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I aim to find the exact form solution to the this ODE:

$$\frac{dS}{dw}S = \frac{a}{w}S^2 + \frac{b}{w}S - c$$

where S is a continuous differentiable function of w, real positive and a, b, c are positive, non zero, real values.

I follow the procedure in:

Panayotounakos, D. E. and Zarmpoutis, T. I. (2011). Construction of Exact Parametric or Closed Form Solutions of Some Unsolvable Classes of Nonlinear ODEs (Abel's Nonlinear ODEs of the First Kind and Relative Degenerate Equations). International Journal of Mathematics and Mathematical Sciences, 2011.

In particular I move from eq. 4.3 in the paper. I obtain a particular solution of the form:

$$S(w) = K w^{-1/2} $$

where K is a combination of the parameters a, b, c. But this form is not a solution to the ODE I started from.

Could you please help me to find the solution? Where am I wrong? Thanks.

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  • $\begingroup$ I followed through some of the working in the paper, the solution looks like it will be hideous to me, and not at all expressible in terms of elementary functions. For instance, if a=b=c=1, Wolfram gives the solution in terms of the LambertW function. Obviously if some of the constants are zero it the result will be simpler. $\endgroup$ – Bennett Gardiner Apr 15 '14 at 6:45
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    $\begingroup$ Thanks Bennett, that is an improvement to get how the solution could look like. Unfortunately the a=b=c=1 is not possible for my concerns. Working out the steps of the paper, the fact that the coefficients are of the type 1/w simplifies things a lot and I have the feeling that an analytic closed form could come out. Thanks! Davide $\endgroup$ – user85884 Apr 15 '14 at 7:17
  • $\begingroup$ I thought they would simplify things as well, but I never got to something that looked simple enough to solve analytically. I barely made it to halfway though, so I could be wrong, but doesn't the fact that the Lambert function appears in that case make you doubtful? Perhaps I'll try again later. $\endgroup$ – Bennett Gardiner Apr 15 '14 at 11:54
  • $\begingroup$ It strikes me that perhaps you don't care if the solution is of the form $w=f(S)$, in which case $a=1$ might be doable. $\endgroup$ – Bennett Gardiner Apr 15 '14 at 13:04
  • $\begingroup$ That is a good news. If it is possible to work out a solution of the form w = f(S) even just for the case a=1, that would be a substantial contribution. Thanks. $\endgroup$ – user85884 Apr 15 '14 at 13:13
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Approach $1$:

Let $S=w^aT$ ,

Then $\dfrac{dS}{dw}=w^a\dfrac{dT}{dw}+aw^{a-1}T$

$\therefore w^aT\left(w^a\dfrac{dT}{dw}+aw^{a-1}T\right)=\dfrac{aw^{2a}T^2}{w}+\dfrac{bw^aT}{w}-c$

$w^{2a}T\dfrac{dT}{dw}+aw^{2a-1}T^2=aw^{2a-1}T^2+bw^{a-1}T-c$

$w^{2a}T\dfrac{dT}{dw}=bw^{a-1}T-c$

$T\dfrac{dT}{dw}=bw^{-a-1}T-cw^{-2a}$

Let $x=-\dfrac{bw^{-a}}{a}$ ,

Then $w=\dfrac{(-1)^\frac{1}{a}b^\frac{1}{a}}{a^\frac{1}{a}x^\frac{1}{a}}$

$\dfrac{dT}{dw}=\dfrac{dT}{dx}\dfrac{dx}{dw}=bw^{-a-1}\dfrac{dT}{dx}$

$\therefore bw^{-a-1}T\dfrac{dT}{dx}=bw^{-a-1}T-cw^{-2a}$

$T\dfrac{dT}{dx}=T-\dfrac{cw^{1-a}}{b}$

$T\dfrac{dT}{dx}-T=\dfrac{(-1)^\frac{1}{a}b^{\frac{1}{a}-2}c}{a^{\frac{1}{a}-1}x^{\frac{1}{a}-1}}$

This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf

According to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf, the ODE has simpler form of the general solution when $1-\dfrac{1}{a}=0,-1,-2,-\dfrac{1}{2}$ , i.e. when $a=1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{2}{3}$

Approach $2$:

$S\dfrac{dS}{dw}=\dfrac{a}{w}S^2+\dfrac{b}{w}S-c$

$S\dfrac{dS}{dw}=\dfrac{aS^2+bS-cw}{w}$

$(aS^2+bS-cw)\dfrac{dw}{dS}=Sw$

Let $X=aS^2+bS-cw$ ,

Then $w=\dfrac{aS^2+bS-X}{c}$

$\dfrac{dw}{dS}=\dfrac{2aS+b}{c}-\dfrac{1}{c}\dfrac{dX}{dS}$

$\therefore X\left(\dfrac{2aS+b}{c}-\dfrac{1}{c}\dfrac{dX}{dS}\right)=S\dfrac{aS^2+bS-X}{c}$

$\dfrac{(2aS+b)X}{c}-\dfrac{X}{c}\dfrac{dX}{dS}=\dfrac{aS^3+bS^2-SX}{c}$

$\dfrac{X}{c}\dfrac{dX}{dS}=\dfrac{((2a+1)S+b)X-aS^3-bS^2}{c}$

$X\dfrac{dX}{dS}=((2a+1)S+b)X-aS^3-bS^2$

Assume $a\neq-\dfrac{1}{2}$ , as the case $a=-\dfrac{1}{2}$ had already been solved in Approach $1$ :

Let $R=S+\dfrac{b}{2a+1}$ ,

Then $X\dfrac{dX}{dR}=RX-a\left(R-\dfrac{b}{2a+1}\right)^3-b\left(R-\dfrac{b}{2a+1}\right)^2$

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This is just a special case solution!!

Using a substitution of the form $S = w^{\alpha}V$, we can transform the equation to a form of $$ \alpha w^{2\alpha-1}V^{2} +w^{2\alpha}VV' = aw^{2\alpha-1}V^{2} + bw^{\alpha-1}V - C $$ Clearly setting $\alpha = a$ is the easiest starting point. This reduces the equation to $$ w^{2\alpha}VV' = bw^{\alpha-1}V - C $$ Re-arranging leads to $$ VV' = bw^{-1-\alpha}V - Cw^{-2\alpha} $$ The next restriction is $\alpha=1$ and then the equation becomes separable and has a solution of the form $$ \left(b\frac{S}{w}-C\right) +C\ln\left(b\frac{S}{w}-C\right) = -\frac{b^{3}}{w} + \lambda $$

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  • $\begingroup$ Thanks. The equation you come after the substitution involves the same solution technique of the original one, as it is specified in the Handbook of exact solutions for ODE, section about Abel type. $\endgroup$ – user85884 Apr 24 '14 at 7:53
  • $\begingroup$ Nice! I didn't know that. I have been looking for other substitutions :). $\endgroup$ – Chinny84 Apr 24 '14 at 10:12

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