2.7.2 Problem (b.1)

Internal problem ID [20987]
Book : Ordinary Differential Equations. By Wolfgang Walter. Graduate texts in Mathematics. Springer. NY. QA372.W224 1998
Section : Chapter 2. Theory of First order differential equations. Excercises IV at page 89
Problem number : (b.1)
Date solved : Saturday, November 29, 2025 at 01:26:37 AM
CAS classification : [_Abel]

\begin{align*} y^{\prime }&=x^{3}+y^{3} \\ y \left (0\right ) &= 1 \\ \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 &= x^{3}+y^{3}\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial y} F_0 \\ &= 3 y^{5}+3 y^{2} x^{3}+3 x^{2}\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial y} F_1 \\ &= 15 y^{7}+21 y^{4} x^{3}+6 y x^{6}+9 y^{2} x^{2}+6 x\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_2}{\partial x}+ \frac {\partial F_2}{\partial y} F_2 \\ &= 105 y^{9}+189 x^{3} y^{6}+90 y^{3} x^{6}+6 x^{9}+81 x^{2} y^{4}+54 x^{5} y+18 x y^{2}+6\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_3}{\partial x}+ \frac {\partial F_3}{\partial y} F_3 \\ &= 945 y^{11}+2079 y^{8} x^{3}+891 x^{2} y^{6}+1404 y^{5} x^{6}+198 y^{4} x +918 x^{5} y^{3}+\left (270 x^{9}+18\right ) y^{2}+306 x^{4} y+108 x^{8} \end{align*}

And so on. Evaluating all the above at initial conditions \(x \left (0\right ) = 0\) and \(y \left (0\right ) = 1\) gives

\begin{align*} F_0 &= 1\\ F_1 &= 3\\ F_2 &= 15\\ F_3 &= 111\\ F_4 &= 963 \end{align*}

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

\[ y = x +1+\frac {3 x^{2}}{2}+\frac {5 x^{3}}{2}+\frac {37 x^{4}}{8}+\frac {321 x^{5}}{40}+O\left (x^{6}\right ) \]

Solution plot	Slope field \(y^{\prime } = x^{3}+y^{3}\)

✓ Maple. Time used: 0.003 (sec). Leaf size: 20

Order:=6; 
ode:=diff(y(x),x) = x^3+y(x)^3; 
ic:=[y(0) = 1]; 
dsolve([ode,op(ic)],y(x),type='series',x=0);
 

\[ y = 1+x +\frac {3}{2} x^{2}+\frac {5}{2} x^{3}+\frac {37}{8} x^{4}+\frac {321}{40} x^{5}+\operatorname {O}\left (x^{6}\right ) \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
trying Abel 
Looking for potential symmetries 
Looking for potential symmetries 
Looking for potential symmetries 
trying inverse_Riccati 
trying an equivalence to an Abel ODE 
differential order: 1; trying a linearization to 2nd order 
--- trying a change of variables {x -> y(x), y(x) -> x} 
differential order: 1; trying a linearization to 2nd order 
trying 1st order ODE linearizable_by_differentiation 
--- Trying Lie symmetry methods, 1st order --- 
   -> Computing symmetries using: way = 3 
   -> Computing symmetries using: way = 4 
   -> Computing symmetries using: way = 2 
trying symmetry patterns for 1st order ODEs 
-> trying a symmetry pattern of the form [F(x)*G(y), 0] 
-> trying a symmetry pattern of the form [0, F(x)*G(y)] 
-> trying symmetry patterns of the forms [F(x),G(y)] and [G(y),F(x)] 
-> trying a symmetry pattern of the form [F(x),G(x)] 
-> trying a symmetry pattern of the form [F(y),G(y)] 
-> trying a symmetry pattern of the form [F(x)+G(y), 0] 
-> trying a symmetry pattern of the form [0, F(x)+G(y)] 
-> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
-> trying a symmetry pattern of conformal type
 

✓ Mathematica. Time used: 0.017 (sec). Leaf size: 34

ode=D[y[x],x]==x^3+y[x]^3; 
ic={y[0]==1}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

\[ y(x)\to \frac {321 x^5}{40}+\frac {37 x^4}{8}+\frac {5 x^3}{2}+\frac {3 x^2}{2}+x+1 \]

✓ Sympy. Time used: 0.193 (sec). Leaf size: 36

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x**3 - y(x)**3 + Derivative(y(x), x),0) 
ics = {y(0): 1} 
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
 

\[ y{\left (x \right )} = 1 + x + \frac {3 x^{2}}{2} + \frac {5 x^{3}}{2} + \frac {37 x^{4}}{8} + \frac {321 x^{5}}{40} + O\left (x^{6}\right ) \]