22.14 problem 14

Internal problem ID [2377]
Internal file name [OUTPUT/2377_Tuesday_February_27_2024_08_36_31_AM_55098066/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 40, page 186
Problem number: 14.
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]]

\[ \boxed {y^{\prime \prime }+\frac {{y^{\prime }}^{2}}{2}-y=0} \] 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 &= -\frac {{y^{\prime }}^{2}}{2}+y\\ 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 \\ &= \frac {y^{\prime } \left ({y^{\prime }}^{2}-2 y+2\right )}{2}\\ 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 \\ &= -\frac {\left (3 {y^{\prime }}^{2}-2 y\right ) \left ({y^{\prime }}^{2}-2 y+2\right )}{4}\\ 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 \\ &= \frac {\left (3 {y^{\prime }}^{2}-4 y+1\right ) \left ({y^{\prime }}^{2}-2 y+2\right ) y^{\prime }}{2}\\ 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 \\ &= 4 \left (-\frac {{y^{\prime }}^{2}}{2}+y-1\right ) \left (\frac {15 {y^{\prime }}^{4}}{8}+\left (-\frac {15 y}{4}+\frac {11}{8}\right ) {y^{\prime }}^{2}+y^{2}-\frac {y}{4}\right )\\ F_5 &= \frac {d F_4}{dx} \\ &= \frac {\partial F_{4}}{\partial x}+ \frac {\partial F_{4}}{\partial y} y^{\prime }+ \frac {\partial F_{4}}{\partial y^{\prime }} F_4 \\ &= -34 \left (-\frac {{y^{\prime }}^{2}}{2}+y-1\right ) \left (\frac {1}{34}+\frac {45 {y^{\prime }}^{4}}{68}+\frac {\left (13-30 y\right ) {y^{\prime }}^{2}}{17}+y^{2}-\frac {10 y}{17}\right ) y^{\prime } \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 &= -{\frac {1}{2}}\\ F_1 &= {\frac {3}{2}}\\ F_2 &= -{\frac {9}{4}}\\ F_3 &= 6\\ F_4 &= -{\frac {39}{2}}\\ F_5 &= {\frac {297}{4}} \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y = x -\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {3 x^{4}}{32}+\frac {x^{5}}{20}-\frac {13 x^{6}}{480}+\frac {33 x^{7}}{2240}+O\left (x^{7}\right ) \] \[ y = x -\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {3 x^{4}}{32}+\frac {x^{5}}{20}-\frac {13 x^{6}}{480}+\frac {33 x^{7}}{2240}+O\left (x^{7}\right ) \] 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 &= x -\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {3 x^{4}}{32}+\frac {x^{5}}{20}-\frac {13 x^{6}}{480}+\frac {33 x^{7}}{2240}+O\left (x^{7}\right ) \\ \end{align*}

Verification of solutions

\[ y = x -\frac {x^{2}}{4}+\frac {x^{3}}{4}-\frac {3 x^{4}}{32}+\frac {x^{5}}{20}-\frac {13 x^{6}}{480}+\frac {33 x^{7}}{2240}+O\left (x^{7}\right ) \] 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)+(1/2)*_b(_a)^2-_a = 0, _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 20

Order:=7; 
dsolve([diff(y(x),x$2)+1/2*diff(y(x),x)^2-y(x)=0,y(0) = 0, D(y)(0) = 1],y(x),type='series',x=0);
 

\[ y \left (x \right ) = x -\frac {1}{4} x^{2}+\frac {1}{4} x^{3}-\frac {3}{32} x^{4}+\frac {1}{20} x^{5}-\frac {13}{480} x^{6}+\operatorname {O}\left (x^{7}\right ) \]

✓ Solution by Mathematica

Time used: 0.033 (sec). Leaf size: 40

AsymptoticDSolveValue[{y''[x]+1/2*y'[x]^2-y[x]==0,{y[0]==0,y'[0]==1}},y[x],{x,0,6}]
 

\[ y(x)\to -\frac {13 x^6}{480}+\frac {x^5}{20}-\frac {3 x^4}{32}+\frac {x^3}{4}-\frac {x^2}{4}+x \]