Internal problem ID [11725]
Internal file name [OUTPUT/11735_Thursday_April_11_2024_08_49_14_PM_91859302/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 4, Section 4.1. Basic theory of linear differential equations. Exercises page
124
Problem number: 4.
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"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x^{2}-x +1\right ) y^{\prime \prime }-\left (x^{2}+x \right ) y^{\prime }+y \left (x +1\right )=0} \] Given that one solution of the ode is \begin {align*} y_1 &= 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 \[ p \left (x \right ) = \frac {-x^{2}-x}{x^{2}-x +1} \] Therefore \begin{align*} y_{2}\left (x \right ) &= x \left (\int \frac {{\mathrm e}^{-\left (\int \frac {-x^{2}-x}{x^{2}-x +1}d x \right )}}{x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= x \int \frac {{\mathrm e}^{x +\ln \left (x^{2}-x +1\right )}}{x^{2}} , dx \\ y_{2}\left (x \right ) &= x \left (\int \frac {{\mathrm e}^{x} \left (x^{2}-x +1\right )}{x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= \left (x -1\right ) {\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 ) \\ &= c_{1} x +c_{2} \left (x -1\right ) {\mathrm e}^{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +c_{2} \left (x -1\right ) {\mathrm e}^{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +c_{2} \left (x -1\right ) {\mathrm e}^{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-x +1\right ) y^{\prime \prime }+\left (-x^{2}-x \right ) y^{\prime }+y \left (x +1\right )=0 \\ \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 {\left (x +1\right ) y}{x^{2}-x +1}+\frac {x \left (x +1\right ) y^{\prime }}{x^{2}-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 {x \left (x +1\right ) y^{\prime }}{x^{2}-x +1}+\frac {\left (x +1\right ) y}{x^{2}-x +1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {\left (x +1\right ) x}{x^{2}-x +1}, P_{3}\left (x \right )=\frac {x +1}{x^{2}-x +1}\right ] \\ {} & \circ & \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}}}=0 \\ {} & \circ & \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left (\left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}}}=0 \\ {} & \circ & x =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x^{2}-x +1\right ) y^{\prime \prime }-x \left (x +1\right ) y^{\prime }+y \left (x +1\right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-\mathrm {I} u \sqrt {3}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-u^{2}-2 u +\mathrm {I} u \sqrt {3}+\mathrm {I} \sqrt {3}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u +\frac {3}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\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..2 \\ {} & {} & 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^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -\mathrm {I} \sqrt {3}\, r \left (r -2\right ) a_{0} u^{r -1}+\left (-\mathrm {I} \sqrt {3}\, \left (1+r \right ) \left (r -1\right ) a_{1}+\frac {\left (3+2 \,\mathrm {I} r \sqrt {3}-\mathrm {I} \sqrt {3}+2 r^{2}-6 r \right ) a_{0}}{2}\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-\mathrm {I} \sqrt {3}\, \left (k +1+r \right ) \left (k +r -1\right ) a_{k +1}+\frac {\left (2 \,\mathrm {I} \sqrt {3}\, k +2 \,\mathrm {I} r \sqrt {3}-\mathrm {I} \sqrt {3}+2 k^{2}+4 k r +2 r^{2}-6 k -6 r +3\right ) a_{k}}{2}-a_{k -1} \left (k -2+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \mathrm {-I} \sqrt {3}\, r \left (r -2\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} \\ {} & {} & -\mathrm {I} \sqrt {3}\, \left (1+r \right ) \left (r -1\right ) a_{1}+\frac {\left (3+2 \,\mathrm {I} r \sqrt {3}-\mathrm {I} \sqrt {3}+2 r^{2}-6 r \right ) a_{0}}{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {\mathrm {I} \left (\left (-1+2 r +2 k \right ) a_{k}-2 a_{k +1} \left (k +r -1\right ) \left (k +1+r \right )\right ) \sqrt {3}}{2}+\frac {\left (3+2 k^{2}+2 \left (-3+2 r \right ) k +2 r^{2}-6 r \right ) a_{k}}{2}-a_{k -1} \left (k -2+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {\mathrm {I} \left (\left (1+2 r +2 k \right ) a_{k +1}-2 a_{k +2} \left (k +r \right ) \left (k +2+r \right )\right ) \sqrt {3}}{2}+\frac {\left (3+2 \left (k +1\right )^{2}+2 \left (-3+2 r \right ) \left (k +1\right )+2 r^{2}-6 r \right ) a_{k +1}}{2}-a_{k} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {-\frac {\mathrm {I}}{6} \left (2 \,\mathrm {I} \sqrt {3}\, r a_{k +1}+2 \,\mathrm {I} \sqrt {3}\, k a_{k +1}+\mathrm {I} \sqrt {3}\, a_{k +1}+2 k^{2} a_{k +1}+4 k r a_{k +1}+2 r^{2} a_{k +1}-2 k a_{k}-2 a_{k +1} k -2 r a_{k}-2 a_{k +1} r +2 a_{k}-a_{k +1}\right ) \sqrt {3}}{k^{2}+2 k r +r^{2}+2 k +2 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {-\frac {\mathrm {I}}{6} \left (2 \,\mathrm {I} \sqrt {3}\, k a_{k +1}+\mathrm {I} \sqrt {3}\, a_{k +1}+2 k^{2} a_{k +1}-2 k a_{k}-2 a_{k +1} k +2 a_{k}-a_{k +1}\right ) \sqrt {3}}{k^{2}+2 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 {-\frac {\mathrm {I}}{6} \left (2 \,\mathrm {I} \sqrt {3}\, k a_{k +1}+\mathrm {I} \sqrt {3}\, a_{k +1}+2 k^{2} a_{k +1}-2 k a_{k}-2 a_{k +1} k +2 a_{k}-a_{k +1}\right ) \sqrt {3}}{k^{2}+2 k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {-\frac {\mathrm {I}}{6} \left (5 \,\mathrm {I} \sqrt {3}\, a_{k +1}+2 \,\mathrm {I} \sqrt {3}\, k a_{k +1}+2 k^{2} a_{k +1}+6 a_{k +1} k +3 a_{k +1}-2 k a_{k}-2 a_{k}\right ) \sqrt {3}}{k^{2}+6 k +8} \\ \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 {-\frac {\mathrm {I}}{6} \left (5 \,\mathrm {I} \sqrt {3}\, a_{k +1}+2 \,\mathrm {I} \sqrt {3}\, k a_{k +1}+2 k^{2} a_{k +1}+6 a_{k +1} k +3 a_{k +1}-2 k a_{k}-2 a_{k}\right ) \sqrt {3}}{k^{2}+6 k +8}, -3 \,\mathrm {I} \sqrt {3}\, a_{1}+\frac {\left (-1+3 \,\mathrm {I} \sqrt {3}\right ) a_{0}}{2}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{k +2}, a_{k +2}=\frac {-\frac {\mathrm {I}}{6} \left (5 \,\mathrm {I} \sqrt {3}\, a_{k +1}+2 \,\mathrm {I} \sqrt {3}\, k a_{k +1}+2 k^{2} a_{k +1}+6 a_{k +1} k +3 a_{k +1}-2 k a_{k}-2 a_{k}\right ) \sqrt {3}}{k^{2}+6 k +8}, -3 \,\mathrm {I} \sqrt {3}\, a_{1}+\frac {\left (-1+3 \,\mathrm {I} \sqrt {3}\right ) a_{0}}{2}=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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 15
dsolve([(x^2-x+1)*diff(y(x),x$2)-(x^2+x)*diff(y(x),x)+(x+1)*y(x)=0,x],singsol=all)
\[ y \left (x \right ) = c_{1} x +c_{2} {\mathrm e}^{x} \left (-1+x \right ) \]
✓ Solution by Mathematica
Time used: 0.093 (sec). Leaf size: 19
DSolve[(x^2-x+1)*y''[x]-(x^2+x)*y'[x]+(x+1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x+c_2 e^x (x-1) \]