Internal problem ID [12048]
Internal file name [OUTPUT/10701_Sunday_September_03_2023_12_36_42_PM_1254071/index.tex]
Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C.
ROBINSON. Cambridge University Press 2004
Section: Chapter 17, Reduction of order. Exercises page 162
Problem number: 17.2.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_change_of_variable_on_y_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x -1\right ) y^{\prime \prime }-x y^{\prime }+y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{x} \end {align*}
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 \[ p \left (x \right ) = -\frac {x}{x -1} \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{x} \left (\int {\mathrm e}^{-\left (\int -\frac {x}{x -1}d x \right )} {\mathrm e}^{-2 x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{x} \int \frac {{\mathrm e}^{x +\ln \left (x -1\right )}}{{\mathrm e}^{2 x}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{x} \left (\int \left (x -1\right ) {\mathrm e}^{-x}d x \right ) \\ y_{2}\left (x \right ) &= -{\mathrm e}^{x} x \,{\mathrm e}^{-x} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= {\mathrm e}^{x} c_{1} -c_{2} {\mathrm e}^{x} x \,{\mathrm e}^{-x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x} c_{1} -c_{2} {\mathrm e}^{x} x \,{\mathrm e}^{-x} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x} c_{1} -c_{2} {\mathrm e}^{x} x \,{\mathrm e}^{-x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )-x y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {y}{x -1}+\frac {x y^{\prime }}{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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {x y^{\prime }}{x -1}+\frac {y}{x -1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {x}{x -1}, P_{3}\left (x \right )=\frac {1}{x -1}\right ] \\ {} & \circ & \left (x -1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =1 \\ {} & {} & \left (\left (x -1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}1}}}=-1 \\ {} & \circ & \left (x -1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =1 \\ {} & {} & \left (\left (x -1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}1}}}=0 \\ {} & \circ & x =1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )-x y^{\prime }+y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & u \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-u -1\right ) \left (\frac {d}{d u}y \left (u \right )\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 \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}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}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}{d u}\frac {d}{d u}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} \\ {} & {} & a_{0} r \left (-2+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )-a_{k} \left (k +r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 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 in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (a_{k +1} \left (k +1+r \right )-a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +1+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {a_{k}}{k +1}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -1\right )^{k}, a_{k +1}=\frac {a_{k}}{k +1}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +3} \\ \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 +1}=\frac {a_{k}}{k +3}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -1\right )^{k +2}, a_{k +1}=\frac {a_{k}}{k +3}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x -1\right )^{k +2}\right ), a_{k +1}=\frac {a_{k}}{k +1}, b_{k +1}=\frac {b_{k}}{k +3}\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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 12
dsolve([(x-1)*diff(y(x),x$2)-x*diff(y(x),x)+y(x)=0,exp(x)],singsol=all)
\[ y \left (x \right ) = c_{2} {\mathrm e}^{x}+c_{1} x \]
✓ Solution by Mathematica
Time used: 0.052 (sec). Leaf size: 17
DSolve[(x-1)*y''[x]-x*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 e^x-c_2 x \]