 I need to solve the following differential equation x^2y''+(ax-b)y'-ay=0 with a,b>0,x\geq0 and veksetz 2021-12-17 Answered
I need to solve the following differential equation
$$\displaystyle{x}^{{2}}{y}{''}+{\left({a}{x}-{b}\right)}{y}'-{a}{y}={0}$$
with $$\displaystyle{a},{b}{>}{0},{x}\geq{0}$$ and $$\displaystyle{y}{\left({0}\right)}={0}$$. 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.

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$$\displaystyle{x}^{{2}}{y}{''}+{\left({a}{x}-{b}\right)}{y}'-{a}{y}={0}$$
Start by removing the leading behavior for small x. Near x=0 the leading terms in the equation are
$$\displaystyle-{b}{y}'-{a}{y}\approx{0}$$
(this is valid as long as in the end, as $$\displaystyle{x}\to{0},{y}{''}$$ does not grow as fast as $$\displaystyle{y}\frac{'}{{x}^{{2}}}$$ or $$\displaystyle\frac{{y}}{{x}^{{2}}}$$, does not grow as fast $$\displaystyle{\frac{{y}}{{x}}}$$).
The solution to the approximate equation is $$\displaystyle{y}={e}^{{-{\frac{{a}}{{b}}}{x}}}$$ which motivates the substitution
$$\displaystyle{y}={u}{\left({x}\right)}{e}^{{-{\frac{{a}}{{b}}}{x}}}$$
$$\displaystyle{y}'={u}'{e}^{{-{\frac{{a}}{{b}}}{x}}}-{\frac{{a}}{{b}}}{u}{e}^{{-{\frac{{a}}{{b}}}{x}}}$$
$$\displaystyle{y}{''}={u}{''}{e}^{{-{\frac{{{a}}}{{{b}}}}{x}}}-{2}{\frac{{{a}{b}}}{{u}}}'{e}^{{-{\frac{{a}}{{b}}}{x}}}+{\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{u}{e}^{{-{\frac{{a}}{{b}}}{x}}}$$
and the differential equation becomes
$$\displaystyle{x}^{{2}}{u}{''}-{\left({2}{\frac{{{a}}}{{{b}}}}{x}^{{2}}-{a}{x}+{b}\right)}{u}'+{\left({\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{x}^{{2}}-{\frac{{{a}^{{2}}}}{{{b}}}}{x}+{a}-{a}\right)}{u}={0}$$
with u(0)=0
Now if we let $$\displaystyle{u}{\left({x}\right)}={\sum_{{0}}^{\infty}}{u}_{{k}}{x}^{{k}}$$
$$\displaystyle{\sum_{{0}}^{{\infty}}}{k}{\left({k}-{1}\right)}{u}_{{k}}{x}^{{k}}-{2}{\frac{{a}}{{b}}}{\sum_{{0}}^{\infty}}{k}{u}_{{k}}{x}^{{{k}+{1}}}+{a}{\sum_{{0}}^{{\infty}}}{k}{u}_{{k}}{x}^{{k}}$$-$$b\sum ku_kx^{k-1}+\frac{a^2}{b^2}\sum_0^\infty u_kx^{k+2}-\frac{a^2}{b}\sum_0^\infty u_kx^{k+1}=0$$
And in each term we can substitute an offset index to always get $$\displaystyle{x}^{{k}}$$ this gives
$$\displaystyle{\left[{k}{\left({k}-{1}\right)}+{a}{k}\right)}{u}_{{k}}-{\left[{2}{\frac{{{a}}}{{{b}}}}{\left({k}-{1}\right)}+{\frac{{{a}^{{2}}}}{{{b}}}}\right]}{u}_{{{k}-{1}}}-{b}{\left({k}+{1}\right)}{u}_{{{k}+{1}}}+{\frac{{{a}^{{2}}}}{{{b}^{{2}}}}}{u}_{{{k}-{2}}}={0}$$
and this is to be solved with $$\displaystyle{u}_{{-{1}}}={u}_{{0}}={0}$$
there is on free parameter $$\displaystyle{u}_{{1}}$$
$$\displaystyle{u}_{{2}}={\frac{{{a}{u}_{{1}}}}{{{b}}}}$$
and so forth. So the solutions are this series, which is well behaved near zero, times $$\displaystyle{e}^{{-{\frac{{a}}{{b}}}{x}}}$$
All of the coefficients are proportional to $$\displaystyle{u}_{{1}}$$. The overall solution has one free parameter and is then
$$\displaystyle{y}={u}_{{1}}{e}^{{-{a}\frac{{x}}{{b}}}}{\left[{x}+{\frac{{a}}{{b}}}{x}^{{2}}+{\frac{{23}}{{\frac{{{2}^{{2}}}}{{{b}^{{2}}}}}}}{x}^{{3}}+{\frac{{{5}{a}^{{3}}+{3}{a}^{{2}}}}{{{12}{b}^{{2}}}}}{x}^{{4}}+{\frac{{{17}{a}^{{4}}+{39}{a}^{{3}}+{36}}}{{{60}{b}^{{4}}}}}{x}^{{5}}+\ldots\right]}$$
$$\displaystyle={u}_{{1}}{\left[{x}+{\frac{{{a}^{{2}}}}{{{6}{b}^{{2}}}}}{x}^{{3}}+{\frac{{{a}^{{3}}+{3}{a}^{{2}}}}{{{12}{b}^{{3}}}}}{x}^{{4}}+{\frac{{{3}{a}^{{4}}+{16}{a}^{{3}}+{24}{a}^{{2}}}}{{{60}{b}^{{4}}}}}{x}^{{5}}+\ldots\right]}$$
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.

