problem 732

Internal problem ID [15475]
Internal ﬁle name [OUTPUT/15476_Friday_May_10_2024_05_47_24_PM_54640770/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: 732.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "linear", "separable", "homogeneousTypeD2", "ﬁrst_order_ode_lie_symmetry_lookup", "ﬁrst order ode series method. Ordinary point", "ﬁrst order ode series method. Taylor series method"

\[ \boxed {y^{\prime }-2 x y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

23.9.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

23.9.2 Solving as series ode

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 &= 2 x y\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial y} F_0 \\ &= \left (4 x^{2}+2\right ) y\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial y} F_1 \\ &= \left (8 x^{3}+12 x \right ) y\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_2}{\partial x}+ \frac {\partial F_2}{\partial y} F_2 \\ &= 4 y \left (4 x^{4}+12 x^{2}+3\right )\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_3}{\partial x}+ \frac {\partial F_3}{\partial y} F_3 \\ &= 32 \left (x^{4}+5 x^{2}+\frac {15}{4}\right ) y x \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 &= 0\\ F_1 &= 2\\ F_2 &= 0\\ F_3 &= 12\\ F_4 &= 0 \end {align*}

Substituting all the above in (6) and simplifying gives the solution as \[ y = x^{2}+1+\frac {x^{4}}{2}+O\left (x^{6}\right ) \] 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}+1+\frac {1}{2} x^{4} \end {align*}

Hence the solution can be written as \begin {gather*} y = x^{2}+1+\frac {x^{4}}{2}+O\left (x^{6}\right ) \end {gather*} which simpliﬁes to \begin {gather*} y = x^{2}+1+\frac {x^{4}}{2}+O\left (x^{6}\right ) \end {gather*} Since \(x = 0\) is also an ordinary point, then standard power series can also be used. Writing the ODE as \begin {align*} y^{\prime } + q(x)y &= p(x) \\ y^{\prime }-2 x y &= 0 \end {align*}

Next, the type of the expansion point \(x = 0\) is determined. This point can be an ordinary point, a regular singular point (also called removable singularity), or irregular singular point (also called non-removable singularity or essential singularity). When \(x = 0\) is an ordinary point, then the standard power series is used. If the point is a regular singular point, Frobenius series is used instead. Irregular singular point requires more advanced methods (asymptotic methods) and is not supported now. Hopefully this will be added in the future. \(x = 0\) is called an ordinary point \(q(x)\) has a Taylor series expansion around the point \(x = 0\). \(x = 0\) is called a regular singular point if \(q(x)\) is not not analytic at \(x = 0\) but \(x q(x)\) has Taylor series expansion. And ﬁnally, \(x = 0\) is an irregular singular point if the point is not ordinary and not regular singular. This is the most complicated case. Now the expansion point \(x = 0\) is checked to see if it is an ordinary point or not. Let the solution be represented as power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} \] Then \begin {align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1} \end {align*}

Substituting the above back into the ode gives \begin {align*} \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )-2 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0\tag {1} \end {align*}

Which simpliﬁes to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (1+n \right ) a_{1+n} x^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n}\right ) \\ \end{align*} Substituting all the above in Eq (2) gives the following equation where now all powers of \(x\) are the same and equal to \(n\). \begin{equation} \tag{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (1+n \right ) a_{1+n} x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n}\right ) = 0 \end{equation} For \(1\le n\), the recurrence equation is \begin{equation} \tag{4} \left (1+n \right ) a_{1+n}-2 a_{n -1} = 0 \end{equation} Solving for \(a_{1+n}\), gives \begin{equation} \tag{5} a_{1+n} = \frac {2 a_{n -1}}{1+n} \end{equation} For \(n = 1\) the recurrence equation gives \[ 2 a_{2}-2 a_{0} = 0 \] Which after substituting the earlier terms found becomes \[ a_{2} = a_{0} \] For \(n = 2\) the recurrence equation gives \[ 3 a_{3}-2 a_{1} = 0 \] Which after substituting the earlier terms found becomes \[ a_{3} = 0 \] For \(n = 3\) the recurrence equation gives \[ 4 a_{4}-2 a_{2} = 0 \] Which after substituting the earlier terms found becomes \[ a_{4} = \frac {a_{0}}{2} \] For \(n = 4\) the recurrence equation gives \[ 5 a_{5}-2 a_{3} = 0 \] Which after substituting the earlier terms found becomes \[ a_{5} = 0 \] For \(n = 5\) the recurrence equation gives \[ 6 a_{6}-2 a_{4} = 0 \] Which after substituting the earlier terms found becomes \[ a_{6} = \frac {a_{0}}{6} \] And so on. Therefore the solution is \begin {align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ &= a_{3} x^{3}+a_{2} x^{2}+a_{1} x +a_{0} + \dots \end {align*}

Substituting the values for \(a_{n}\) found above, the solution becomes \[ y = a_{0}+a_{0} x^{2}+\frac {1}{2} a_{0} x^{4}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y = \left (x^{2}+1+\frac {1}{2} x^{4}\right ) a_{0}+O\left (x^{6}\right ) \end{equation} At \(x = 0\) the solution above becomes \[ y(0) = a_{0} \] Therefore the solution in Eq(3) now can be written as \[ y = \left (x^{2}+1+\frac {1}{2} x^{4}\right ) y \left (0\right )+O\left (x^{6}\right ) \] 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 ) = 1 \end {align*}

Therefore the solution becomes \begin {align*} y = x^{2}+1+\frac {1}{2} x^{4} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2}+1+\frac {x^{4}}{2}+O\left (x^{6}\right ) \\ \tag{2} y &= x^{2}+1+\frac {x^{4}}{2}+O\left (x^{6}\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = x^{2}+1+\frac {x^{4}}{2}+O\left (x^{6}\right ) \] Veriﬁed OK.

23.9.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 x y=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 x y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=2 x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int 2 x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=x^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{x^{2}+c_{1}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{x^{2}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{x^{2}} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`

23.9 problem 732

23.9.1 Existence and uniqueness analysis

23.9.2 Solving as series ode

23.9.3 Maple step by step solution