Solution to a nonlinear ODE

$$a_1 y'''''+ a_2 y'''+\left(a_3 + y^2 \right) y' = 0$$

where $$a_1, a_2, a_3$$ are constants with $$a_1>0$$ and $$a_2,a_3 \in \mathbb{R}$$.

Is there a general solution $$y(x)$$ to the above differential equation? I am aware that there is an easy solution to the linearized version of the above equation $$(a_1 y'''''+ a_2 y'''+ a_3 y' = 0)$$ and want to know if there is a solution to the nonlinear equation too.

The equation arises in a mechanics problem. We have an unevenly pre-stretched ribbon (narrow plate). The constants $$a_1,a_2$$ depend on prestretch, and $$a_3$$ is a regularizing parameter. We are interested in the mechanics of twisting this pre-stretched ribbon. The mechanical system can be seen in figure-1 of this arxiv paper

• Can you briefly describe a mechanics problem? May 15 at 14:23
• Thank you for your reply @ArcticChar. We have an unevenly pre-stretched ribbon (narrow plate). The constants $a_1, a_2$ depend on prestretch, and $a_3$ is a regularizing parameter. We are interested in the mechanics of twisting this pre-stretched ribbon. The mechanical system can be seen in figure-1 of this paper arxiv.org/pdf/2304.01291.pdf.
– akr
May 15 at 14:32
• Your ODE has quite a few invariances that may be helpful to note: $x \to x + \Delta x$, $x \to -x$ and $y \to -y$. This is a problem probably very suitable to perturbative methods. Is $a_1$ very small or large compared to the other scales in the system? Regardless an expansion of $y \sim \sum y_k \epsilon^k$ where $\epsilon \ll 1$ may get you quite far. May 15 at 14:35
• What are the boundary conditions? May 15 at 19:57
• Well, in the particular case that $\int_0^1y(x) dx = constant = 0$ the solution is $y=0$... May 19 at 9:09

Beside a series solution, what else could we do for such a monster ?

Having the feeling that there is a problem with the boundary conditions, I give the Mathematica solution with none of them for $$A y^{(5)}+B y^{(3)}+\left(C+y^2\right) y'=0$$ $$y=c_1+c_2 x+c_3 x^2+c_4 x^3+c_5 x^4-\frac{ 6 B c_4+C c_2+c_2 c_1{}^2}{120 A}x^5-$$ $$\frac{12 B c_5+C c_3+c_3 c_1{}^2+c_2{}^2 c_1}{360 A}x^6+$$ $$\frac{-6 A C c_4-2 A c_2{}^3-12 A c_1 c_3 c_2-6 A c_1{}^2 c_4+6 B^2 c_4+B C c_2+B c_1{}^2 c_2}{5040 A^2}x^7+\cdots$$

The next coefficients are too long to be typed here but, if you e-mail (my address is in my profile), I shall send you an output file.

Considering that the boundary conditions are all initial, follows a MATHEMATICA script which calculates as many series coefficients as needed

Here the series is given by $$Y = \sum_{k=0}^{k=n}a[k] x^k$$ with the initial conditions associated to $$a[k], \ \ k = 0,\cdots,4$$

d[y_, x_] := a1 D[y, {x, 5}] + a2 D[y, {x, 3}] + (a3 + y^2) D[y, x]
n = 8;
Y = Sum[a[k] x^k, {k, 0, n}];
res = d[Y, x]
rels = CoefficientList[res, x];
A = Table[a[k], {k, 5, n}];
rels0 = Take[rels, {1, n - 4}];
sol = Solve[rels0 == 0, A][[1]];
Yx = Y /. sol


The shown results are for $$n = 8$$

$$Y_8 =a[0]+x a[1]+x^2 a[2]+x^3 a[3]+x^4 a[4]-\frac{x^5 \left(\text{a3} a[1]+a[0]^2 a[1]+6 \text{a2} a[3]\right)}{120 \text{a1}}-\frac{x^7 \left(-\text{a2} \text{a3} a[1]-\text{a2} a[0]^2 a[1]+2 \text{a1} a[1]^3+12 \text{a1} a[0] a[1] a[2]-6 \text{a2}^2 a[3]+6 \text{a1} \text{a3} a[3]+6 \text{a1} a[0]^2 a[3]\right)}{5040 \text{a1}^2}-\frac{x^6 \left(a[0] a[1]^2+\text{a3} a[2]+a[0]^2 a[2]+12 \text{a2} a[4]\right)}{360 \text{a1}}+x^8 \left(\frac{\text{a2} \left(a[0] a[1]^2+\text{a3} a[2]+a[0]^2 a[2]+12 \text{a2} a[4]\right)}{20160 \text{a1}^2}-\frac{a[1]^2 a[2]+a[0] a[2]^2+2 a[0] a[1] a[3]+\text{a3} a[4]+a[0]^2 a[4]}{1680 \text{a1}}\right)$$