Nonlinear ODE Help: Strategies to Solve a Challenging 1st Order PDE

[BobbyBear]

Haiya :P

In the process of trying to find the solution of a 1st order PDE, I've reached a point where I have to solve the following ode:

$$\frac{dy}{dx} = \frac{x^2y-4(y-x)^3}{xy^2+4(y-x)^3}$$

and I am stuck here :( It's not separable, homogeneous, or exact and I really don't know how to tackle this, is it most likely there is no analytical way to solve? Please suggest a strategy or help in any way, if you can :P:P

Thanks x

 

[CFDFEAGURU]

Hello,

Have you tried a numerical solution? Do you have any intial conditions for this ODE?

Thanks
Matt

 

[HallsofIvy]

Why do you say it is not homogenous? If you replace x by $\lambda x$ and y by $\lambda y$ the right side becomes
$$\frac{(\lambda x)^2(\lambda y)- 4(\lambda y- \lamba x)^3}{(\lambda x)(\lambda y)^2+ 4(\lambda y- \lambda x)^3}$$$$= \frac{\lambda^3(x^2y- (y- x)^3}{\lambda^3(xy^2+ (y- x)^3}= \frac{x^2y- (y- x)^3}{xy^2+ (y-x)^3}$$
so this equation certainly is homogenous.

 

[BobbyBear]

Thank you very much for your replies.

Oooh Ivy, I didn't realize that xD The numerator and denominator are both homogeneous of 3rd degree :) So then, what I do is:

$$\frac{dy}{dx} = \frac{x^2y-4(y-x)^3}{xy^2+4(y-x)^3} = \frac{x^3[\frac{y}{x}-4(\frac{y}{x} -1)^3]}{x^3[(\frac{y}{x})^2+4(\frac{y}{x} -1)^3]} = \frac{\frac{y}{x}-4(\frac{y}{x} -1)^3}{(\frac{y}{x})^2+4(\frac{y}{x} -1)^3}$$

and using the change of variable z=y/x, I'm left with an separable equation: xD

$$\frac{dy}{dx} = x \frac{dz}{dx} + z = \frac{z-4(z-1)^3}{z^2+4(z-1)^3}$$

$$\frac{dz}{\frac{z-4(z-1)^3}{z^2+4(z-1)^3} -1} = \frac{dx}{x}$$

which after some manipulation leaves me with:

$$\left[-\frac{1}{2} +\frac{1}{2}\left(\frac{z^2+z}{-8z^3+23z^2-23z+8}\right) \right] dz=\frac{dx}{x}$$

Um, I need help once again integrating the left hand side, I mean, I can only obtain the roots of the 3rd degree polynomial numerically, so should I use those (approximate roots) to carry out the partial fraction expansion of the polynomial quotient so that I can integrate? I'm not sure what to do, I want to end up with a relationship between the x and y variables; I don't have intial values because this ode came up while trying to solve a 1st order linear PDE (in fact, I'm trying to solve the characteristic system:

$$\frac{dx}{xy^2+4(y-x)^3} = \frac{dy}{x^2y-4(y-x)^3} = \frac{du}{-1}$$

so that's why I needed to solve that ode.

If you have any suggestions on how to go about this I'd appreciate it! Thank you! :)

 

[tanujkush]

Thank you very much for your replies.

Oooh Ivy, I didn't realize that xD The numerator and denominator are both homogeneous of 3rd degree :) So then, what I do is:

$$\frac{dy}{dx} = \frac{x^2y-4(y-x)^3}{xy^2+4(y-x)^3} = \frac{x^3[\frac{y}{x}-4(\frac{y}{x} -1)^3]}{x^3[(\frac{y}{x})^2+4(\frac{y}{x} -1)^3]} = \frac{\frac{y}{x}-4(\frac{y}{x} -1)^3}{(\frac{y}{x})^2+4(\frac{y}{x} -1)^3}$$

and using the change of variable z=y/x, I'm left with an separable equation: xD

$$\frac{dy}{dx} = x \frac{dz}{dx} + z = \frac{z-4(z-1)^3}{z^2+4(z-1)^3}$$

$$\frac{dz}{\frac{z-4(z-1)^3}{z^2+4(z-1)^3} -1} = \frac{dx}{x}$$

which after some manipulation leaves me with:

$$\left[-\frac{1}{2} +\frac{1}{2}\left(\frac{z^2+z}{-8z^3+23z^2-23z+8}\right) \right] dz=\frac{dx}{x}$$

Um, I need help once again integrating the left hand side, I mean, I can only obtain the roots of the 3rd degree polynomial numerically, so should I use those (approximate roots) to carry out the partial fraction expansion of the polynomial quotient so that I can integrate? I'm not sure what to do, I want to end up with a relationship between the x and y variables; I don't have intial values because this ode came up while trying to solve a 1st order linear PDE (in fact, I'm trying to solve the characteristic system:

$$\frac{dx}{xy^2+4(y-x)^3} = \frac{dy}{x^2y-4(y-x)^3} = \frac{du}{-1}$$

so that's why I needed to solve that ode.

If you have any suggestions on how to go about this I'd appreciate it! Thank you! :)

Hi BobbyBear,

just by visually inspecting the algebraic third order equation in the denominator, i can tell that z=1 is one root of this equation.

 

[BobbyBear]

Hi BobbyBear,

just by visually inspecting the algebraic third order equation in the denominator, i can tell that z=1 is one root of this equation.

Oooh tanujkush, I love you! :)
*hugs*

I guess I am quite short sighted :P:P