I am trying to solve this third order differential equation with non-constant coefficients $$(x^3)y'''+6 x^2 y''+[6+(1+a-bx^2)x^2]x y'+[1+3 a-5 b x^2]x^2 y=0$$ where $a$ and $b$ are constants and $y$ is a function of $x$ only.

The problem originated form a Micropolar fluid flow problem involving longitudinal and torsional oscillations. The original differential equation was 4th order and a combination of the Laplace transform, a change in variables as well as multiple integral transforms were used to bring the form given below.

I have tried the standard methods for solving differential equations with variable coefficients, as well as the Frobenius power series method, using x =0 as a regular singular point, however it became too complicated for a pattern and hence solutions to be formed.

I would be grateful if anyone had any further ideas on how to solve this differential equation.

  • $\begingroup$ Is there some reason you did not add the 6 to the $1+a$? $\endgroup$ Feb 4 '19 at 17:43
  • $\begingroup$ @JohnWaylandBales ty I will correct the issue now, that term is multiplied by x squared $\endgroup$
    – user447631
    Feb 4 '19 at 17:45
  • $\begingroup$ It looked promising, but didn't quite work. I'll post in case it offers inspiration. Can re-write as $(x^3 y'')' + 3(x^2y')' +((ax^3+bx^5)y)'+x^3y' + x^2 y=0$. If only it was $3x^2 y$ at the end. $\endgroup$
    – user121049
    Feb 4 '19 at 18:15
  • $\begingroup$ @user121049 thanks so much for this- to get to this differential equation I had about 50 pages of pen to paper working, I will redo and make sure I do not have any errors that would allow for the 3 x-squared y- much appreciated! $\endgroup$
    – user447631
    Feb 4 '19 at 19:14

The indicial roots are $0$, $-1$, $-2$. There is a series solution of the form $\sum_{k=0}^\infty c_{2k} x^{2k}$ with $c_0 = 1$, $c_2 = -a/8 - 1/24$, and

$$ -(n+5) b c_n + ((a+1)n + 5a + 3) c_{n+2} + + (n+4)(n+5)(n+6) c_{n+4} = 0 $$

and a series solution of the form $\sum_{k=0}^\infty c_{2k-1} x^{2k-1}$ with $c_{-1}=1$, $c_1 = -a/3$, and this same recurrence. A third fundamental solution involves $x^n$ for even $n \ge -2$ and $x^n \ln(x)$ for even $n \ge 0$.

  • $\begingroup$ thank you for your help! I got to the indicial roots as well but the pattern became too hard to spot for me when I listed out the terms odd and evens . I will go through this again, much appreciated! $\endgroup$
    – user447631
    Feb 4 '19 at 19:15

$$(x^3)y'''+6 x^2 y''+[6+(1+a-bx^2)x^2]x y'+[1+3 a-5 b x^2]x^2 y=0$$ This is not a final answer, only a way to simplify the problem.

One observe that the change of $x$ to $-x$ doesn't change the equation. This draw us to a change of variable in order to simplify : $$X=x^2\quad;\quad X'=2x \quad;\quad y(x)=u(X)$$ $y'= 2xu'\quad;\quad y''=4x^2u''+2u' \quad;\quad y'''=8x^3u'''+12xu''$

The derivatives of $y$ are with respect to $x$. The derivatives of $u$ are with respect to $X$.

$x^3(8x^3u'''+12xu'')+6 x^2 (4x^2u''+2u') +[6+(1+a-bx^2)x^2]x (2xu')+[1+3 a-5 b x^2]x^2 y=0$

After simplification :

$$8X^2u'''+36Xu'' +[24+2(1+a)X-2bX^2]u'+[1+3 a-5 bX] y=0$$

In the general case, I don't think that a closed form solution exists with the available standard functions.

In the case $a=0$ and $b\neq 0$ WolframAlpha cannot find a closed form solution.

In the case $b=0$ and $a\neq 0$ WolframAlpha gives a very complicated closed form solution involving hypergeometric functions and a Meijer-G function.

In the particular case $a=0$ and $b=0$ WolframAlpha gives the solution on the form $u=c_1\frac{1}{\sqrt{X}}+c_2\:_1F_2(\frac12,\frac32;2;-\frac{X}{4})+c_3$(Meijer-G function).

  • $\begingroup$ thank you very much, i will definitely look into this technique today- in the case where both a and b are zero will turn the problem from a non-newtonian fluid to a newtonian one, which i know has been solved so it makes sense you got a solution there. This opened my eyes a lot on how to approach the problem and I appreciate you taking the time to help me with this. $\endgroup$
    – user447631
    Feb 5 '19 at 15:34


Let $r=x^2$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=2x\dfrac{dy}{dr}$







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