34.3.1 problem 1
Internal
problem
ID
[6040]
Book
:
A
treatise
on
ordinary
and
partial
differential
equations
by
William
Woolsey
Johnson.
1913
Section
:
Chapter
VII,
Solutions
in
series.
Examples
XIV.
page
177
Problem
number
:
1
Date
solved
:
Monday, January 27, 2025 at 01:34:15 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
\begin{align*} x y^{\prime \prime }+\left (x +n \right ) y^{\prime }+\left (n +1\right ) y&=0 \end{align*}
Using series method with expansion around
\begin{align*} 0 \end{align*}
✓ Solution by Maple
Time used: 0.013 (sec). Leaf size: 164
Order:=6;
dsolve(x*diff(y(x),x$2)+(x+n)*diff(y(x),x)+(n+1)*y(x)=0,y(x),type='series',x=0);
\[
y = c_1 \,x^{-n +1} \left (1+2 \frac {1}{n -2} x +3 \frac {1}{\left (n -3\right ) \left (n -2\right )} x^{2}+4 \frac {1}{\left (n -4\right ) \left (n -3\right ) \left (n -2\right )} x^{3}+5 \frac {1}{\left (n -5\right ) \left (n -4\right ) \left (n -3\right ) \left (n -2\right )} x^{4}+6 \frac {1}{\left (n -6\right ) \left (n -5\right ) \left (n -4\right ) \left (n -3\right ) \left (n -2\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_2 \left (1+\frac {-n -1}{n} x +\frac {1}{2} \frac {n +2}{n} x^{2}-\frac {1}{6} \frac {n +3}{n} x^{3}+\frac {1}{24} \frac {n +4}{n} x^{4}-\frac {1}{120} \frac {n +5}{n} x^{5}+\operatorname {O}\left (x^{6}\right )\right )
\]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 519
AsymptoticDSolveValue[x*D[y[x],{x,2}]+(x+n)*D[y[x],x]+(n+1)*y[x]==0,y[x],{x,0,"6"-1}]
\[
y(x)\to c_2 \left (\frac {(-n-1) (n+2) (n+3) (n+4) (n+5) x^5}{n (2 n+2) (3 n+6) (4 n+12) (5 n+20)}-\frac {(-n-1) (n+2) (n+3) (n+4) x^4}{n (2 n+2) (3 n+6) (4 n+12)}+\frac {(-n-1) (n+2) (n+3) x^3}{n (2 n+2) (3 n+6)}-\frac {(-n-1) (n+2) x^2}{n (2 n+2)}+\frac {(-n-1) x}{n}+1\right )+c_1 \left (-\frac {720 x^5}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n)) ((4-n) (5-n)+n (5-n)) ((5-n) (6-n)+n (6-n))}+\frac {120 x^4}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n)) ((4-n) (5-n)+n (5-n))}-\frac {24 x^3}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n))}+\frac {6 x^2}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n))}-\frac {2 x}{(1-n) (2-n)+n (2-n)}+1\right ) x^{1-n}
\]