# Differential equation with nasty coefficients $x^2(1-x)^2 y'' + (Ax + b)y = 0$

I have encountered a differential equation on the form

$$x^2(1-x)^2 y'' + (Ax + b)y = 0$$

My math background is too limited to even know where to begin, so any help of solving the equation (if a solution exist?) would be greatly appreciated!

• This is a second order homogenous linear equation with continuous coefficient functions. Therefore the initial value problem has a unique solution on intervals which do not contain zeroes of $x^2(1-x)^2$. A series solution can be computed with the Frobenius method, you can get further information from Trench, William F., "Elementary Differential Equations with Boundary Value Problems" (2013). Books and Monographs. Book 9., digitalcommons.trinity.edu/mono/9. Wolfram Alpha gives a solution in terms of the Gauss hypergeometric function Commented Mar 29, 2016 at 11:38
• Thank you! I forgot to mention $x \in (0,1)$, so the interval should be ok
Commented Mar 29, 2016 at 11:42
• Even for $a=0$, $b=1$ the solution seems to be a nightmare ! Commented Mar 29, 2016 at 12:41
• That is discouraging, but thanks anyways.
Commented Mar 29, 2016 at 13:45

Let $$y=x^p(1-x)^qu$$ ,

Then $$y'=x^p(1-x)^qu'+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u$$

$$y''=x^p(1-x)^qu''+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u=x^p(1-x)^qu''+2x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u$$

$$\therefore x^2(1-x)^2\left(x^p(1-x)^qu''+2x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u\right)+(Ax+b)x^p(1-x)^qu=0$$

$$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u+\dfrac{Ax+b}{x^2(1-x)^2}u=0$$

$$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u+\left(\dfrac{A+2b}{x}+\dfrac{b}{x^2}+\dfrac{A+2b}{1-x}+\dfrac{A+b}{(1-x)^2}\right)u=0$$

$$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)+b}{x^2}+\dfrac{A+2b-2pq}{x(1-x)}+\dfrac{q(q-1)+A+b}{(1-x)^2}\right)u=0$$

Choose $$p(p-1)+b=0$$ and $$q(q-1)+A+b=0$$ , the ODE reduces to the Gaussian hypergeometric equation.

Note that the special case of $$A=0$$ the ODE reduces to the form of http://eqworld.ipmnet.ru/en/solutions/ode/ode0224.pdf.

The solution to this ODE is obtained by applying a Moebius transformation to hypergeometric functions and then by multiplying the result by another function. See case (D) in my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. .