14.14.24 problem 24

Internal problem ID [2661]
Book : Differential equations and their applications, 4th ed., M. Braun
Section : Chapter 2. Second order differential equations. Section 2.8.2, Regular singular points, the method of Frobenius. Excercises page 216
Problem number : 24
Date solved : Monday, January 27, 2025 at 06:05:36 AM
CAS classification : [_Bessel]

\begin{align*} t^{2} y^{\prime \prime }+t y^{\prime }+\left (t^{2}-v^{2}\right ) y&=0 \end{align*}

Using series method with expansion around

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

✓ Solution by Maple

Time used: 0.010 (sec). Leaf size: 73

Order:=6; 
dsolve(t^2*diff(y(t),t$2)+t*diff(y(t),t)+(t^2-v^2)*y(t)=0,y(t),type='series',t=0);
 

\[ y = c_1 \,t^{-v} \left (1+\frac {1}{-4+4 v} t^{2}+\frac {1}{32} \frac {1}{\left (v -2\right ) \left (v -1\right )} t^{4}+\operatorname {O}\left (t^{6}\right )\right )+c_2 \,t^{v} \left (1-\frac {1}{4 v +4} t^{2}+\frac {1}{32} \frac {1}{\left (v +2\right ) \left (v +1\right )} t^{4}+\operatorname {O}\left (t^{6}\right )\right ) \]

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 160

AsymptoticDSolveValue[t^2*D[y[t],{t,2}]+t*D[y[t],t]+(t^2-v^2)*y[t]==0,y[t],{t,0,"6"-1}]
 

\[ y(t)\to c_2 \left (\frac {t^4}{\left (-v^2-v+(1-v) (2-v)+2\right ) \left (-v^2-v+(3-v) (4-v)+4\right )}-\frac {t^2}{-v^2-v+(1-v) (2-v)+2}+1\right ) t^{-v}+c_1 \left (\frac {t^4}{\left (-v^2+v+(v+1) (v+2)+2\right ) \left (-v^2+v+(v+3) (v+4)+4\right )}-\frac {t^2}{-v^2+v+(v+1) (v+2)+2}+1\right ) t^v \]