14.12.8 problem 8

Internal problem ID [2618]
Book : Differential equations and their applications, 4th ed., M. Braun
Section : Chapter 2. Second order differential equations. Section 2.8. Series solutions. Excercises page 197
Problem number : 8
Date solved : Monday, January 27, 2025 at 06:04:40 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} y^{\prime \prime }+\left (t^{2}+2 t +1\right ) y^{\prime }-\left (4+4 t \right ) y&=0 \end{align*}

Using series method with expansion around

\begin{align*} -1 \end{align*}

With initial conditions

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

✓ Solution by Maple

Time used: 0.002 (sec). Leaf size: 12

Order:=6; 
dsolve([diff(y(t),t$2)+(t^2+2*t+1)*diff(y(t),t)-(4+4*t)*y(t)=0,y(-1) = 0, D(y)(-1) = 1],y(t),type='series',t=-1);
 

\[ y = \left (t +1\right )+\frac {1}{4} \left (t +1\right )^{4}+\operatorname {O}\left (\left (t +1\right )^{6}\right ) \]

✓ Solution by Mathematica

Time used: 0.001 (sec). Leaf size: 15

AsymptoticDSolveValue[{D[y[t],{t,2}]+(t^2+2*t+1)*D[y[t],t]-(4+4*t)*y[t]==0,{y[-1]==0,Derivative[1][y][-1] ==1}},y[t],{t,-1,"6"-1}]
 

\[ y(t)\to \frac {1}{4} (t+1)^4+t+1 \]