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I am working on more series solutions for ODEs and am struggling using the Frobenius method to solve a problem however I have managed to get part way through just question but am unable to continue.

$$y'' + (x-6)y = 0 $$ Is the given DE and I am asked to find a basis of solutions using the Frobenius method and there is a hint to try and identify the series expansions of known functions.

So far I have calculated the indicial equation which I believe to be: $r(r-1) + b_0r + c_0 = 0$

From here I believe the solutions to this are zero and one which I have then substituted into $$\sum_{m=0}^\infty a_m(m+r)(m+r-1)x^{m+r-2} + a_mx^{m+1} - 6a_mx^m = 0$$

But here I believe I have done something wrong somewhere as I do not have an $a_{m+1}$ term which I could then solve for and start equation coefficients. I tried to continue here however I tried solving for $a_m$ but did not come up with anything that makes sense.

If possible could we please go through the solution to this problem as I have written answers in my book however there are no solutions or explanations as to how the answers were formulated.

Thank you for any help or feedback,

Michael

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The biggest problem is that you are going to have three different subscripts on $a$ for each power of $x$. Let's fix this by the substitution $z=x-6$. Then $\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{dy}{dz}$ and similarly $\frac{d^2y}{dx^2}=\frac{d^2y}{dz^2}$ so $$\frac{d^2y}{dz^2}+zy=0$$ $z=0$ is an ordinary point of the differential equation and we don't need to mess around with the indicial equation because a Taylor series will solve it. Accordingly, we let $$y=\sum_{m=0}^{\infty}a_mz^m$$ On substitution into the differential equation we get $$\begin{align}\sum_{m=0}^{\infty}m(m-1)a_mz^{m-2}+\sum_{m=0}^{\infty}a_mz^{m+1}&=\sum_{m=2}^{\infty}m(m-1)a_mz^{m-2}+\sum_{m=0}^{\infty}a_mz^{m+1}\\ &=\sum_{m=0}^{\infty}(m+2)(m+1)a_{m+2}z^m+\sum_{m=1}^{\infty}a_{m-1}z^m\\ &=2a_2+\sum_{m=1}^{\infty}\left[(m+2)(m+1)a_{m+2}+a_{m-1}\right]z^m\\ &=0\end{align}$$ Note how we strove to get to a common power of $z$ rather than a common subscript of $a$. Matching coefficients for each power of $z$, we see firt off the $a-2=0$ and have a relationship between $a_n$ and $a_{n+3}$. For this reason we want to consider separately each congruence class $\pmod{3}$. If $m\equiv0\pmod{3}$, then $$\begin{align}(3m+2)(3m+1)a_{3m+2}&=3\left(m+\frac23\right)\cdot3\left(m+\frac13\right)a_{3m+2}\\ &=\frac{3^{m+1}}{3^m}\frac{\Gamma\left(m+\frac53\right)}{\Gamma\left(m+\frac23\right)}\frac{3^{m+1}}{3^m}\frac{\Gamma\left(m+\frac43\right)}{\Gamma\left(m+\frac13\right)}a_{3m+2}\\ &=-a_{3m-1}=\frac{(-1)^{m+1}}{(-1)^m}a_{3m-1}\end{align}$$ $$\begin{align}\frac{3^{2m+2}\Gamma\left(m+\frac53\right)\Gamma\left(m+\frac43\right)}{(-1)^{m+1}}a_{3m+2}&=\frac{3^{2m}\Gamma\left(m+\frac23\right)\Gamma\left(m+\frac13\right)}{(-1)^{m}}a_{3m-1}\\ &=-9\Gamma\left(\frac53\right)\Gamma\left(\frac43\right)a_2=0\end{align}$$ If $m\equiv1\pmod{3}$ then $$\begin{align}(3m+3)(3m+2)a_{3m+3}&=3(m+1)\cdot3\left(m+\frac23\right)a_{3m+3}\\ &=\frac{3^{m+1}}{3^m}\frac{(m+1)!}{m!}\frac{3^{m+1}}{3^m}\frac{\Gamma\left(m+\frac53\right)}{\Gamma\left(m+\frac23\right)}a_{3m+3}\\ &=-a_{3m}=\frac{(-1)^{m+1}}{(-1)^m}a_{3m}\end{align}$$ $$\frac{3^{2m+2}(m+1)!\Gamma\left(m+\frac53\right)}{(-1)^{m+1}}a_{3m+3}=\frac{3^{2m}m!\Gamma\left(m+\frac23\right)}{(-1)^{m}}a_{3m}=\Gamma\left(\frac23\right)a_0$$ So that gives us one solution $$y_1=a_0\sum_{m=0}^{\infty}\frac{(-1)^m\Gamma\left(\frac23\right)}{3^{2m}m!\Gamma\left(m+\frac23\right)}(x-6)^{3m}$$ If $m\equiv2\pmod{3}$ then $$\begin{align}(3m+4)(3m+3)a_{3m+4}&=3\left(m+\frac43\right)\cdot3(m+1)a_{3m+4}\\ &=\frac{3^{m+1}}{3^m}\frac{\Gamma\left(m+\frac73\right)}{\Gamma\left(m+\frac43\right)}\frac{3^{m+1}}{3^m}\frac{(m+1)!}{m!}a_{3m+4}\\ &=-a_{3m+1}=\frac{(-1)^{m+1}}{(-1)^m}a_{3m+1}\end{align}$$ $$\frac{3^{2m+2}\Gamma\left(m+\frac73\right)(m+1)!}{(-1)^{m+1}}a_{3m+4}=\frac{3^{2m}\Gamma\left(m+\frac43\right)m!}{(-1)^m}a_{3m+1}=\Gamma\left(\frac43\right)a_1$$ So that is our second linearly independent solution $$y_1=a_1\sum_{m=0}^{\infty}\frac{(-1)^m\Gamma\left(\frac43\right)}{3^{2m}m!\Gamma\left(m+\frac43\right)}(x-6)^{3m+1}$$ So once you get a two-term recurrence relation and recall that $x\Gamma(x)=\Gamma(x+1)$ so you can write all the factors as ratios of funciont of $m$ and $m+1$, you can solve the recurrence relation to get a closed form for the general term.

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