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

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SUMMARY

The discussion centers on solving the first-order nonlinear ordinary differential equation (ODE) given by \(\frac{dy}{dx} = \frac{x^2 - y^2}{x^2 + 5yx}\). The original poster attempted to use exact differentials and integrating factors but found that the equation was not exact, as \(M_y \neq N_x\). A suggested substitution is \(v = \frac{y}{x}\), transforming the ODE into a separable form, which simplifies the solution process.

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LawlQuals
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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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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.
 

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