6.8 problem 8

Internal problem ID [5026]
Internal file name [OUTPUT/4519_Sunday_June_05_2022_03_00_03_PM_91206933/index.tex]

Book: Fundamentals of Differential Equations. By Nagle, Saff and Snider. 9th edition. Boston. Pearson 2018.
Section: Chapter 8, Series solutions of differential equations. Section 8.4. page 449
Problem number: 8.
ODE order: 1.
ODE degree: 1.

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

Maple gives the following as the ode type

[_separable]

\[ \boxed {y^{\prime }-2 x y=0} \] With the expansion point for the power series method at \(x = 1\).

6.8.1 Solving as series ode

The ode does not have its expansion point at \(x = 0\), therefore to simplify the computation of power series expansion, change of variable is made on the independent variable to shift the initial conditions and the expasion point back to zero. The new ode is then solved more easily since the expansion point is now at zero. The solution converted back to the original independent variable. Let \[ t = x -1 \] The ode is converted to be in terms of the new independent variable \(t\). This results in \[ \frac {d}{d t}y \left (t \right )-2 \left (t +1\right ) y \left (t \right ) = 0 \] With its expansion point and initial conditions now at \(t = 0\). The transformed ODE is now solved.

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 \left (t +1\right ) y \left (t \right )\\ F_1 &= \frac {d F_0}{dt} \\ &= \frac {\partial F_0}{\partial t}+ \frac {\partial F_0}{\partial y} F_0 \\ &= \left (4 t^{2}+8 t +6\right ) y \left (t \right )\\ F_2 &= \frac {d F_1}{dt} \\ &= \frac {\partial F_1}{\partial t}+ \frac {\partial F_1}{\partial y} F_1 \\ &= 8 \left (t +1\right ) \left (t^{2}+2 t +\frac {5}{2}\right ) y \left (t \right )\\ F_3 &= \frac {d F_2}{dt} \\ &= \frac {\partial F_2}{\partial t}+ \frac {\partial F_2}{\partial y} F_2 \\ &= 16 \left (t^{4}+4 t^{3}+9 t^{2}+10 t +\frac {19}{4}\right ) y \left (t \right )\\ F_4 &= \frac {d F_3}{dt} \\ &= \frac {\partial F_3}{\partial t}+ \frac {\partial F_3}{\partial y} F_3 \\ &= 32 \left (t +1\right ) y \left (t \right ) \left (t^{4}+4 t^{3}+11 t^{2}+14 t +\frac {39}{4}\right ) \end {align*}

And so on. Evaluating all the above at initial conditions \(t \left (0\right ) = 0\) and \(y \left (0\right ) = y \left (0\right )\) gives \begin {align*} F_0 &= 2 y \left (0\right )\\ F_1 &= 6 y \left (0\right )\\ F_2 &= 20 y \left (0\right )\\ F_3 &= 76 y \left (0\right )\\ F_4 &= 312 y \left (0\right ) \end {align*}

Substituting all the above in (6) and simplifying gives the solution as \[ y \left (t \right ) = \left (1+2 t +3 t^{2}+\frac {10}{3} t^{3}+\frac {19}{6} t^{4}+\frac {13}{5} t^{5}\right ) y \left (0\right )+O\left (t^{6}\right ) \] Since \(t = 0\) is also an ordinary point, then standard power series can also be used. Writing the ODE as \begin {align*} \frac {d}{d t}y \left (t \right ) + q(t)y \left (t \right ) &= p(t) \\ \frac {d}{d t}y \left (t \right )+\left (-2 t -2\right ) y \left (t \right ) &= 0 \end {align*}

Where \begin {align*} q(t) &= -2 t -2\\ p(t) &= 0 \end {align*}

Next, the type of the expansion point \(t = 0\) 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 \(t = 0\) 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. \(t = 0\) is called an ordinary point \(q(t)\) has a Taylor series expansion around the point \(t = 0\). \(t = 0\) is called a regular singular point if \(q(t)\) is not not analytic at \(t = 0\) but \(t q(t)\) has Taylor series expansion. And finally, \(t = 0\) 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 \(t = 0\) is checked to see if it is an ordinary point or not. Let the solution be represented as power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n} \] Then \begin {align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1} \end {align*}

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

Which simplifies to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 t^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} t^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (1+n \right ) a_{1+n} t^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 t^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} t^{n}\right ) \\ \end{align*} Substituting all the above in Eq (2) gives the following equation where now all powers of \(t\) 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} t^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} t^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} t^{n}\right ) = 0 \end{equation} \(n=0\) gives \begin {align*} a_{1}-2 a_{0}&=0 \\ a_{1} &= 2 a_{0} \end {align*}

