2.7.1 Problem (a)

Internal problem ID [20986]
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 : (a)
Date solved : Saturday, November 29, 2025 at 01:26:21 AM
CAS classification : [`y=_G(x,y')`]

\begin{align*} y^{\prime }&={\mathrm e}^{x}+\cos \left (y\right ) x \\ y \left (0\right ) &= 0 \\ \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 &= {\mathrm e}^{x}+\cos \left (y\right ) x\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial y} F_0 \\ &= \left (-x^{2} \cos \left (y\right )-x \,{\mathrm e}^{x}\right ) \sin \left (y\right )+\cos \left (y\right )+{\mathrm e}^{x}\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial y} F_1 \\ &= -\cos \left (y\right )^{3} x^{3}-2 \,{\mathrm e}^{x} \cos \left (y\right )^{2} x^{2}+x \left (x^{2} \sin \left (y\right )^{2}-{\mathrm e}^{2 x}-3 \sin \left (y\right )\right ) \cos \left (y\right )+\left (1+x^{2} \sin \left (y\right )^{2}+\left (-x -2\right ) \sin \left (y\right )\right ) {\mathrm e}^{x}\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_2}{\partial x}+ \frac {\partial F_2}{\partial y} F_2 \\ &= \left (\left (\sin \left (y\right ) x^{2}-2 x -1\right ) \cos \left (y\right )+{\mathrm e}^{x} \sin \left (y\right ) x \right ) {\mathrm e}^{2 x}+6 \left (x^{4} \sin \left (y\right )-2 x^{2}\right ) \cos \left (y\right )^{3}+12 \left (\sin \left (y\right ) x^{2}-\frac {x}{3}-\frac {7}{6}\right ) {\mathrm e}^{x} x \cos \left (y\right )^{2}+\left (\left (-x^{4}+6 x^{2} {\mathrm e}^{2 x}-3\right ) \sin \left (y\right )+\left (-x -2\right ) {\mathrm e}^{2 x}+6 x^{2}\right ) \cos \left (y\right )-\left (\left (x^{3}+x +3\right ) \sin \left (y\right )-x^{2}-5 x -1\right ) {\mathrm e}^{x}\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_3}{\partial x}+ \frac {\partial F_3}{\partial y} F_3 \\ &= \left (14 \cos \left (y\right )^{3} x^{3}+14 \,{\mathrm e}^{x} \cos \left (y\right )^{2} x^{2}+\left (-9+17 \left (x^{2}+x \right ) \sin \left (y\right )-7 x^{3}-6 x \right ) \cos \left (y\right )-7 \left (-\frac {3 \left (x +1\right ) \sin \left (y\right )}{7}+x^{2}\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{2 x}+\left (x^{2} \cos \left (y\right )^{2}+x \,{\mathrm e}^{x} \cos \left (y\right )+\sin \left (y\right ) \left (3 x +1\right )\right ) {\mathrm e}^{3 x}+24 \cos \left (y\right )^{5} x^{5}+60 \,{\mathrm e}^{x} \cos \left (y\right )^{4} x^{4}+2 \left (-10 x^{5}+18 \,{\mathrm e}^{2 x} x^{3}+30 x^{3} \sin \left (y\right )-15 x \right ) \cos \left (y\right )^{3}-45 \,{\mathrm e}^{x} \left (-\frac {4 x^{2} \left (x +5\right ) \sin \left (y\right )}{9}+x^{4}+\frac {x^{2}}{9}+\frac {5 x}{9}+\frac {4}{9}\right ) \cos \left (y\right )^{2}+\left (2 \left (2 \left (2 x^{2}+7 x \right ) {\mathrm e}^{2 x}-5 x^{3}\right ) \sin \left (y\right )+\left (-25 x^{3}-x -3\right ) {\mathrm e}^{2 x}+x^{5}+15 x \right ) \cos \left (y\right )+\left (\left (-x^{3}-9 x^{2}-x -4\right ) \sin \left (y\right )+x^{4}+x^{2}+7 x +9\right ) {\mathrm e}^{x} \end{align*}

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

\begin{align*} F_0 &= 1\\ F_1 &= 2\\ F_2 &= 1\\ F_3 &= -2\\ F_4 &= -23 \end{align*}

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

\[ y = x^{2}+x +\frac {x^{3}}{6}-\frac {x^{4}}{12}-\frac {23 x^{5}}{120}+O\left (x^{6}\right ) \]

Solution plot	Slope field \(y^{\prime } = {\mathrm e}^{x}+\cos \left (y\right ) x\)

✓ Maple. Time used: 0.003 (sec). Leaf size: 18

Order:=6; 
ode:=diff(y(x),x) = exp(x)+x*cos(y(x)); 
ic:=[y(0) = 0]; 
dsolve([ode,op(ic)],y(x),type='series',x=0);
 

\[ y = x +x^{2}+\frac {1}{6} x^{3}-\frac {1}{12} x^{4}-\frac {23}{120} x^{5}+\operatorname {O}\left (x^{6}\right ) \]

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 
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 = 5 
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: 0.081 (sec). Leaf size: 29

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

\[ y(x)\to -\frac {23 x^5}{120}-\frac {x^4}{12}+\frac {x^3}{6}+x^2+x \]

✓ Sympy. Time used: 0.399 (sec). Leaf size: 27

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

\[ y{\left (x \right )} = x + x^{2} + \frac {x^{3}}{6} - \frac {x^{4}}{12} - \frac {23 x^{5}}{120} + O\left (x^{6}\right ) \]