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,



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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