First order nonlinear ODE - Integrating factor + exact differentials, or not?

In summary, the conversation discusses a first order nonlinear ODE and potential methods for solving it, including integrating factors and exact differentials. However, the author has not been able to find a solution using these methods and is seeking suggestions. One potential solution method involving a substitution is mentioned.
  • #1
LawlQuals
30
0
First order nonlinear ODE -- Integrating factor + exact differentials, or not?

Hello everyone,

(I apologize if this did not format properly, if not I will attempt to edit it if that functionality is available upon submitting a question).

I recently came across the following nonlinear ODE that I am having difficulty in solving.

Homework Statement



Determine the general solution to the differential equation:

[tex]\frac{dy}{dx} = \frac{x^2 - y^2}{x^2 + 5yx} = \frac{(x+y)(x-y)}{x(x + 5y)}[/tex]

Homework Equations



Any methods up to and including a typical advanced engineering mathematics (undergrad) course (integrating factors, variation of parameters, transforms, exact differentials, etc.)

The Attempt at a Solution



My only ideas have involved forcing the equation to be an exact differential. However, I have not been able to come to a solution by this route. To demonstrate, I pursued putting the equation in the form of conventional notation for exact differentials,

[tex]
\underbrace{(x^2 - y^2)}_{=\, M}dx - \underbrace{(x^2 + 5yx)}_{=\, -N}dy = 0
[/tex]

[tex]
Mdx + Ndy = 0
[/tex]

where

[tex]M = x^2 - y^2 \Rightarrow M_x = 2x, \quad M_y = -2y[/tex]
[tex]N = -(x^2 + 5yx) \Rightarrow N_x = -2x - 5x \quad N_y = 5x[/tex]

and the subscripts denote differentiation with respect to the labeled parameter. It is evident it is not exact ([tex]M_y \neq N_x[/tex]). Further, I was unable to implement an integrating factor [tex]\sigma[/tex] such that [tex]\frac{\partial}{\partial y}(\sigma M) = \frac{\partial}{\partial x} (\sigma N)[/tex] upon multiplication of the entire equation by [tex]\sigma[/tex]. It can readily be seen that the integrating factor [tex]\sigma[/tex] is not only a function of [tex]x[/tex] or [tex]y[/tex], e.g. writing out the condition for exact differentials involving the factor [tex]\sigma[/tex], and enforcing [tex]\sigma[/tex] to be a function of [tex]x[/tex] or [tex]y[/tex] alone involves results of the form

[tex]\frac{\sigma_y}{\sigma} \sim \frac{M_y - N_x}{M}[/tex]

or

[tex]\frac{\sigma_x}{\sigma} \sim \frac{M_y - N_x}{N}[/tex]

which demonstrates that [tex]\sigma[/tex] is a function of both variables in either case (in contradiction to the posits that lead to these equations).

Choosing a form [tex]\sigma (x,y) = x^a y^b[/tex], I could not discern proper constants [tex]a[/tex], and [tex]b[/tex].

Does anyone have any suggestions regarding solving this equation? Thanks very much for any insight, and take care.
 
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  • #2


LawlQuals said:
Hello everyone,

(I apologize if this did not format properly, if not I will attempt to edit it if that functionality is available upon submitting a question).
No need for apologies - what you have looks great.
LawlQuals said:
I recently came across the following nonlinear ODE that I am having difficulty in solving.

Homework Statement



Determine the general solution to the differential equation:

[tex]\frac{dy}{dx} = \frac{x^2 - y^2}{x^2 + 5yx} = \frac{(x+y)(x-y)}{x(x + 5y)}[/tex]

Homework Equations



Any methods up to and including a typical advanced engineering mathematics (undergrad) course (integrating factors, variation of parameters, transforms, exact differentials, etc.)

The Attempt at a Solution



My only ideas have involved forcing the equation to be an exact differential. However, I have not been able to come to a solution by this route. To demonstrate, I pursued putting the equation in the form of conventional notation for exact differentials,

[tex]
\underbrace{(x^2 - y^2)}_{=\, M}dx - \underbrace{(x^2 + 5yx)}_{=\, -N}dy = 0
[/tex]

[tex]
Mdx + Ndy = 0
[/tex]

where

[tex]M = x^2 - y^2 \Rightarrow M_x = 2x, \quad M_y = -2y[/tex]
[tex]N = -(x^2 + 5yx) \Rightarrow N_x = -2x - 5x \quad N_y = 5x[/tex]

and the subscripts denote differentiation with respect to the labeled parameter. It is evident it is not exact ([tex]M_y \neq N_x[/tex]). Further, I was unable to implement an integrating factor [tex]\sigma[/tex] such that [tex]\frac{\partial}{\partial y}(\sigma M) = \frac{\partial}{\partial x} (\sigma N)[/tex] upon multiplication of the entire equation by [tex]\sigma[/tex]. It can readily be seen that the integrating factor [tex]\sigma[/tex] is not only a function of [tex]x[/tex] or [tex]y[/tex], e.g. writing out the condition for exact differentials involving the factor [tex]\sigma[/tex], and enforcing [tex]\sigma[/tex] to be a function of [tex]x[/tex] or [tex]y[/tex] alone involves results of the form

[tex]\frac{\sigma_y}{\sigma} \sim \frac{M_y - N_x}{M}[/tex]

or

[tex]\frac{\sigma_x}{\sigma} \sim \frac{M_y - N_x}{N}[/tex]

which demonstrates that [tex]\sigma[/tex] is a function of both variables in either case (in contradiction to the posits that lead to these equations).

Choosing a form [tex]\sigma (x,y) = x^a y^b[/tex], I could not discern proper constants [tex]a[/tex], and [tex]b[/tex].

Does anyone have any suggestions regarding solving this equation? Thanks very much for any insight, and take care.
Try this substitution: Let v = y/x ==> y = vx ==> y' = v'x + v

Then the DE y' = (x^2 - y^2)/(x^2 + 5xy) becomes
v'x + v = (x^2 - v^2 x)/(x^2 + 5vx^2)
[tex]v'x + v = \frac{x^2 - v^2x^2}{x^2 + 5vx^2}[/tex]

After a bit of algebraic manipulation, you get a DE in v and x that is separable. Undo the substitution and you have your solution.
 

1. What is a first order nonlinear ODE?

A first order nonlinear ODE (ordinary differential equation) is an equation that involves an unknown function and its derivatives up to the first order, where the function is nonlinear. This means that the function cannot be written as a linear combination of its derivatives.

2. What is an integrating factor in the context of first order nonlinear ODEs?

An integrating factor is a function that is multiplied to both sides of a first order nonlinear ODE in order to make it exact. It is used to simplify the equation and make it easier to solve.

3. How do you determine if a first order nonlinear ODE is exact or not?

A first order nonlinear ODE is exact if it can be written as the total differential of a function. This means that the equation can be expressed as the derivative of a single function with respect to one of its variables. If the equation cannot be written in this form, it is not exact and an integrating factor may be needed to solve it.

4. What is the role of exact differentials in solving first order nonlinear ODEs?

Exact differentials are used to determine if a first order nonlinear ODE is exact or not. If the equation can be expressed as the total differential of a function, it is considered exact and can be solved using integration. If not, an integrating factor may be used to make it exact and easier to solve.

5. What are some applications of first order nonlinear ODEs?

First order nonlinear ODEs have many applications in the fields of physics, engineering, and biology. They are used to model a wide range of phenomena, including population growth, chemical reactions, electrical circuits, and fluid dynamics. They are also essential in the development of mathematical models and simulations in scientific research.

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