2.7.3 Problem (b.2)

Internal problem ID [20988]
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.2)
Date solved : Saturday, November 29, 2025 at 01:26:52 AM
CAS classification : [_quadrature]

\begin{align*} u^{\prime }&=u^{3} \\ u \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 &= u^{3}\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial u} F_0 \\ &= 3 u^{5}\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial u} F_1 \\ &= 15 u^{7}\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_2}{\partial x}+ \frac {\partial F_2}{\partial u} F_2 \\ &= 105 u^{9}\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_3}{\partial x}+ \frac {\partial F_3}{\partial u} F_3 \\ &= 945 u^{11} \end{align*}

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

\begin{align*} F_0 &= 1\\ F_1 &= 3\\ F_2 &= 15\\ F_3 &= 105\\ F_4 &= 945 \end{align*}

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

\[ u = 1+x +\frac {3 x^{2}}{2}+\frac {5 x^{3}}{2}+\frac {35 x^{4}}{8}+\frac {63 x^{5}}{8}+O\left (x^{6}\right ) \]

Solution plot	Slope field \(u^{\prime } = u^{3}\)

✓ Maple. Time used: 0.002 (sec). Leaf size: 20

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

\[ u = 1+x +\frac {3}{2} x^{2}+\frac {5}{2} x^{3}+\frac {35}{8} x^{4}+\frac {63}{8} x^{5}+\operatorname {O}\left (x^{6}\right ) \]

Maple trace

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

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}u \left (x \right )=u \left (x \right )^{3}, u \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}u \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=u \left (x \right )^{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )^{3}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )^{3}}d x =\int 1d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 u \left (x \right )^{2}}=x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {1}{\sqrt {-2 \mathit {C1} -2 x}}, u \left (x \right )=-\frac {1}{\sqrt {-2 \mathit {C1} -2 x}}\right \} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{u \left (x \right )=\frac {1}{\sqrt {\mathit {C1} -2 x}}, u \left (x \right )=-\frac {1}{\sqrt {\mathit {C1} -2 x}}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} u \left (0\right )=1 \\ {} & {} & 1=\frac {1}{\sqrt {\mathit {C1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=\frac {1}{\sqrt {1-2 x}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} u \left (0\right )=1 \\ {} & {} & 1=-\frac {1}{\sqrt {\mathit {C1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \textrm {No solution}\hspace {3pt} \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & u \left (x \right )=\frac {1}{\sqrt {1-2 x}} \end {array} \]

✓ Mathematica. Time used: 0.018 (sec). Leaf size: 34

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

\[ u(x)\to \frac {63 x^5}{8}+\frac {35 x^4}{8}+\frac {5 x^3}{2}+\frac {3 x^2}{2}+x+1 \]

✓ Sympy. Time used: 0.138 (sec). Leaf size: 36

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

\[ u{\left (x \right )} = 1 + x + \frac {3 x^{2}}{2} + \frac {5 x^{3}}{2} + \frac {35 x^{4}}{8} + \frac {63 x^{5}}{8} + O\left (x^{6}\right ) \]