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tse8682

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- #1

tse8682

- 30

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- #2

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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##.

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:

- #3

tse8682

- 30

- 1

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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