do we have other ways of determining the nature of solutions other than first determining the integral curves?

The other aspect of checking for stability of soluitions ( asymptotically or unstable) is pretty straightforward. Cheers.

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chwala said:
I noted that, i shall have a separation of variables, that leads to

##\left[\int \dfrac{1}{y(y-1)} dy\right]= \int \dfrac {1}{6} dx##
Can you show how you get that?

chwala
Philip Koeck said:
Can you show how you get that?
@Philip Koeck Boss are you saying that you cannot see this line,

##6\dfrac{dy}{dx} = y^2-y## ?

A study on separation of variables? then from there ...it is pretty straightforward to what i have.

chwala said:
are you saying that you cannot see this line,

##6\dfrac{dy}{dx} = y^2-y## ?
No, I can't. Explain how you get that.

chwala
It ois not necessary to solve the ODE.

For first-order autonomous ODEs, $y$ is strictly increasing when $y' > 0$ and strictly decreasing when $y' < 0$. Here $y'= y^2 - y - 6 = (y + 2)(y - 3)$ is a quadratic with two distinct real roots and a positive leading coefficient, so it is positive outside the roots and negative between them. Thus the lower root -2 is a stable equilibrium solution and the upper root 3 is unstable.

chwala
pasmith said:
It ois not necessary to solve the ODE.

For first-order autonomous ODEs, $y$ is strictly increasing when $y' > 0$ and strictly decreasing when $y' < 0$. Here $y'= y^2 - y - 6 = (y + 2)(y - 3)$ is a quadratic with two distinct real roots and a positive leading coefficient, so it is positive outside the roots and negative between them. Thus the lower root -2 is a stable equilibrium solution and the upper root 3 is unstable.
I see ... with the condition that ##f(y)## and ##\frac{∂f}{∂y}## have no discontinuities.

chwala said:
I noted that, i shall have a separation of variables, that leads to
##\left[\int \dfrac{1}{y(y-1)} dy\right]= \int \dfrac {1}{6} dt##
No it doesn't.
chwala said:
Boss are you saying that you cannot see this line, ##6\dfrac{dy}{dx} = y^2-y## ?
Not only is @Philip Koeck unable to follow your work, it's incorrect as well.
Starting with the equation ##y' = y^2 - y - 6##, you apparently added 6 to the right side, but multiplied the left side by 6. That ain't kosher, my friend.

Also, as already noted, you aren't asked to solve the ODE -- just find the equilibrium points, points at which y' = 0 . Solving the DE is icing on the cake.

chwala
Mark44 said:
No it doesn't.

Not only is @Philip Koeck unable to follow your work, it's incorrect as well.
Starting with the equation ##y' = y^2 - y - 6##, you apparently added 6 to the right side, but multiplied the left side by 6. That ain't kosher, my friend.

Also, as already noted, you aren't asked to solve the ODE -- just find the equilibrium points, points at which y' = 0 . Solving the DE is icing on the cake.
I will check my working later today...yeah mistake there. I am now informed that the problem does not require one to solve the non linear differential equation but I would be interested in now trying to find a way of how to solve the differential equation...if you may point me in right direction then it would be nice boss.

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Mark44 said:
No it doesn't.

Not only is @Philip Koeck unable to follow your work, it's incorrect as well.
Starting with the equation ##y' = y^2 - y - 6##, you apparently added 6 to the right side, but multiplied the left side by 6. That ain't kosher, my friend.

Also, as already noted, you aren't asked to solve the ODE -- just find the equilibrium points, points at which y' = 0 . Solving the DE is icing on the cake.
So you're implying this solving ##\frac {dy}{dx} =0## applies to all orders? ...first order ode, second order...( needs to be transformed to 1st order?)... or there is a different approach for differential equations with more than first order. Thanks.

chwala said:
So you're implying this solving ##\frac {dy}{dx} =0## applies to all orders? ...first order ode, second order...( needs to be transformed to 1st order?)... or there is a different approach for differential equations with more than first order.
No, I'm not implying anything other than what I wrote, which was about first order DEs only.

chwala
Mark44 said:
No, I'm not implying anything other than what I wrote, which was about first order DEs only.
It would have been easier if he had pointed out the error on the ##6## part.
Of late, I have realized that the smarter I become, the more I make this kinda of mistakes ...