Not exactly what you’re looking for? Gerald Lopez
It has a trivial solution $$\displaystyle{y}={a}{x}-{b}$$
Let $$\displaystyle{y}={\left({a}{x}-{b}\right)}{u}$$
Then $$\displaystyle{y}'={\left({a}{x}-{b}\right)}{u}'+{a}{u}$$
$$\displaystyle{y}{''}={\left({a}{x}-{b}\right)}{u}{''}+{a}{u}'+{a}{u}'={\left({a}{x}-{b}\right)}{u}{''}+{2}{a}{u}'$$
$$\displaystyle{x}^{{2}}{\left({\left({a}{x}-{b}\right)}{u}{''}+{2}{a}{u}'\right)}+{\left({a}{x}-{b}\right)}{\left({\left({a}{x}-{b}\right)}{u}'+{a}{u}\right)}-{a}{\left({a}{x}-{b}\right)}{u}={0}$$
$$\displaystyle{x}^{{2}}{\left({a}{x}-{b}\right)}{u}{''}+{2}{a}{x}^{{2}}{u}'+{\left({a}{x}-{b}\right)}^{{2}}{u}'+{a}{\left({a}{x}-{b}\right)}{u}-{a}{\left({a}{x}-{b}\right)}{u}={0}$$
$$\displaystyle{x}^{{2}}{\left({a}{x}-{b}\right)}{u}{''}=-{\left({2}{a}{x}^{{2}}+{\left({a}{x}-{b}\right)}^{{2}}\right)}{u}'$$
$$\displaystyle{\frac{{{u}{''}}}{{{u}}}}=-{\frac{{{2}{a}}}{{{a}{x}-{b}}}}-{\frac{{{a}}}{{{x}}}}+{\frac{{{b}}}{{{x}^{{2}}}}}$$
$$\displaystyle{\ln{{u}}}'=-{2}{\ln{{\left({a}{x}-{b}\right)}}}-{a}{\ln{{x}}}-{\frac{{{b}}}{{{x}}}}+{c}$$
$$\displaystyle{u}'={\frac{{{C}{e}^{{-{\frac{{{b}}}{{{x}}}}}}}}{{{x}^{{a}}{\left({a}{x}-{b}\right)}^{{2}}}}}$$
$$\displaystyle{u}={C}_{{1}}+{C}_{{2}}\int^{{x}}{\frac{{{e}^{{-{\frac{{{b}}}{{{x}}}}}}}}{{{x}^{{a}}{\left({a}{x}-{b}\right)}^{{2}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{y}={C}_{{1}}{\left({a}{x}-{b}\right)}+{C}_{{2}}{\left({a}{x}-{b}\right)}$$
$$\displaystyle\int^{{x}}{\frac{{{e}^{{-{\frac{{{b}}}{{{x}}}}}}}}{{{x}^{{a}}{\left({a}{x}-{b}\right)}^{{2}}}}}{\left.{d}{x}\right.}$$ RizerMix

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.