Existence of closed form solution of the below DE

I have a problem finding an analytical closed form solution of the following DE.

$$\displaystyle \left( \frac{y(x)+2}{3} \right) \frac{dy}{dx} + y(x) +\frac{1}{x} \left( a y(x)^2 + b y(x) -c \right) =0. \\ \{a,b,c\} \in \mathbb{R}, \qquad \{a,b,c\}>0.$$

I encountered this differential equation in my research in fluid dynamics. I failed to reduce it to some standard form of DE for which solutions are possible. Mathematica and Maple are unable to find analytical solutions for this.

Can anyone point out any method that can be used to know if a solution for this DE exists?

And if a solution exists, can anyone point out a method to solve the DE?

• How about using a power series? Commented Feb 26, 2019 at 14:01
• @Mattos Power series solution for this DE diverges. Also I am looking for a closed form solution if possible. Commented Feb 27, 2019 at 9:43
• I'm confused as to why you write the 'power series solution for this DE diverges'. The power series solution is the solution, hence if the power series diverges, then the solution diverges. You've already computed a power series solution, so you've shown a solution exists, but you can check to see if a solution exists and is unique using the Picard-Lindelof theorem i.e show that $\partial f(x,y)/\partial y$ exists somewhere for the ODE $y' = f(x,y(x))$, though this existence theorem might be too strenuous. Commented Feb 27, 2019 at 9:56

Hint:

Approach $$1$$:

This belongs to an Abel equation of the second kind.

Let $$u=y+2$$ ,

Then $$\dfrac{du}{dx}=\dfrac{dy}{dx}$$

$$\therefore\dfrac{u}{3}\dfrac{du}{dx}+u-2+\dfrac{1}{x}(a(u-2)^2+b(u-2)-c)=0$$

$$\dfrac{u}{3}\dfrac{du}{dx}+u-2+\dfrac{1}{x}(au^2-(4a-b)u+4a-2b-c)=0$$

$$u\dfrac{du}{dx}=-\dfrac{3au^2}{x}-3\left(1-\dfrac{4a-b}{x}\right)u+6-\dfrac{3(4a-2b-c)}{x}$$

Let $$u=x^{-3a}v$$ ,

Then $$\dfrac{du}{dx}=x^{-3a}\dfrac{dv}{dx}-3ax^{-3a-1}v$$

$$\therefore x^{-3a}v\left(x^{-3a}\dfrac{dv}{dx}-3ax^{-3a-1}v\right)=-\dfrac{3ax^{-6a}v^2}{x}-3\left(1-\dfrac{4a-b}{x}\right)x^{-3a}v+6-\dfrac{3(4a-2b-c)}{x}$$

$$x^{-6a}v\dfrac{dv}{dx}-3ax^{-6a-1}v^2=-3ax^{-6a-1}v^2-3\left(1-\dfrac{4a-b}{x}\right)x^{-3a}v+6-\dfrac{3(4a-2b-c)}{x}$$

$$x^{-6a}v\dfrac{dv}{dx}=-3\left(1-\dfrac{4a-b}{x}\right)x^{-3a}v+6-\dfrac{3(4a-2b-c)}{x}$$

$$v\dfrac{dv}{dx}=-3(x^{3a}-(4a-b)x^{3a-1})v+6x^{6a}-3(4a-2b-c)x^{6a-1}$$

Approach $$2$$:

$$\dfrac{y+2}{3}\dfrac{dy}{dx}+y+\dfrac{ay^2+by-c}{x}=0$$

$$y+\dfrac{ay^2+by-c}{x}=-\dfrac{y+2}{3}\dfrac{dy}{dx}$$

$$\left(x+ay+b-\dfrac{c}{y}\right)\dfrac{dx}{dy}=-\left(\dfrac{1}{3}+\dfrac{2}{3y}\right)x$$

This belongs to an Abel equation of the second kind.

Let $$u=x+ay+b-\dfrac{c}{y}$$ ,

Then $$x=u-ay-b+\dfrac{c}{y}$$

$$\dfrac{dx}{dy}=\dfrac{du}{dy}-a-\dfrac{c}{y^2}$$

$$\therefore u\left(\dfrac{du}{dy}-a-\dfrac{c}{y^2}\right)=-\left(\dfrac{1}{3}+\dfrac{2}{3y}\right)\left(u-ay-b+\dfrac{c}{y}\right)$$

$$u\dfrac{du}{dy}-\left(a+\dfrac{c}{y^2}\right)u=-\left(\dfrac{1}{3}+\dfrac{2}{3y}\right)u+\dfrac{(y+2)(ay^2+by-c)}{3y^2}$$

$$u\dfrac{du}{dy}=\dfrac{(3a-1)y^2-2y+3c}{3y^2}u+\dfrac{(y+2)(ay^2+by-c)}{3y^2}$$

• Thank you! That's extremely helpful. Commented Mar 5, 2019 at 10:08
• I tried reducing the form given in approach 1 in normal form. Doing a variable transform, $$z=x^{3 \gamma} \left( \frac{4 a - b}{a} - \frac{3}{1+3 a} x \right)$$ to the result obtained from approach 1, the DE reduces to- $$v \frac{dv}{dz} -v = - x^{3a} \frac{(c + 2b -4a) +x}{(b - 4a) + x}.$$ RHS can't be made a function of just z. I am confused here because I thought Able equation of second kind can always be reduced to canonical form. The value of constants in my case is $a=4/3, b=10/21, c=4/15$. Commented Apr 13, 2019 at 11:07