2.4.1 Problem 1

Internal problem ID [10346]
Book : First order enumerated odes
Section : section 4. First order odes solved using series method
Problem number : 1
Date solved : Thursday, November 27, 2025 at 10:34:51 AM
CAS classification : [_separable]

\begin{align*} y^{\prime }+2 x y&=x \\ \end{align*}

Series expansion around \(x=0\).

Solving ode using Taylor series method. This gives review on how the Taylor series method works for solving first order ode. Let

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

Where \(f\left ( x,y\right ) \) is analytic at expansion point \(x_{0}\). We can always shift to \(x_{0}=0\) if \(x_{0}\) is not zero. So from now we assume \(x_{0}=0\,\). Assume also that \(y\left ( x_{0}\right ) =y_{0}\). Using Taylor series

\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 }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xf+\frac {x^{2}}{2}\left . \frac {df}{dx}\right \vert _{x_{0},y_{0}}+\frac {x^{3}}{3!}\left . \frac {d^{2}f}{dx^{2}}\right \vert _{x_{0},y_{0}}+\cdots \\ & =y_{0}+\sum _{n=0}^{\infty }\frac {x^{n+1}}{\left ( n+1\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0}}\end{align*}

But

\begin{align} \frac {df}{dx} & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f\tag {1}\\ \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 ) 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 ) f\tag {3}\\ & \vdots \nonumber \end{align}

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

\begin{align} F_{0} & =f\left ( x,y\right ) \tag {4}\\ 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 ) F_{0} \tag {5}\end{align}

For example, for \(n=1\,\) we see that

\begin{align*} F_{1} & =\frac {d}{dx}\left ( F_{0}\right ) \\ & =\frac {\partial }{\partial x}F_{0}+\left ( \frac {\partial F_{0}}{\partial y}\right ) F_{0}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f \end{align*}

Which is (1). And when \(n=2\)

\begin{align*} F_{2} & =\frac {d}{dx}\left ( F_{1}\right ) \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) F_{0}\\ & =\frac {\partial }{\partial x}\left ( \frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f\right ) +\frac {\partial }{\partial y}\left ( \frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f\right ) f\\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) f \end{align*}

Which is (2) and so on. Therefore (4,5) can be used from now on along with

\begin{equation} y\left ( x\right ) =y_{0}+\sum _{n=0}^{\infty }\frac {x^{n+1}}{\left ( n+1\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0}} \tag {6}\end{equation}

Hence

\begin{align*} F_0 &= -2 x y+x\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial y} F_0 \\ &= 4 y x^{2}-2 x^{2}-2 y+1\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial y} F_1 \\ &= -2 \left (2 x^{2}-3\right ) x \left (2 y-1\right )\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_2}{\partial x}+ \frac {\partial F_2}{\partial y} F_2 \\ &= 2 \left (4 x^{4}-12 x^{2}+3\right ) \left (2 y-1\right )\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_3}{\partial x}+ \frac {\partial F_3}{\partial y} F_3 \\ &= -4 \left (4 x^{4}-20 x^{2}+15\right ) x \left (2 y-1\right ) \end{align*}

And so on. Evaluating all the above at initial conditions \(x \left (0\right ) = 0\) and \(y \left (0\right ) = y \left (0\right )\) gives

\begin{align*} F_0 &= 0\\ F_1 &= -2 y \left (0\right )+1\\ F_2 &= 0\\ F_3 &= 12 y \left (0\right )-6\\ F_4 &= 0 \end{align*}

Substituting all the above in (6) and simplifying gives the solution as

\[ y = \left (1-x^{2}+\frac {1}{2} x^{4}\right ) y \left (0\right )+\frac {x^{2}}{2}-\frac {x^{4}}{4}+O\left (x^{6}\right ) \]

Since \(x = 0\) is also an ordinary point, then standard power series can also be used. Writing the ODE as

\begin{align*} y^{\prime } + q(x)y &= p(x) \\ y^{\prime }+2 x y &= x \end{align*}