For \(1\le n\), the recurrence equation is \begin{equation} \tag{4} \left (1+n \right ) a_{1+n}-2 a_{n -1}-2 a_{n} = 0 \end{equation} Solving for \(a_{1+n}\), gives \begin{equation} \tag{5} a_{1+n} = \frac {2 a_{n -1}+2 a_{n}}{1+n} \end{equation} For \(n = 1\) the recurrence equation gives \[ 2 a_{2}-2 a_{0}-2 a_{1} = 0 \] Which after substituting the earlier terms found becomes \[ a_{2} = 3 a_{0} \] For \(n = 2\) the recurrence equation gives \[ 3 a_{3}-2 a_{1}-2 a_{2} = 0 \] Which after substituting the earlier terms found becomes \[ a_{3} = \frac {10 a_{0}}{3} \] For \(n = 3\) the recurrence equation gives \[ 4 a_{4}-2 a_{2}-2 a_{3} = 0 \] Which after substituting the earlier terms found becomes \[ a_{4} = \frac {19 a_{0}}{6} \] For \(n = 4\) the recurrence equation gives \[ 5 a_{5}-2 a_{3}-2 a_{4} = 0 \] Which after substituting the earlier terms found becomes \[ a_{5} = \frac {13 a_{0}}{5} \] For \(n = 5\) the recurrence equation gives \[ 6 a_{6}-2 a_{4}-2 a_{5} = 0 \] Which after substituting the earlier terms found becomes \[ a_{6} = \frac {173 a_{0}}{90} \] And so on. Therefore the solution is \begin {align*} y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ &= a_{3} t^{3}+a_{2} t^{2}+a_{1} t +a_{0} + \dots \end {align*}

Substituting the values for \(a_{n}\) found above, the solution becomes \[ y \left (t \right ) = a_{0}+2 a_{0} t +3 a_{0} t^{2}+\frac {10}{3} a_{0} t^{3}+\frac {19}{6} a_{0} t^{4}+\frac {13}{5} a_{0} t^{5}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y \left (t \right ) = \left (1+2 t +3 t^{2}+\frac {10}{3} t^{3}+\frac {19}{6} t^{4}+\frac {13}{5} t^{5}\right ) a_{0}+O\left (t^{6}\right ) \end{equation} Replacing \(t\) in the above with the original independent variable \(xs\) using \(t = x -1\) results in \begin {gather*} y = \left (-1+2 x +3 \left (x -1\right )^{2}+\frac {10 \left (x -1\right )^{3}}{3}+\frac {19 \left (x -1\right )^{4}}{6}+\frac {13 \left (x -1\right )^{5}}{5}\right ) y \left (1\right )+O\left (\left (x -1\right )^{6}\right ) \end {gather*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (-1+2 x +3 \left (x -1\right )^{2}+\frac {10 \left (x -1\right )^{3}}{3}+\frac {19 \left (x -1\right )^{4}}{6}+\frac {13 \left (x -1\right )^{5}}{5}\right ) y \left (1\right )+O\left (\left (x -1\right )^{6}\right ) \\ \end{align*}

Figure 145: Slope field plot

Verification of solutions

\[ y = \left (-1+2 x +3 \left (x -1\right )^{2}+\frac {10 \left (x -1\right )^{3}}{3}+\frac {19 \left (x -1\right )^{4}}{6}+\frac {13 \left (x -1\right )^{5}}{5}\right ) y \left (1\right )+O\left (\left (x -1\right )^{6}\right ) \] Verified OK.

6.8.2 Maple step by step solution

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

Maple trace

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

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 36

Order:=6; 
dsolve(diff(y(x),x)-2*x*y(x)=0,y(x),type='series',x=1);
 

\[ y \left (x \right ) = \left (-1+2 x +3 \left (x -1\right )^{2}+\frac {10 \left (x -1\right )^{3}}{3}+\frac {19 \left (x -1\right )^{4}}{6}+\frac {13 \left (x -1\right )^{5}}{5}\right ) y \left (1\right )+O\left (x^{6}\right ) \]

✓ Solution by Mathematica

Time used: 0.001 (sec). Leaf size: 47

AsymptoticDSolveValue[y'[x]-2*x*y[x]==0,y[x],{x,1,5}]
 

\[ y(x)\to c_1 \left (\frac {13}{5} (x-1)^5+\frac {19}{6} (x-1)^4+\frac {10}{3} (x-1)^3+3 (x-1)^2+2 (x-1)+1\right ) \]