Internal problem ID [14794]
Internal file name [OUTPUT/14474_Monday_April_08_2024_06_25_39_AM_92220687/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.8, page 203
Problem number: 21.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second order series method. Taylor series method"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Van_der_Pol]
\[ \boxed {y^{\prime \prime }+\left (-1+y^{2}\right ) y^{\prime }+y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}
With the expansion point for the power series method at \(x = 0\).
Solving 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 &= -y^{2} y^{\prime }-y+y^{\prime }\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_{0}}{\partial x}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= \left (-2 {y^{\prime }}^{2}+\left (y^{3}-2 y\right ) y^{\prime }+y^{2}-1\right ) y\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_{1}}{\partial x}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= -2 {y^{\prime }}^{3}+\left (8 y^{3}-8 y\right ) {y^{\prime }}^{2}+\left (-y^{6}+3 y^{4}+5 y^{2}-1\right ) y^{\prime }-y^{5}+2 y^{3}\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_{2}}{\partial x}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= \left (30 y^{2}-14\right ) {y^{\prime }}^{3}+\left (-22 y^{5}+44 y^{3}\right ) {y^{\prime }}^{2}+\left (y^{8}-4 y^{6}-23 y^{4}+28 y^{2}-1\right ) y^{\prime }+y^{7}-3 y^{5}-5 y^{3}+y\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_{3}}{\partial x}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= 60 y {y^{\prime }}^{4}+\left (-200 y^{4}+264 y^{2}-42\right ) {y^{\prime }}^{3}+\left (52 y^{7}-156 y^{5}-94 y^{3}+98 y\right ) {y^{\prime }}^{2}+\left (-y^{10}+5 y^{8}+70 y^{6}-154 y^{4}+14 y^{2}\right ) y^{\prime }-y^{9}+4 y^{7}+23 y^{5}-28 y^{3}+y \end {align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = 0\) and \(y^{\prime }\left (0\right ) = 1\) gives \begin {align*} F_0 &= 1\\ F_1 &= 0\\ F_2 &= -3\\ F_3 &= -15\\ F_4 &= -42 \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[ y = \frac {x^{2}}{2}+x -\frac {x^{4}}{8}-\frac {x^{5}}{8}-\frac {7 x^{6}}{120}+O\left (x^{6}\right ) \] \[ y = \frac {x^{2}}{2}+x -\frac {x^{4}}{8}-\frac {x^{5}}{8}-\frac {7 x^{6}}{120}+O\left (x^{6}\right ) \] Unable to also solve using normal power series since not linear ode. Not currently supported.
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2}}{2}+x -\frac {x^{4}}{8}-\frac {x^{5}}{8}-\frac {7 x^{6}}{120}+O\left (x^{6}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2}}{2}+x -\frac {x^{4}}{8}-\frac {x^{5}}{8}-\frac {7 x^{6}}{120}+O\left (x^{6}\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_a^2*_b(_a)-_b(_a)+_a = 0, _b(_a)` *** Sublevel 2 *** 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 -> trying 2nd order, dynamical_symmetries, fully reducible to Abel through one integrating factor of the form G(x,y)/(1+H(x,y)*y)^2 --- trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for dynamical symmetries --- -> trying 2nd order, dynamical_symmetries, fully reducible to Abel through one integrating factor of the form G(x,y)/(1+H(x,y)*y) trying 2nd order, integrating factors of the form mu(x,y)/(y)^n, only the singular cases trying differential order: 2; exact nonlinear trying 2nd order, integrating factor of the form mu(x,y) -> trying 2nd order, the S-function method -> trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for the S-function -> trying 2nd order, the S-function method -> trying 2nd order, No Point Symmetries Class V --- trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for dynamical symmetries --- -> trying 2nd order, No Point Symmetries Class V -> trying 2nd order, No Point Symmetries Class V --- trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for dynamical symmetries --- -> trying 2nd order, No Point Symmetries Class V -> trying 2nd order, No Point Symmetries Class V --- trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for dynamical symmetries --- -> trying 2nd order, No Point Symmetries Class V trying 2nd order, integrating factor of the form mu(x,y)/(y)^n, only the general case -> trying 2nd order, dynamical_symmetries, only a reduction of order through one integrating factor of the form G(x,y)/(1+H(x,y)*y)^ --- trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for dynamical symmetries --- -> trying 2nd order, dynamical_symmetries, only a reduction of order through one integrating factor of the form G(x,y)/(1+H(x,y)* --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 5 `, `-> Computing symmetries using: way = formal *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful -> Calling odsolve with the ODE`, diff(y(x), x), y(x)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 16
Order:=6; dsolve([diff(y(x),x$2)+(y(x)^2-1)*diff(y(x),x)+y(x)=0,y(0) = 0, D(y)(0) = 1],y(x),type='series',x=0);
\[ y \left (x \right ) = x +\frac {1}{2} x^{2}-\frac {1}{8} x^{4}-\frac {1}{8} x^{5}+\operatorname {O}\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.021 (sec). Leaf size: 26
AsymptoticDSolveValue[{y''[x]+(y[x]^2-1)*y'[x]+y[x]==0,{y[0]==0,y'[0]==1}},y[x],{x,0,5}]
\[ y(x)\to -\frac {x^5}{8}-\frac {x^4}{8}+\frac {x^2}{2}+x \]