2.12.1 Problem 1

Internal problem ID [2610]
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 : 1
Date solved : Saturday, December 06, 2025 at 09:20:00 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

\begin{align*} y^{\prime \prime }+t y^{\prime }+y&=0 \\ \end{align*}

Series expansion around \(t=0\).

Entering second order ode series solverEntering second order ode series solver taylor solverSolving ode using Taylor series method. This gives review on how the Taylor series method works for solving second order ode.

Let

\[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \]

Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives

\begin{align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\end{align*}

But

\begin{align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end{align}

And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as

\begin{align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6}\end{align}

Therefore (6) can be used from now on along with

\begin{equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7}\end{equation}

To find \(y\left ( x\right ) \) series solution around \(x=0\). Hence

\begin{align*} F_0 &= -t y^{\prime }-y\\ F_1 &= \frac {d F_0}{dt} \\ &= \frac {\partial F_{0}}{\partial t}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= y^{\prime } t^{2}+t y-2 y^{\prime }\\ F_2 &= \frac {d F_1}{dt} \\ &= \frac {\partial F_{1}}{\partial t}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= -y^{\prime } t^{3}-t^{2} y+5 t y^{\prime }+3 y\\ F_3 &= \frac {d F_2}{dt} \\ &= \frac {\partial F_{2}}{\partial t}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= \left (t^{4}-9 t^{2}+8\right ) y^{\prime }+t y \left (t^{2}-7\right )\\ F_4 &= \frac {d F_3}{dt} \\ &= \frac {\partial F_{3}}{\partial t}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= \left (-t^{5}+14 t^{3}-33 t \right ) y^{\prime }-y \left (t^{4}-12 t^{2}+15\right ) \end{align*}

And so on. Evaluating all the above at initial conditions \(t = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives

\begin{align*} F_0 &= -y \left (0\right )\\ F_1 &= -2 y^{\prime }\left (0\right )\\ F_2 &= 3 y \left (0\right )\\ F_3 &= 8 y^{\prime }\left (0\right )\\ F_4 &= -15 y \left (0\right ) \end{align*}

Substituting all the above in (7) and simplifying gives the solution as

\[ y = \left (1-\frac {1}{2} t^{2}+\frac {1}{8} t^{4}\right ) y \left (0\right )+\left (t -\frac {1}{3} t^{3}+\frac {1}{15} t^{5}\right ) y^{\prime }\left (0\right )+O\left (t^{6}\right ) \]

Since the expansion point \(t = 0\) is an ordinary point, then this can also be solved using the standard power series method. Entering second order ode series solver ordinary point solverLet the solution be represented as power series of the form

\[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n} \]

Then

\begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2} \end{align*}

Substituting the above back into the ode gives

\begin{align*} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+t \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0\tag {1} \end{align*}

Which simplifies to

\begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n \,t^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0 \end{equation}

The next step is to make all powers of \(t\) be \(n\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n}\) and adjusting the power and the corresponding index gives

\begin{align*} \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} t^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) t^{n} \\ \end{align*}

Substituting all the above in Eq (2) gives the following equation where now all powers of \(t\) are the same and equal to \(n\).

\begin{equation} \tag{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) t^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n \,t^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 0 \end{equation}

\(n=0\) gives

\[ 2 a_{2}+a_{0}=0 \]

\[ a_{2} = -\frac {a_{0}}{2} \]

For \(1\le n\), the recurrence equation is

\begin{equation} \tag{4} \left (n +2\right ) a_{n +2} \left (n +1\right )+n a_{n}+a_{n} = 0 \end{equation}

Solving for \(a_{n +2}\), gives

\begin{equation} \tag{5} a_{n +2} = -\frac {a_{n}}{n +2} \end{equation}

For \(n = 1\) the recurrence equation gives

\[ 6 a_{3}+2 a_{1} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{3} = -\frac {a_{1}}{3} \]

For \(n = 2\) the recurrence equation gives

\[ 12 a_{4}+3 a_{2} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{4} = \frac {a_{0}}{8} \]

For \(n = 3\) the recurrence equation gives

\[ 20 a_{5}+4 a_{3} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{5} = \frac {a_{1}}{15} \]

For \(n = 4\) the recurrence equation gives

\[ 30 a_{6}+5 a_{4} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{6} = -\frac {a_{0}}{48} \]

For \(n = 5\) the recurrence equation gives

\[ 42 a_{7}+6 a_{5} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{7} = -\frac {a_{1}}{105} \]

And so on. Therefore the solution is

\begin{align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ &= a_{3} t^{3}+a_{2} t^{2}+a_{1} t +a_{0} + \dots \end{align*}

Substituting the values for \(a_{n}\) found above, the solution becomes

\[ y = a_{0}+a_{1} t -\frac {1}{2} a_{0} t^{2}-\frac {1}{3} a_{1} t^{3}+\frac {1}{8} a_{0} t^{4}+\frac {1}{15} a_{1} t^{5}+\dots \]

Collecting terms, the solution becomes

\begin{equation} \tag{3} y = \left (1-\frac {1}{2} t^{2}+\frac {1}{8} t^{4}\right ) a_{0}+\left (t -\frac {1}{3} t^{3}+\frac {1}{15} t^{5}\right ) a_{1}+O\left (t^{6}\right ) \end{equation}

At \(t = 0\) the solution above becomes

\[ y = \left (1-\frac {1}{2} t^{2}+\frac {1}{8} t^{4}\right ) y \left (0\right )+\left (t -\frac {1}{3} t^{3}+\frac {1}{15} t^{5}\right ) y^{\prime }\left (0\right )+O\left (t^{6}\right ) \]

2.12.1.1 ✓ Maple. Time used: 0.003 (sec). Leaf size: 39

Order:=6; 
ode:=diff(diff(y(t),t),t)+diff(y(t),t)*t+y(t) = 0; 
dsolve(ode,y(t),type='series',t=0);
 

\[ y = \left (1-\frac {1}{2} t^{2}+\frac {1}{8} t^{4}\right ) y \left (0\right )+\left (t -\frac {1}{3} t^{3}+\frac {1}{15} t^{5}\right ) y^{\prime }\left (0\right )+O\left (t^{6}\right ) \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )+t \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k} \\ \square & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot \left (\frac {d}{d t}y \left (t \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot \left (\frac {d}{d t}y \left (t \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} k \,t^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d^{2}}{d t^{2}}y \left (t \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) t^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) t^{k} \\ & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & {} & \moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )+a_{k} \left (k +1\right )\right ) t^{k}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1\right ) \left (a_{k +2} \left (k +2\right )+a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y \left (t \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +2}=-\frac {a_{k}}{k +2}\right ] \end {array} \]

2.12.1.2 ✓ Mathematica. Time used: 0.001 (sec). Leaf size: 42

ode=D[y[t],{t,2}]+t*D[y[t],t]+y[t]==0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[t],{t,0,5}]
 

\[ y(t)\to c_2 \left (\frac {t^5}{15}-\frac {t^3}{3}+t\right )+c_1 \left (\frac {t^4}{8}-\frac {t^2}{2}+1\right ) \]

2.12.1.3 ✓ Sympy. Time used: 0.181 (sec). Leaf size: 29

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t*Derivative(y(t), t) + y(t) + Derivative(y(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
 

\[ y{\left (t \right )} = C_{2} \left (\frac {t^{4}}{8} - \frac {t^{2}}{2} + 1\right ) + C_{1} t \left (1 - \frac {t^{2}}{3}\right ) + O\left (t^{6}\right ) \]