78.19.18 problem 7 (a)

Internal problem ID [18437]
Book : DIFFERENTIAL EQUATIONS WITH APPLICATIONS AND HISTORICAL NOTES by George F. Simmons. 3rd edition. 2017. CRC press, Boca Raton FL.
Section : Chapter 5. Power Series Solutions and Special Functions. Section 29. Regular singular Points. Problems at page 227
Problem number : 7 (a)
Date solved : Tuesday, January 28, 2025 at 11:50:02 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} y^{\prime \prime }+\frac {n y^{\prime }}{x^{2}}+\frac {q y}{x^{3}}&=0 \end{align*}

Using series method with expansion around

\begin{align*} 0 \end{align*}

Solution by Maple

Order:=6; 
dsolve(diff(y(x),x$2)+n/x^2*diff(y(x),x)+q/x^3*y(x)=0,y(x),type='series',x=0);
 
\[ \text {No solution found} \]

Solution by Mathematica

Time used: 0.756 (sec). Leaf size: 790

AsymptoticDSolveValue[D[y[x],{x,2}]+n/x^2*D[y[x],x]+q/x^3*y[x]==0,y[x],{x,0,"6"-1}]
 
\[ y(x)\to e^{n/x} \left (\frac {n}{x}\right )^{-\frac {q}{n}} \left (-\frac {c_2 x^5 (n+q) (2 n+q)^2 (3 n+q)^2 (4 n+q) \left (-\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (-\frac {n}{x}\right )+\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (\frac {n}{x}\right )+\pi \cos \left (\frac {\pi (n+q)}{n}\right )\right )}{6 \pi n^{11}}-\frac {c_2 x^4 (n+q) (2 n+q)^2 (3 n+q) \left (-\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (-\frac {n}{x}\right )+\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (\frac {n}{x}\right )+\pi \cos \left (\frac {\pi (n+q)}{n}\right )\right )}{2 \pi n^8}-\frac {c_2 x^3 (n+q) (2 n+q) \left (-\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (-\frac {n}{x}\right )+\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (\frac {n}{x}\right )+\pi \cos \left (\frac {\pi (n+q)}{n}\right )\right )}{\pi n^5}-\frac {c_2 x^2 \left (-\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (-\frac {n}{x}\right )+\sin \left (\frac {\pi (n+q)}{n}\right ) \log \left (\frac {n}{x}\right )+\pi \cos \left (\frac {\pi (n+q)}{n}\right )\right )}{\pi n^2}\right )+\left (-\frac {n}{x}\right )^{\frac {q}{n}} \left (\frac {c_1 q^2 x^5 (n-q)^2 (2 n-q)^2 (3 n-q)^2 (4 n-q) (n+q)}{120 n^{15} \operatorname {Gamma}\left (\frac {q}{n}+2\right )}-\frac {c_1 q^2 x^4 (n-q)^2 (2 n-q)^2 (3 n-q) (n+q)}{24 n^{12} \operatorname {Gamma}\left (\frac {q}{n}+2\right )}+\frac {c_1 q^2 x^3 (n-q)^2 (2 n-q) (n+q)}{6 n^9 \operatorname {Gamma}\left (\frac {q}{n}+2\right )}-\frac {c_1 q^2 x^2 (n-q) (n+q)}{2 n^6 \operatorname {Gamma}\left (\frac {q}{n}+2\right )}-\frac {c_1 q x (n+q)}{n^3 \operatorname {Gamma}\left (\frac {q}{n}+2\right )}+\frac {c_1}{\operatorname {Gamma}\left (\frac {q}{n}+2\right )}\right )+e^{n/x} \left (\frac {n}{x}\right )^{-\frac {q}{n}-2} \left (\frac {c_1 x^5 (n+q) (2 n+q)^2 (3 n+q)^2 (4 n+q)^2 (5 n+q)^2 (6 n+q)}{120 n^{15} \operatorname {Gamma}\left (-\frac {q}{n}\right )}+\frac {c_1 x^4 (n+q) (2 n+q)^2 (3 n+q)^2 (4 n+q)^2 (5 n+q)}{24 n^{12} \operatorname {Gamma}\left (-\frac {q}{n}\right )}+\frac {c_1 x^3 (n+q) (2 n+q)^2 (3 n+q)^2 (4 n+q)}{6 n^9 \operatorname {Gamma}\left (-\frac {q}{n}\right )}+\frac {c_1 x^2 (n+q) (2 n+q)^2 (3 n+q)}{2 n^6 \operatorname {Gamma}\left (-\frac {q}{n}\right )}+\frac {c_1 x (n+q) (2 n+q)}{n^3 \operatorname {Gamma}\left (-\frac {q}{n}\right )}+\frac {c_1}{\operatorname {Gamma}\left (-\frac {q}{n}\right )}\right ) \]