16.4 problem 4

Internal problem ID [15418]
Book : INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section : Chapter 4. Higher Order Equations. Exercises 4.8, page 203
Problem number : 4
Date solved : Friday, October 18, 2024 at 01:36:49 PM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} y^{\prime \prime }+3 y^{\prime }-18 y&=0 \end{align*}

Using series expansion around \(x=0\)

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 &= -3 y^{\prime }+18 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 \\ &= 27 y^{\prime }-54 y\\ 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 \\ &= -135 y^{\prime }+486 y\\ 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 \\ &= 891 y^{\prime }-2430 y\\ 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 \\ &= -5103 y^{\prime }+16038 y \end{align*}

And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives

\begin{align*} F_0 &= -3 y^{\prime }\left (0\right )+18 y \left (0\right )\\ F_1 &= 27 y^{\prime }\left (0\right )-54 y \left (0\right )\\ F_2 &= -135 y^{\prime }\left (0\right )+486 y \left (0\right )\\ F_3 &= 891 y^{\prime }\left (0\right )-2430 y \left (0\right )\\ F_4 &= -5103 y^{\prime }\left (0\right )+16038 y \left (0\right ) \end{align*}

Substituting all the above in (7) and simplifying gives the solution as

\[ y = \left (1+9 x^{2}-9 x^{3}+\frac {81}{4} x^{4}-\frac {81}{4} x^{5}\right ) y \left (0\right )+\left (x -\frac {3}{2} x^{2}+\frac {9}{2} x^{3}-\frac {45}{8} x^{4}+\frac {297}{40} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \]

Since the expansion point \(x = 0\) is an ordinary point, then this can also be solved using the standard power series method. 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}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} \end{align*}

Substituting the above back into the ode gives

\begin{align*} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+3 \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )-18 \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 =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 n a_{n} x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-18 a_{n} x^{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 =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n} \\ \moverset {\infty }{\munderset {n =1}{\sum }}3 n a_{n} x^{n -1} &= \moverset {\infty }{\munderset {n =0}{\sum }}3 \left (n +1\right ) a_{n +1} x^{n} \\ \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 (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \left (n +1\right ) a_{n +1} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-18 a_{n} x^{n}\right ) = 0 \end{equation}

For \(0\le n\), the recurrence equation is

\begin{equation} \tag{4} \left (n +2\right ) a_{n +2} \left (n +1\right )+3 \left (n +1\right ) a_{n +1}-18 a_{n} = 0 \end{equation}

Solving for \(a_{n +2}\), gives

\begin{align*} \tag{5} a_{n +2}&= -\frac {3 \left (n a_{n +1}-6 a_{n}+a_{n +1}\right )}{\left (n +2\right ) \left (n +1\right )} \\ &= \frac {18 a_{n}}{\left (n +2\right ) \left (n +1\right )}-\frac {3 a_{n +1}}{n +2} \\ \end{align*}

For \(n = 0\) the recurrence equation gives

\[ 2 a_{2}+3 a_{1}-18 a_{0} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{2} = -\frac {3 a_{1}}{2}+9 a_{0} \]

For \(n = 1\) the recurrence equation gives

\[ 6 a_{3}+6 a_{2}-18 a_{1} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{3} = \frac {9 a_{1}}{2}-9 a_{0} \]

For \(n = 2\) the recurrence equation gives

\[ 12 a_{4}+9 a_{3}-18 a_{2} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{4} = -\frac {45 a_{1}}{8}+\frac {81 a_{0}}{4} \]

For \(n = 3\) the recurrence equation gives

\[ 20 a_{5}+12 a_{4}-18 a_{3} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{5} = \frac {297 a_{1}}{40}-\frac {81 a_{0}}{4} \]

For \(n = 4\) the recurrence equation gives

\[ 30 a_{6}+15 a_{5}-18 a_{4} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{6} = -\frac {567 a_{1}}{80}+\frac {891 a_{0}}{40} \]

For \(n = 5\) the recurrence equation gives

\[ 42 a_{7}+18 a_{6}-18 a_{5} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{7} = \frac {3483 a_{1}}{560}-\frac {729 a_{0}}{40} \]

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

\[ y = a_{0}+a_{1} x +\left (-\frac {3 a_{1}}{2}+9 a_{0}\right ) x^{2}+\left (\frac {9 a_{1}}{2}-9 a_{0}\right ) x^{3}+\left (-\frac {45 a_{1}}{8}+\frac {81 a_{0}}{4}\right ) x^{4}+\left (\frac {297 a_{1}}{40}-\frac {81 a_{0}}{4}\right ) x^{5}+\dots \]

Collecting terms, the solution becomes

\begin{equation} \tag{3} y = \left (1+9 x^{2}-9 x^{3}+\frac {81}{4} x^{4}-\frac {81}{4} x^{5}\right ) a_{0}+\left (x -\frac {3}{2} x^{2}+\frac {9}{2} x^{3}-\frac {45}{8} x^{4}+\frac {297}{40} x^{5}\right ) a_{1}+O\left (x^{6}\right ) \end{equation}

At \(x = 0\) the solution above becomes

\[ y = \left (1+9 x^{2}-9 x^{3}+\frac {81}{4} x^{4}-\frac {81}{4} x^{5}\right ) c_1 +\left (x -\frac {3}{2} x^{2}+\frac {9}{2} x^{3}-\frac {45}{8} x^{4}+\frac {297}{40} x^{5}\right ) c_2 +O\left (x^{6}\right ) \]

Figure 1554: Slope field plot
\(y^{\prime \prime }+3 y^{\prime }-18 y = 0\)

16.4.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )+3 \frac {d}{d x}y \left (x \right )-18 y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+3 r -18=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +6\right ) \left (r -3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-6, 3\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{-6 x} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{3 x} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} \,{\mathrm e}^{-6 x}+\mathit {C2} \,{\mathrm e}^{3 x} \end {array} \]

16.4.2 Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

16.4.3 Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 59

dsolve(diff(diff(y(x),x),x)+3*diff(y(x),x)-18*y(x) = 0,y(x), 
       series,x=0)
 

\[ y \left (x \right ) = \left (1+9 x^{2}-9 x^{3}+\frac {81}{4} x^{4}-\frac {81}{4} x^{5}\right ) y \left (0\right )+\left (x -\frac {3}{2} x^{2}+\frac {9}{2} x^{3}-\frac {45}{8} x^{4}+\frac {297}{40} x^{5}\right ) D\left (y \right )\left (0\right )+O\left (x^{6}\right ) \]

16.4.4 Mathematica DSolve solution

Solving time : 0.001 (sec)
Leaf size : 66

AsymptoticDSolveValue[{D[y[x],{x,2}]+3*D[y[x],x]-18*y[x]==0,{}}, 
       y[x],{x,0,5}]
 

\[ y(x)\to c_1 \left (-\frac {81 x^5}{4}+\frac {81 x^4}{4}-9 x^3+9 x^2+1\right )+c_2 \left (\frac {297 x^5}{40}-\frac {45 x^4}{8}+\frac {9 x^3}{2}-\frac {3 x^2}{2}+x\right ) \]