14.9.11 problem 14

Internal problem ID [2577]
Book : Differential equations and their applications, 4th ed., M. Braun
Section : Chapter 2. Second order differential equations. Section 2.2.2. Equal roots, reduction of order. Excercises page 149
Problem number : 14
Date solved : Monday, January 27, 2025 at 06:01:09 AM
CAS classification : [_Gegenbauer]

\begin{align*} \left (-t^{2}+1\right ) y^{\prime \prime }-2 t y^{\prime }+6 y&=0 \end{align*}

Using reduction of order method given that one solution is

\begin{align*} y&=3 t^{2}-1 \end{align*}

✓ Solution by Maple

Time used: 0.004 (sec). Leaf size: 44

dsolve([(1-t^2)*diff(y(t),t$2)-2*t*diff(y(t),t)+6*y(t)=0,3*t^2-1],singsol=all)
 

\[ y = \frac {c_2 \left (3 t^{2}-1\right ) \ln \left (-1+t \right )}{2}+\frac {\left (-3 t^{2}+1\right ) c_2 \ln \left (t +1\right )}{2}-3 c_1 \,t^{2}+3 c_2 t +c_1 \]

✓ Solution by Mathematica

Time used: 0.025 (sec). Leaf size: 55

DSolve[(1-t^2)*D[y[t],{t,2}]-2*t*D[y[t],t]+6*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{2} c_1 \left (3 t^2-1\right )-\frac {1}{4} c_2 \left (\left (3 t^2-1\right ) \log (1-t)+\left (1-3 t^2\right ) \log (t+1)+6 t\right ) \]