Internal problem ID [4187]
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.
ODE order: 2.
ODE degree: 1.
CAS Maple gives this as type [[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x y^{\prime \prime }+\left (x +n \right ) y^{\prime }+\left (n +1\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 248
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);
\[ y \left (x \right ) = c_{1} x^{-n +1} \left (1+2 \frac {1}{-2+n} x +3 \frac {1}{\left (-3+n \right ) \left (-2+n \right )} x^{2}+4 \frac {1}{\left (n -4\right ) \left (-3+n \right ) \left (-2+n \right )} x^{3}+5 \frac {1}{\left (n -5\right ) \left (n -4\right ) \left (-3+n \right ) \left (-2+n \right )} x^{4}+6 \frac {1}{\left (n -6\right ) \left (n -5\right ) \left (n -4\right ) \left (-3+n \right ) \left (-2+n \right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+\frac {-n -1}{n} x +\frac {1}{2} \frac {n +2}{n} x^{2}-\frac {1}{6} \frac {n +3}{n} x^{3}+\frac {1}{24} \frac {n +4}{n} x^{4}-\frac {1}{120} \frac {n +5}{n} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 519
AsymptoticDSolveValue[x*y''[x]+(x+n)*y'[x]+(n+1)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {(-n-1) (n+2) (n+3) (n+4) (n+5) x^5}{n (2 n+2) (3 n+6) (4 n+12) (5 n+20)}-\frac {(-n-1) (n+2) (n+3) (n+4) x^4}{n (2 n+2) (3 n+6) (4 n+12)}+\frac {(-n-1) (n+2) (n+3) x^3}{n (2 n+2) (3 n+6)}-\frac {(-n-1) (n+2) x^2}{n (2 n+2)}+\frac {(-n-1) x}{n}+1\right )+c_1 \left (-\frac {720 x^5}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n)) ((4-n) (5-n)+n (5-n)) ((5-n) (6-n)+n (6-n))}+\frac {120 x^4}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n)) ((4-n) (5-n)+n (5-n))}-\frac {24 x^3}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n))}+\frac {6 x^2}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n))}-\frac {2 x}{(1-n) (2-n)+n (2-n)}+1\right ) x^{1-n} \]