...just yesterday I finished studying 2 chapters on vector calculus without losing focus...a feat I could not do for many years...

Philip Koeck
chwala said:
It would have been easier if he had pointed out the error on the ##6## part.
Of late, I have realized that the smarter I become, the more I make this kinda of mistakes ...

...just yesterday I finished studying 2 chapters on vector calculus without losing focus...a feat I could not do for many years...
Aaaaah! I think I've seen it will post working later as I have work now,

We shall have

$$\dfrac{dy}{dx}= (y+2)(y-3)$$

$$\left[\int \dfrac{dy}{(y+2)(y-3)} \right] = \int dx$$

$$\left[\int \dfrac{1}{5(y-3)} dy - \int \dfrac{1}{5(y+2)} dy\right] =\int dx$$

$$\dfrac{1}{5} ln|y-3|-\dfrac{1}{5} \ln |y+2|=x +k$$

$$\dfrac{1}{5}\ln \left(\dfrac{y-3}{y+2}\right)= x+k$$

$$ln \left(\dfrac{y-3}{y+2}\right)=5(x+k)$$

$$e^{5(x+k)}=\dfrac{y-3}{y+2}$$

$$y=\dfrac{2e^{5(x+k)} +3}{1-e^{5(x+k)}}$$

I proceeded to check the graph with different ##k## values, graph is similar to waht we have on original post further i noted that the lines ##y=-2## and ##y=3## are tangents to the curve having gradient,## \frac{dy}{dx}=0## cheers.

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chwala said:
We shall have <snip> ##e^{5(x+k)}=\dfrac{y-3}{y+2}##
You can simplify things a bit in the subsequent lines by recognizing that ##e^{5(x + k)}## can be rewritten as ##Ae^{5x}##, where ##A = e^{5k}##.

chwala
pasmith said:
It ois not necessary to solve the ODE.

For first-order autonomous ODEs, $y$ is strictly increasing when $y' > 0$ and strictly decreasing when $y' < 0$. Here $y'= y^2 - y - 6 = (y + 2)(y - 3)$ is a quadratic with two distinct real roots and a positive leading coefficient, so it is positive outside the roots and negative between them. Thus the lower root -2 is a stable equilibrium solution and the upper root 3 is unstable.
Meaning that if the graph was inverted i.e ##\dfrac{dy}{dx}= -(y+2)(y-3)## then the lower root ##-2## would be unstable as it is decreasing below and increasing above and the other root ##3## would be semi-stable as it is decreasing below and above?
cheers.

No. The sign of $y' = -(y+2)(y-3)$ does not change between -2 and 3; accordingly the fixed point at -2 is unstable (decreasing below, increasing above) and the fixed point at 3 is stable (increasing below, decreasing above).

chwala said:
Meaning that if the graph was inverted i.e ##\dfrac{dy}{dx}= -(y+2)(y-3)## then the lower root ##-2## would be unstable as it is decreasing below and increasing above and the other root ##3## would be semi-stable as it is decreasing below and above?
cheers.
I think my problem is with the language, in my understanding -decreasing below means ##y## values are decreasing that is, ##-1,-2,-3 ...## which is wrong; it ought to be increasing below, that is ##y## values increasing (but in negative direction) as per your rightful insight...

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chwala said:
I think my problem is with the language, in my understanding -decreasing below means ##y## values are decreasing that is, ##-1,-2,-3 ...## which is wrong; it ought to be increasing below, that is ##y## values increasing (but in negative direction) as per your rightful insight...
I think i now get it!

Following my post ##14##.

Where i have ##\dfrac{dy}{dx}= -(y+2)(y-3).## I shall proceed as follows:
The roots are at ##y=-2## and ##y=3##. Now for ##y<-2##, ##\dfrac{dy}{dx}## is negative therefore ##y## is decreasing. For ##-2<y<3##, ##\dfrac{dy}{dx}## is positive, therefore ##y## is increasing. For ##y>3##, ##\dfrac{dy}{dx}## is negative therefore ##y## is decreasing thus,
##y=-2## is unstable point and ##y=3## is a Stable point.
Cheers.

∫ 1/(y(y -1)) = Γ(1) + c

y' = -(y + Γ(1))