Second-order linear differential equation of the form $x^2 y'' + (ax-b)y' - ay =0$

Question

I need to solve the following differential equation \begin{equation} x^2 y'' + (ax-b)y' - ay =0 \end{equation} with $a,b>0$, $x\geq 0$ and $y(0)=0$. The power series method will fail since there is a singularity at $x=0$, while the form of the equation does not conform with the Frobenius method.

What other methods can I try in order to solve this?

Answer 1

With Laplace transforms this equation will be $$s^2F''(s)+(4-a-b)sF'(s)+2(1-a)F(s)=0$$ which is Euler's equation.

Answer 2

$$x^2 y'' +(ax-b)y'-ay=0 $$ Start by removing the leading behavior for small $x$. Near $x=0$ the leading terms in the equation are $$ -by'-ay \approx 0 $$ (this is valid as long as in the end, as $x\to 0$, $y''$ does not grow as fast as $y'/x^2$ or $y/x^2$, and $y'$ does not grow as fast as $\frac{y}x$).

The solution to the approximate equation is $y=e^{-\frac{a}{b}x}$ which motivates the substitution $$ y = u(x) e^{-\frac{a}{b}x} \\ y' = u' e^{-\frac{a}{b}x} - \frac{a}{b}ue^{-\frac{a}{b}x} \\ y'' = u''e^{-\frac{a}{b}x}-2\frac{a}{b}u'e^{-\frac{a}{b}x}+\frac{a^2}{b^2}ue^{-\frac{a}{b}x} $$ and the differential equation becomes $$ x^2u'' - \left( 2\frac{a}{b}x^2 -ax+b\right)u' +\left( \frac{a^2}{b^2}x^2-\frac{a^2}{b}x+a-a \right)u=0 $$ with $u(0)=0$.

Now if we let $u(x) = \sum_0^\infty u_kx^k$ $$ \sum_0^\infty k(k-1)u_kx^k-2\frac{a}{b}\sum_0^\infty ku_kx^{k+1} + a\sum_0^\infty ku_kx^k-b\sum_0^\infty ku_kx^{k-1} \\ +\frac{a^2}{b^2}\sum_0^\infty u_kx^{k+2}-\frac{a^2}{b}\sum_0^\infty u_kx^{k+1}=0 $$ And in each term we can substitute an offset index to always get $x^k$; this gives $$ \left[ k(k-1) +ak\right]u_k -\left[2\frac{a}{b}(k-1)+\frac{a^2}{b}\right]u_{k-1} -b(k+1)u_{k+1}+ \frac{a^2}{b^2}u_{k-2}=0 $$ and this is to be solved with $u_{-1}=u_0=0$, there is one free parameter $u_1$. $$ u_2 = \frac{au_1}{b} \\ $$ and so forth. So the solutions are this series, which is well behaved near zero, times $e^{-\frac{a}{b}x}$.

All of the coefficients are proportional to $u_1$. The overall solution has one free parameter (a scale factor that we will continue to call $u_1$) and is then $$ y =u_1 e^{-ax/b} \left[ x+\frac{a}{b}x^2+\frac23 \frac{2^2}{b^2} x^3+ \frac{5a^3+3a^2}{12b^2}x^4 +\frac{17a^4+39a^3+36}{60b^4}x^5 + \cdots\right]\\ = u_1 \left[ x + \frac{a^2}{6b^2}x^3+\frac{a^3+3a^2}{12b^3}x^4 + \frac{3a^4+16a^3+24a^2}{60b^4}x^5+ \right] $$ As to whether this series is convergent for all positive $x$, or has a finite radius of convergence, that depends on the values of $a$ and $b$.