Undergrad Nonlinear Second Order ODE: Can We Find an Analytical Solution?

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SUMMARY

The discussion focuses on solving the nonlinear second order ordinary differential equation (ODE) given by $$\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}-\frac{y}{ay+b}=0$$ where ##a## and ##b## are constants. It is established that when ##b=0## and ##a\neq 0##, the ODE simplifies to a linear form with a known analytical solution: $$y=\frac{x^2}{4a}+C_1\ln{x}+C_2$$. Similarly, when ##a=0## and ##b\neq 0##, the equation reduces to the modified Bessel equation of order zero, yielding the solution $$y=C_1I_0\left(\frac{x}{\sqrt{b}}\right)+C_2K_0\left(\frac{x}{\sqrt{b}}\right)$$. However, the case where both ##a## and ##b## are non-zero remains unsolved, and participants express the need for effective methods to tackle this scenario.

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  • Understanding of nonlinear second order ordinary differential equations (ODEs)
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  • Knowledge of analytical solution techniques for differential equations
  • Experience with variable transformations in ODEs
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I'm trying to solve the following nonlinear second order ODE where ##a## and ##b## are constants: $$\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}-\frac{y}{ay+b}=0$$ It looks somewhat like the modified Bessel equation, except the third term on the left makes it nonlinear. I've been trying to determine some way to find an analytical solution but haven't been able to come up with anything. It doesn't help much but it can also be written:$$\frac{1}{x}\frac{d}{dx}\left(x\frac{dy}{dx}\right)=\frac{y}{ay+b}$$Any suggestions would be greatly appreciated, thanks!
 
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Looks like for the case that ##b=0,a\neq 0## there is the analytical solution because then the ODE becomes linear with non constant coefficients (and I think the solution is a polynomial of 2nd order).

The case that ##a=0,b\neq 0## also seems to fallback to linear ODE as well so there should be an analytical solution.

But I am all out of ideas how to effectively treat the case ##a,b\neq 0##.
 
Last edited:
Delta2 said:
Looks like for the case that ##b=0,a\neq 0## there is the analytical solution because then the ODE becomes linear with non constant coefficients (and I think the solution is a polynomial of 2nd order).

The case that ##a=0,b\neq 0## also seems to fallback to linear ODE as well so there should be an analytical solution.

But I am all out of ideas how to effectively treat the case ##a,b\neq 0##.

Yeah, if ##b=0,a\neq 0## then the solution is ##y=\frac{x^2}{4a}+C_1\ln{x}+C_2##. If ##a=0,b\neq 0##, then it becomes the modified Bessel equation of order zero and the solution is ##y=C_1I_0\left(\frac{x}{\sqrt{b}}\right)+C_2K_0\left(\frac{x}{\sqrt{b}}\right)##.

It can be transformed if ##x=e^t## so ##t=\ln{x}##. With that it becomes: $$\frac{d^2y}{dt^2}=\frac{e^{2t}y}{ay+b}$$ They have another transformation here for equations that kinda look like that but I haven't been able to get that transformation to work.
 
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