Solving the differential $\frac{y'y'''}{y''} = x$

Question

I've been trying to solve the differential equation: $\frac{f'(x)f'''(x)}{f''(x)} = x$, $x\in \mathbb{R}$

My initial attempt was to integrate by parts, something like this:

$$\int_{}^{} \frac{f'(x)f'''(x)}{f''(x)} dx = \int_{}^{}x dx \Leftrightarrow \\ ln(f''(x))f'(x) - \int_{}^{} \ln(f''(x)) f''(x)dx = \frac{x^2}{2} + C1$$

but I soon came to realize that there was absolutely no way to proceed. Even if I try to reduce it to a second-order nonlinear ODE by letting $u(x) = f'(x)$, I can't seem to solve it by simple integration.

After some trial and error, I managed to figure out that $f(x) = \frac{x^3}{3} + C ,x\in \mathbb{R}$ (as noted by @Vasili) is indeed a solution (if not the only solution) to this differential.

And here I ask you: How do we mathematically prove that $f(x) = \frac{x^3}{3}$ + C is a solution? How do we reach that conclusion without simply making guesses?

Thanks in advance for any insight!

Answer 1

My idea is to assume that $y$ is an $n$-th degree polynomial, which means $$ \deg\left(\frac{y'y'''}{y''}\right) = \deg x \\ \Rightarrow (n-1) + (n-3) - (n-2) = 1 \\ \Rightarrow n = 3 $$ So our $y$ is a 3rd degree polynomial, which means we can write it in this form $$ y = Ax^3 + Bx^2 + Cx + D, A \not = 0 $$ Now what's left to do is to find the coefficients, given that $$ \frac{y'y'''}{y''} = x $$ We can plug in our polynomial expression for $y$ and get $$ 6A\cdot\frac{3Ax^2 + 2Bx + C}{6Ax + 2B} = x \\ \Rightarrow 6A \left(\frac{1}{2}x + \frac{B}{6A} + \frac{C - \frac{B^2}{3A}}{6Ax + 2B}\right) = x \\ \Rightarrow 3Ax + B + \frac{6AC - 2B^2}{6Ax + 2B} = x $$ And from here, it's easy to see that $$ A = \frac{1}{3} \\ B = C = 0 $$ So in conclusion, our solution to this DE is $$ y = \frac{1}{3} x^3 + D $$

Answer 2

The differential equation to be solved is $$ \frac{f'(x)f'''(x)}{f''(x)} = x. \tag1 $$ One way to do this is the method of undetermined coefficients.

Assume that $\,f(x)\,$ has a power series expansion $$ f(x) = c_0 + c_1\frac{x}{1!} +c_2\frac{x^2}{2!} + c_3\frac{x^3}{3!} + \dots \tag2 $$ where $\,c_0, c_1, \dots\,$ are constants to be determined. For convenience, rewrite equation $(1)$ as $$ a_0 + a_1\frac{x}{1!} +a_2\frac{x^2}{2!}+\dots := f'(x)f'''(x) - x\,f''(x) = 0 \tag3 $$ with the proviso that $\,f''(x)\ne 0.\,$ Expand the derivatives of $\,f(x)\,$ in series to get $$ a_0 = c_1c_3,\; a_1 = c_1c_4+ c_2(c_3-1),\; a_2 = c_1c_5 +2c_2c_4 +c_3(c_3-2), \; \dots \tag4 $$ From here it is a case by case analysis. That is, $\,a_0 = 0\,$ iff $\,c_1=0\,$ or $\,c_3=0.\,$ The cases are:

I. $\,c_1=0.\,$ Now $\,a_1 = c_2(c_3-1) = 0\,$ iff $\,c_2=0\,$ or $\,c_3=1.\,$

Ia. $\,c_1=0, c_2=0.\,$ Now $\,a_2 = c_3(c_3-2) = 0\,$ iff $\,c_3=0\,$ or $\,c_3=2.\,$

Iai. $\,c_1=0, c_2=0, c_3=0.\,$ After this point, all other coefficients must equal zero. This implies $\,f(x)\,$ is the zero function which makes equation $(1)$ undefined.

Iaii. $\,c_1=0, c_2=0, c_3=2.\,$ After this point, all other coefficients must equal zero. This implies $\,f(x) = c_0 + x^3/3\,$ which is verified as a solution.

Ib. $\,c_1=0, c_3=1.\,$ Now $\,a_2 = 2c_2c_4-1 = 0\,$ which implies that both $\,c_2\,$ and $\,c_4\,$ are nonzero. After this points all other coefficients are determined. The result is $$ f(x) = c_0 + t\frac{x^2}{2!} + \frac{x^3}{3!} + t^{-1}\frac12 \frac{x^4}{4!} - t^{-2}\frac16\frac{x^5}{5!} - t^{-3}\frac18\frac{x^6}{6!} + t^{-4}\frac25\frac{x^7}{7!} - t^{-5}\frac{67}{144}\frac{x^8}{8!} + \dots \tag5 $$ where $\,t\ne 0\,$ and the pattern of coefficients is not evident.

II. $\,c_3=0.\,$ Now $\,a_1 = c_1c_4-c_2 = 0.\,$ If $\,c_1=0\,$ then $\,c_2=0\,$ and this is case Iai again. Thus, assume $\,c_1\ne0.\,$ After this point all other coefficients are determined. The result is $$ f(x) = c_0 + u^{-1}\frac{x^1}{1!} + t\frac{x^2}{2!} + tu\frac{x^4}{4!} - 2t^2u^2\frac{x^5}{5!} +3u^2 t(1+2ut^2)\frac{x^6}{6!} + \dots \tag6 $$ where $\,u\ne 0\,$ and the pattern of coefficients is not evident.