Internal problem ID [15481]
Internal file name [OUTPUT/15482_Friday_May_10_2024_05_47_26_PM_28168364/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 18.1 Integration of differential equation in
series. Power series. Exercises page 171
Problem number: 738.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first order ode series method. Taylor series method"
Maple gives the following as the ode type
[`y=_G(x,y')`]
\[ \boxed {y^{\prime }-{\mathrm e}^{y}-y x=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \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 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}^{y}+y x\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial y} F_0 \\ &= {\mathrm e}^{2 y}+\left (y+1\right ) x \,{\mathrm e}^{y}+\left (x^{2}+1\right ) y\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial y} F_1 \\ &= x \left (3 y+2\right ) {\mathrm e}^{2 y}+2 \,{\mathrm e}^{3 y}+\left (y^{2} x^{2}+y \left (2 x^{2}+1\right )+x^{2}+2\right ) {\mathrm e}^{y}+\left (x^{3}+3 x \right ) y\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_2}{\partial x}+ \frac {\partial F_2}{\partial y} F_2 \\ &= \left (7 y^{2} x^{2}+11 y x^{2}+3 x^{2}+4 y+5\right ) {\mathrm e}^{2 y}+x \left (12 y+7\right ) {\mathrm e}^{3 y}+6 \,{\mathrm e}^{4 y}+\left (y^{3} x^{3}+\left (4 x^{3}+3 x \right ) y^{2}+\left (3 x^{3}+7 x \right ) y+x^{3}+5 x \right ) {\mathrm e}^{y}+\left (x^{4}+6 x^{2}+3\right ) y\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_3}{\partial x}+ \frac {\partial F_3}{\partial y} F_3 \\ &= \left (15 y^{3} x^{3}+\left (43 x^{3}+25 x \right ) y^{2}+7 \left (4 x^{3}+7 x \right ) y+4 x^{3}+18 x \right ) {\mathrm e}^{2 y}+\left (50 y^{2} x^{2}+69 y x^{2}+17 x^{2}+20 y+21\right ) {\mathrm e}^{3 y}+3 x \left (20 y+11\right ) {\mathrm e}^{4 y}+24 \,{\mathrm e}^{5 y}+\left (x^{4} y^{4}+\left (7 x^{4}+6 x^{2}\right ) y^{3}+\left (11 x^{4}+25 x^{2}+3\right ) y^{2}+\left (4 x^{4}+21 x^{2}+7\right ) y+x^{4}+9 x^{2}+8\right ) {\mathrm e}^{y}+\left (x^{5}+10 x^{3}+15 x \right ) y \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 &= 1\\ F_2 &= 4\\ F_3 &= 11\\ F_4 &= 53 \end {align*}
Substituting all the above in (6) and simplifying gives the solution as \[ y = x +\frac {x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {11 x^{4}}{24}+\frac {53 x^{5}}{120}+O\left (x^{6}\right ) \] Now we substitute the given initial conditions in the above to solve for \(y \left (0\right )\). Solving for \(y \left (0\right )\) from initial conditions gives \begin {align*} y \left (0\right ) = y \left (0\right ) \end {align*}
Therefore the solution becomes \begin {align*} y = x +\frac {1}{2} x^{2}+\frac {2}{3} x^{3}+\frac {11}{24} x^{4}+\frac {53}{120} x^{5} \end {align*}
Hence the solution can be written as \begin {gather*} y = x +\frac {x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {11 x^{4}}{24}+\frac {53 x^{5}}{120}+O\left (x^{6}\right ) \end {gather*} which simplifies to \begin {gather*} y = x +\frac {x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {11 x^{4}}{24}+\frac {53 x^{5}}{120}+O\left (x^{6}\right ) \end {gather*} 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 &= x +\frac {x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {11 x^{4}}{24}+\frac {53 x^{5}}{120}+O\left (x^{6}\right ) \\ \end{align*}
Verification of solutions
\[ y = x +\frac {x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {11 x^{4}}{24}+\frac {53 x^{5}}{120}+O\left (x^{6}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }={\mathrm e}^{y}+y x , y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{y}+y x \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 18
Order:=6; dsolve([diff(y(x),x)=exp(y(x))+x*y(x),y(0) = 0],y(x),type='series',x=0);
\[ y = x +\frac {1}{2} x^{2}+\frac {2}{3} x^{3}+\frac {11}{24} x^{4}+\frac {53}{120} x^{5}+\operatorname {O}\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.039 (sec). Leaf size: 33
AsymptoticDSolveValue[{y'[x]==Exp[y[x]]+x*y[x],{y[0]==0}},y[x],{x,0,5}]
\[ y(x)\to \frac {53 x^5}{120}+\frac {11 x^4}{24}+\frac {2 x^3}{3}+\frac {x^2}{2}+x \]