I am trying to solve an ODE by subbing in a series form and then looking individually at the coefficients of different powers of the variable. I'm looking at a general form of equation:
$$\frac{\mathrm{d}^2w}{\mathrm{d}z^2} + f_1(z) \frac{\mathrm{d}w}{\mathrm{d}z} + f_0(z)=0$$
I want to fill in the solution form $w(z) = e^{\lambda z} z^\mu \sum_{s=0}^\infty a_s z^{-s}$ and then collect coefficients of the various powers of $z$. However when I try to do this the series terms are not being multiplied with any $z$ terms which are multiplying the series from the outside. A simpler example of the problem I'm having is when I input say:
w[z_] = E^(λ z)z^μ Sum[a[s] z^-s, {s, 0, Infinity}] Coefficient[z^5 w[z], z^μ] The output of this is to give the coefficient as: $$e^{\lambda z} z^5\sum_{s=0}^\infty a_s z^{-s}$$
What I want is for the $z^5$ to multiply the $z^\mu$ term and also term-by-term with the series terms, and then give me just the coefficient of the resulting $z^\mu$ term. So the only corresponding term in the sum is the one of the form $e^{\lambda z}z^5z^\mu a_5 z^{-5}$, and therefore the coefficient I should be getting is just: $$e^{\lambda z} a_{5}$$
Can I not do this type of manipulation or have I just done something wrong here?
AsymptoticDSolveValueis sufficient for your use-case. Depending on the specific form of your $f_1$, $f_0$ it may or may not suffice. $\endgroup$