Internal problem ID [15468]
Internal file name [OUTPUT/15469_Wednesday_May_08_2024_04_00_59_PM_27445916/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: 725.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated", "first order ode series method. Taylor series method"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, [_Abel, `2nd type`, `class A`]]
\[ \boxed {y^{\prime }-\frac {y-x}{y+x}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}
With the expansion point for the power series method at \(x = 0\).
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {y -x}{y +x} \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=1\) is \[
\{x <-1\boldsymbol {\lor }-1 The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=1\) is \[
\{x <-1\boldsymbol {\lor }-1
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 &= \frac {y-x}{y+x}\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial y} F_0 \\ &= \frac {-2 x^{2}-2 y^{2}}{\left (y+x \right )^{3}}\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial y} F_1 \\ &= -\frac {4 \left (x -2 y\right ) \left (x^{2}+y^{2}\right )}{\left (y+x \right )^{5}}\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_2}{\partial x}+ \frac {\partial F_2}{\partial y} F_2 \\ &= -\frac {20 \left (x^{2}+y^{2}\right ) \left (x^{2}-2 x y+3 y^{2}\right )}{\left (y+x \right )^{7}}\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_3}{\partial x}+ \frac {\partial F_3}{\partial y} F_3 \\ &= \frac {40 \left (x^{2}+y^{2}\right ) \left (16 y^{3}-13 x y^{2}+10 y x^{2}-3 x^{3}\right )}{\left (y+x \right )^{9}} \end {align*}
And so on. Evaluating all the above at initial conditions \(x \left (0\right ) = 0\) and \(y \left (0\right ) = 1\) gives \begin {align*} F_0 &= 1\\ F_1 &= -2\\ F_2 &= 8\\ F_3 &= -60\\ F_4 &= 640 \end {align*}
Substituting all the above in (6) and simplifying gives the solution as \[
y = -x^{2}+x +1+\frac {4 x^{3}}{3}-\frac {5 x^{4}}{2}+\frac {16 x^{5}}{3}+O\left (x^{6}\right )
\] 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^{2}+x +1+\frac {4}{3} x^{3}-\frac {5}{2} x^{4}+\frac {16}{3} x^{5} \end {align*}
Hence the solution can be written as \begin {gather*} y = -x^{2}+x +1+\frac {4 x^{3}}{3}-\frac {5 x^{4}}{2}+\frac {16 x^{5}}{3}+O\left (x^{6}\right ) \end {gather*} which simplifies to \begin {gather*} y = -x^{2}+x +1+\frac {4 x^{3}}{3}-\frac {5 x^{4}}{2}+\frac {16 x^{5}}{3}+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^{2}+x +1+\frac {4 x^{3}}{3}-\frac {5 x^{4}}{2}+\frac {16 x^{5}}{3}+O\left (x^{6}\right ) \\
\end{align*} Verification of solutions
\[
y = -x^{2}+x +1+\frac {4 x^{3}}{3}-\frac {5 x^{4}}{2}+\frac {16 x^{5}}{3}+O\left (x^{6}\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {y-x}{y+x}, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y-x}{y+x} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 20
\[
y \left (x \right ) = 1+x -x^{2}+\frac {4}{3} x^{3}-\frac {5}{2} x^{4}+\frac {16}{3} x^{5}+\operatorname {O}\left (x^{6}\right )
\]
✓ Solution by Mathematica
Time used: 0.038 (sec). Leaf size: 32
\[
y(x)\to \frac {16 x^5}{3}-\frac {5 x^4}{2}+\frac {4 x^3}{3}-x^2+x+1
\]
23.2.2 Solving as series ode
23.2.3 Maple step by step solution
`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 homogeneous D
<- homogeneous successful`
Order:=6;
dsolve([diff(y(x),x)=(y(x)-x)/(y(x)+x),y(0) = 1],y(x),type='series',x=0);
AsymptoticDSolveValue[{y'[x]==(y[x]-x)/(y[x]+x),{y[0]==1}},y[x],{x,0,5}]