Where

\begin{align*} q(x) &= 2 x\\ p(x) &= x \end{align*}

Next, the type of the expansion point \(x = 0\) is determined. This point can be an ordinary point, a regular singular point (also called removable singularity), or irregular singular point (also called non-removable singularity or essential singularity). When \(x = 0\) is an ordinary point, then the standard power series is used. If the point is a regular singular point, Frobenius series is used instead. Irregular singular point requires more advanced methods (asymptotic methods) and is not supported now. Hopefully this will be added in the future. \(x = 0\) is called an ordinary point \(q(x)\) has a Taylor series expansion around the point \(x = 0\). \(x = 0\) is called a regular singular point if \(q(x)\) is not not analytic at \(x = 0\) but \(x q(x)\) has Taylor series expansion. And finally, \(x = 0\) is an irregular singular point if the point is not ordinary and not regular singular. This is the most complicated case. Now the expansion point \(x = 0\) is checked to see if it is an ordinary point or not. Let the solution be represented as power series of the form

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

Then

\begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1} \end{align*}

Substituting the above back into the ode gives

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

Expanding \(x\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives

\begin{align*} x &= x + \dots \\ &= x \end{align*}

Which simplifies to

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

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

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

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

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

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

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

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

\begin{align*} \left (2 a_{2}+2 a_{0}\right ) x&=x \\ 2 a_{2}+2 a_{0} &= 1 \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{2} = \frac {1}{2}-a_{0} \]

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

\begin{align*} \left (3 a_{3}+2 a_{1}\right ) x^{2}&=0 \\ 3 a_{3}+2 a_{1} &= 0 \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{3} = 0 \]

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

\begin{align*} \left (4 a_{4}+2 a_{2}\right ) x^{3}&=0 \\ 4 a_{4}+2 a_{2} &= 0 \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{4} = -\frac {1}{4}+\frac {a_{0}}{2} \]

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

\begin{align*} \left (5 a_{5}+2 a_{3}\right ) x^{4}&=0 \\ 5 a_{5}+2 a_{3} &= 0 \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{5} = 0 \]

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

\begin{align*} \left (6 a_{6}+2 a_{4}\right ) x^{5}&=0 \\ 6 a_{6}+2 a_{4} &= 0 \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{6} = \frac {1}{12}-\frac {a_{0}}{6} \]

And so on. Therefore the solution is

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

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

\[ y = a_{0}+\left (\frac {1}{2}-a_{0}\right ) x^{2}+\left (-\frac {1}{4}+\frac {a_{0}}{2}\right ) x^{4}+\dots \]

Collecting terms, the solution becomes

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

Figure 2.86: Slope field \(y^{\prime }+2 x y = x\)

✓ Maple. Time used: 0.002 (sec). Leaf size: 34

Order:=6; 
ode:=diff(y(x),x)+2*y(x)*x = x; 
dsolve(ode,y(x),type='series',x=0);
 

\[ y = \left (1-x^{2}+\frac {1}{2} x^{4}\right ) y \left (0\right )+\frac {x^{2}}{2}-\frac {x^{4}}{4}+O\left (x^{6}\right ) \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )+2 x y \left (x \right )=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-2 x y \left (x \right )+x \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{2 y \left (x \right )-1}=-x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{2 y \left (x \right )-1}d x =\int -x d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (2 y \left (x \right )-1\right )}{2}=-\frac {x^{2}}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {{\mathrm e}^{-x^{2}+2 \mathit {C1}}}{2}+\frac {1}{2} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} \,{\mathrm e}^{-x^{2}}+\frac {1}{2} \end {array} \]

✓ Mathematica. Time used: 0.014 (sec). Leaf size: 35

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

\[ y(x)\to -\frac {x^4}{4}+\frac {x^2}{2}+c_1 \left (\frac {x^4}{2}-x^2+1\right ) \]

✓ Sympy. Time used: 0.228 (sec). Leaf size: 27

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x*y(x) - x + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
 

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