# How Do I Solve this 2nd Order Non-Linear ODE and Find Its Equilibrium Points?

• MHB
• metstandard16
In summary: Jacobian matrix to classify the type of equilibrium point. In summary, the 2nd order non-linear ODE can be rewritten as two 1st order ODEs and has equilibrium points at (0,0), (1,0), and (-1,0). To find the phase diagram, the system can be linearized about each equilibrium point and then classified using the Jacobian matrix.
metstandard16
Good morning everyone. First let me thank you for your help in advance!

I have to solve this 2nd order non-linear ODE, and I'm stuck at the beginning. We have to find the equilibrium points, linearize the system, draw the phase portraits and classify the eq points, and solve it numerically.

I think I'm fine with linearizing it and classifying, as well as producing some preliminary phase portraits before I head to matlab.

The ODE:

y''-y'+y2y'+y-y5=0

I've rewritten this 2nd order ODE into (2) 1st order ODE's to start:

u=y
v=y'

u' = y' = v
v'=y'' = v-u2v-u+u5

The second equation can be factored out to = (u-1)(u+1)(u3-v+u)

My confusion lies with the fact that I have a y' in my equation and this presents a v variable subsequently. Do I have to rewrite this equation in terms of v, so it's purely with u's (for the lack of better terms)? Then find where u' = 0? None of my other examples have a (v)/y' in them so it's difficult to reference my notes. Somehow I think I have 3 eq points at (0,0) , (1,0) , & (-1,0) but I wasn't completely sure how that equation panned out.

metstandard16 said:
Good morning everyone. First let me thank you for your help in advance!

I have to solve this 2nd order non-linear ODE, and I'm stuck at the beginning. We have to find the equilibrium points, linearize the system, draw the phase portraits and classify the eq points, and solve it numerically.

I think I'm fine with linearizing it and classifying, as well as producing some preliminary phase portraits before I head to matlab.

The ODE:

y''-y'+y2y'+y-y5=0

I've rewritten this 2nd order ODE into (2) 1st order ODE's to start:

u=y
v=y'

u' = y' = v
v'=y'' = v-u2v-u+u5

The second equation can be factored out to = (u-1)(u+1)(u3-v+u)

My confusion lies with the fact that I have a y' in my equation and this presents a v variable subsequently. Do I have to rewrite this equation in terms of v, so it's purely with u's (for the lack of better terms)? Then find where u' = 0? None of my other examples have a (v)/y' in them so it's difficult to reference my notes. Somehow I think I have 3 eq points at (0,0) , (1,0) , & (-1,0) but I wasn't completely sure how that equation panned out.

Are you trying to solve this exactly, or approximately with a numerical scheme?

Prove It said:
Are you trying to solve this exactly, or approximately with a numerical scheme?

Thanks for your reply. I'm trying to solve it numerically, eventually with the aid of MATLAB and with a few initial conditions based on the phase portraits.

I guess my confusion lies around the equilibrium points and the jacobian matrix system.

To expand on my work in the post above:

v' = y'' = v-u2v-u+u5
v' = v(1-u2)+u(u4-1)
v = u(1-u4)/(1-u2)
= u(1+u2)

v=0
Eq Point: (0,0) seems to be the only real solution, and perhaps a +/- i on the non-real part because of the 1+u^2.

Am I concerned with the imaginary part at this point? Just needed help getting this so I could classify them as (saddle, center, etc) and work up some approximate phase portraits.

Thanks!

Last edited:
metstandard16 said:
Good morning everyone. First let me thank you for your help in advance!

I have to solve this 2nd order non-linear ODE, and I'm stuck at the beginning. We have to find the equilibrium points, linearize the system, draw the phase portraits and classify the eq points, and solve it numerically.

I think I'm fine with linearizing it and classifying, as well as producing some preliminary phase portraits before I head to matlab.

The ODE:

y''-y'+y2y'+y-y5=0

I've rewritten this 2nd order ODE into (2) 1st order ODE's to start:

u=y
v=y'

u' = y' = v
v'=y'' = v-u2v-u+u5

The second equation can be factored out to = (u-1)(u+1)(u3-v+u)

My confusion lies with the fact that I have a y' in my equation and this presents a v variable subsequently. Do I have to rewrite this equation in terms of v, so it's purely with u's (for the lack of better terms)? Then find where u' = 0? None of my other examples have a (v)/y' in them so it's difficult to reference my notes.

And there is no "(v)/y'" here! Your two equations are u'= v and v'= v- u^2v- u+ u^5.

The equilibrium points are where u'= 0 and v'= 0. That is, u'= v= 0 and v'= v- u^2v- u+ u^5= 0. Setting v= 0 in the second equation, we have -u+ u^5= u(u^2+ 1)(u- 1)(u+ 1)=0 so u= 0, u= 1, and u= -1 are the real roots. Those give equilibrium points, (u, v)= (0, 0), (1, 0), and (-1, 0). The simplest way to find the phase diagram is to linearize about each equilibrium point.

## 1. What is a non-linear 2nd order ODE?

A non-linear 2nd order ODE (ordinary differential equation) is a type of mathematical equation that involves a dependent variable, its derivatives, and possibly the independent variable, raised to powers greater than one. This makes the equation non-linear, as the dependent variable is not directly proportional to the independent variable.

## 2. How does a non-linear 2nd order ODE differ from a linear 2nd order ODE?

A linear 2nd order ODE has the dependent variable and its derivatives raised to powers of one. This makes the equation linear, as the dependent variable is directly proportional to the independent variable. In contrast, a non-linear 2nd order ODE involves higher powers, making the equation non-linear.

## 3. What are the applications of non-linear 2nd order ODEs?

Non-linear 2nd order ODEs have various applications in science and engineering, such as in physics, chemistry, biology, and economics. They are used to model complex systems and phenomena, including population growth, chemical reactions, and mechanical oscillations.

## 4. How do you solve a non-linear 2nd order ODE?

The process of solving a non-linear 2nd order ODE involves finding the general solution, which is a function that satisfies the equation for all possible values of the independent variable. This can be done analytically or numerically, depending on the complexity of the equation.

## 5. Can non-linear 2nd order ODEs be solved exactly?

In most cases, non-linear 2nd order ODEs cannot be solved exactly, and an approximate solution must be obtained. However, there are some special cases where exact solutions can be found, such as when the equation is separable or can be transformed into a linear equation. Numerical methods can also be used to obtain more accurate solutions.

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