problem 32

Internal problem ID [1138]
Internal ﬁle name [OUTPUT/1139_Sunday_June_05_2022_02_03_21_AM_93815312/index.tex]

Book: Elementary diﬀerential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.6 Reduction or order. Page 253
Problem number: 32.
ODE order: 2.
ODE degree: 1.

\[ \boxed {\left (3 x -1\right ) y^{\prime \prime }-\left (3 x +2\right ) y^{\prime }-\left (6 x -8\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{2 x} \end {align*}

With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 3] \end {align*}

9.32.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(x) &=\frac {-3 x -2}{3 x -1}\\ q(x) &=\frac {-6 x +8}{3 x -1}\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+\frac {\left (-3 x -2\right ) y^{\prime }}{3 x -1}+\frac {\left (-6 x +8\right ) y}{3 x -1} = 0 \end {align*}

The domain of \(p(x)=\frac {-3 x -2}{3 x -1}\) is \[ \left \{x <\frac {1}{3}\boldsymbol {\lor }\frac {1}{3}

Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coeﬃcient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = \frac {-3 x -2}{3 x -1} \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{2 x} \left (\int {\mathrm e}^{-\left (\int \frac {-3 x -2}{3 x -1}d x \right )} {\mathrm e}^{-4 x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{2 x} \int \frac {{\mathrm e}^{x +\ln \left (3 x -1\right )}}{{\mathrm e}^{4 x}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{2 x} \left (\int \left (3 x -1\right ) {\mathrm e}^{-3 x}d x \right ) \\ y_{2}\left (x \right ) &= -{\mathrm e}^{2 x} x \,{\mathrm e}^{-3 x} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} {\mathrm e}^{2 x}-c_{2} {\mathrm e}^{2 x} x \,{\mathrm e}^{-3 x} \\ \end{align*} Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{2 x}-c_{2} {\mathrm e}^{2 x} x \,{\mathrm e}^{-3 x} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = c_{1}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 c_{1} {\mathrm e}^{2 x}+c_{2} {\mathrm e}^{2 x} x \,{\mathrm e}^{-3 x}-c_{2} {\mathrm e}^{2 x} {\mathrm e}^{-3 x} \end {align*}

substituting \(y^{\prime } = 3\) and \(x = 0\) in the above gives \begin {align*} 3 = 2 c_{1} -c_{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=2\\ c_{2}&=1 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = -{\mathrm e}^{2 x} x \,{\mathrm e}^{-3 x}+2 \,{\mathrm e}^{2 x} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -x \,{\mathrm e}^{-x}+2 \,{\mathrm e}^{2 x} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -x \,{\mathrm e}^{-x}+2 \,{\mathrm e}^{2 x} \] Veriﬁed OK.

9.32.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (3 x -1\right ) y^{\prime \prime }+\left (-3 x -2\right ) y^{\prime }+\left (-6 x +8\right ) y=0, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {2 \left (3 x -4\right ) y}{3 x -1}+\frac {\left (3 x +2\right ) y^{\prime }}{3 x -1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-\frac {\left (3 x +2\right ) y^{\prime }}{3 x -1}-\frac {2 \left (3 x -4\right ) y}{3 x -1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=\frac {1}{3}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {3 x +2}{3 x -1}, P_{3}\left (x \right )=-\frac {2 \left (3 x -4\right )}{3 x -1}\right ] \\ {} & \circ & \left (x -\frac {1}{3}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{3} \\ {} & {} & \left (\left (x -\frac {1}{3}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{3}}}}=-1 \\ {} & \circ & \left (x -\frac {1}{3}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{3} \\ {} & {} & \left (\left (x -\frac {1}{3}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{3}}}}=0 \\ {} & \circ & x =\frac {1}{3}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=\frac {1}{3}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=\frac {1}{3} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (3 x -1\right ) y^{\prime \prime }+\left (-3 x -2\right ) y^{\prime }+\left (-6 x +8\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {1}{3}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & 3 u \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-3 u -3\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-6 u +6\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u \cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 3 a_{0} r \left (-2+r \right ) u^{-1+r}+\left (3 a_{1} \left (1+r \right ) \left (-1+r \right )-3 a_{0} \left (-2+r \right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (3 a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )-3 a_{k} \left (k +r -2\right )-6 a_{k -1}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 3 r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 3 a_{1} \left (1+r \right ) \left (-1+r \right )-3 a_{0} \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+a_{k} \left (-3 k -3 r +6\right )-6 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 a_{k +2} \left (k +2+r \right ) \left (k +r \right )+a_{k +1} \left (-3 k +3-3 r \right )-6 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k a_{k +1}+r a_{k +1}+2 a_{k}-a_{k +1}}{\left (k +2+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {k a_{k +1}+2 a_{k}-a_{k +1}}{\left (k +2\right ) k} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=\frac {k a_{k +1}+2 a_{k}-a_{k +1}}{\left (k +2\right ) k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {k a_{k +1}+2 a_{k}+a_{k +1}}{\left (k +4\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +2}, a_{k +2}=\frac {k a_{k +1}+2 a_{k}+a_{k +1}}{\left (k +4\right ) \left (k +2\right )}, 9 a_{1}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{3}\right )^{k +2}, a_{k +2}=\frac {k a_{k +1}+2 a_{k}+a_{k +1}}{\left (k +4\right ) \left (k +2\right )}, 9 a_{1}=0\right ] \end {array} \]

Maple trace Kovacic algorithm successful

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful`

9.32 problem 32

9.32.1 Existence and uniqueness analysis

9.32.2 Maple step by step solution