problem 33.11 (g)

Internal problem ID [13942]
Internal ﬁle name [OUTPUT/13114_Friday_February_23_2024_06_54_53_AM_93032955/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 33. Power series solutions I: Basic computational methods. Additional Exercises. page 641
Problem number: 33.11 (g).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable", "second order series method. Taylor series method"

\[ \boxed {y^{\prime \prime }-y^{2}=0} \] 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}\\ 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 \\ &= 2 y y^{\prime }\\ 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^{3}+2 {y^{\prime }}^{2}\\ 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 \\ &= 10 y^{2} y^{\prime } \end {align*}

And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives \begin {align*} F_0 &= y \left (0\right )^{2}\\ F_1 &= 2 y^{\prime }\left (0\right ) y \left (0\right )\\ F_2 &= 2 y \left (0\right )^{3}+2 {y^{\prime }\left (0\right )}^{2}\\ F_3 &= 10 y \left (0\right )^{2} y^{\prime }\left (0\right ) \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y = \frac {x^{4} y \left (0\right )^{3}}{12}+\left (\frac {x^{2}}{2}+\frac {y^{\prime }\left (0\right ) x^{5}}{12}\right ) y \left (0\right )^{2}+\left (1+\frac {y^{\prime }\left (0\right ) x^{3}}{3}\right ) y \left (0\right )+x y^{\prime }\left (0\right )+\frac {x^{4} {y^{\prime }\left (0\right )}^{2}}{12}+O\left (x^{5}\right ) \] Unable to also solve using normal power series since not linear ode. Not currently supported.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{4} y \left (0\right )^{3}}{12}+\left (\frac {x^{2}}{2}+\frac {y^{\prime }\left (0\right ) x^{5}}{12}\right ) y \left (0\right )^{2}+\left (1+\frac {y^{\prime }\left (0\right ) x^{3}}{3}\right ) y \left (0\right )+x y^{\prime }\left (0\right )+\frac {x^{4} {y^{\prime }\left (0\right )}^{2}}{12}+O\left (x^{5}\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {x^{4} y \left (0\right )^{3}}{12}+\left (\frac {x^{2}}{2}+\frac {y^{\prime }\left (0\right ) x^{5}}{12}\right ) y \left (0\right )^{2}+\left (1+\frac {y^{\prime }\left (0\right ) x^{3}}{3}\right ) y \left (0\right )+x y^{\prime }\left (0\right )+\frac {x^{4} {y^{\prime }\left (0\right )}^{2}}{12}+O\left (x^{5}\right ) \] Veriﬁed OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
<- 2nd_order WeierstrassP successful`

\[ y \left (x \right ) = \frac {x^{4} y \left (0\right )^{3}}{12}+\frac {y \left (0\right )^{2} x^{2}}{2}+\left (1+\frac {D\left (y \right )\left (0\right ) x^{3}}{3}\right ) y \left (0\right )+x D\left (y \right )\left (0\right )+\frac {D\left (y \right )\left (0\right )^{2} x^{4}}{12}+O\left (x^{5}\right ) \]

\[ y(x)\to \frac {1}{12} \left (c_1{}^3+c_2{}^2\right ) x^4+\frac {1}{3} c_1 c_2 x^3+\frac {c_1{}^2 x^2}{2}+c_2 x+c_1 \]

23.33 problem 33.11 (g)