Internal problem ID [12050]
Internal file name [OUTPUT/10703_Sunday_September_03_2023_12_36_42_PM_95706010/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: 17.4.
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]]
\[ \boxed {\left (-t^{2}+t \right ) x^{\prime \prime }+\left (-t^{2}+2\right ) x^{\prime }+\left (-t +2\right ) x=0} \] Given that one solution of the ode is \begin {align*} x_1 &= {\mathrm e}^{-t} \end {align*}
Given one basis solution \(x_{1}\left (t \right )\), then the second basis solution is given by \[ x_{2}\left (t \right ) = x_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d t \right )}}{x_{1}^{2}}d t \right ) \] Where \(p(x)\) is the coefficient of \(x^{\prime }\) when the ode is written in the normal form \[ x^{\prime \prime }+p \left (t \right ) x^{\prime }+q \left (t \right ) x = f \left (t \right ) \] Looking at the ode to solve shows that \[ p \left (t \right ) = \frac {-t^{2}+2}{-t^{2}+t} \] Therefore \begin{align*} x_{2}\left (t \right ) &= {\mathrm e}^{-t} \left (\int {\mathrm e}^{-\left (\int \frac {-t^{2}+2}{-t^{2}+t}d t \right )} {\mathrm e}^{2 t}d t \right ) \\ x_{2}\left (t \right ) &= {\mathrm e}^{-t} \int \frac {{\mathrm e}^{-t -2 \ln \left (t \right )+\ln \left (t -1\right )}}{{\mathrm e}^{-2 t}} , dt \\ x_{2}\left (t \right ) &= {\mathrm e}^{-t} \left (\int \frac {\left (t -1\right ) {\mathrm e}^{t}}{t^{2}}d t \right ) \\ x_{2}\left (t \right ) &= \frac {{\mathrm e}^{-t} {\mathrm e}^{t}}{t} \\ \end{align*} Hence the solution is \begin{align*} x &= c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right ) \\ &= c_{1} {\mathrm e}^{-t}+\frac {c_{2} {\mathrm e}^{-t} {\mathrm e}^{t}}{t} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} x &= c_{1} {\mathrm e}^{-t}+\frac {c_{2} {\mathrm e}^{-t} {\mathrm e}^{t}}{t} \\ \end{align*}
Verification of solutions
\[ x = c_{1} {\mathrm e}^{-t}+\frac {c_{2} {\mathrm e}^{-t} {\mathrm e}^{t}}{t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-t^{2}+t \right ) x^{\prime \prime }+\left (-t^{2}+2\right ) x^{\prime }+\left (-t +2\right ) x=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & x^{\prime \prime }=-\frac {\left (-2+t \right ) x}{t \left (t -1\right )}-\frac {\left (t^{2}-2\right ) x^{\prime }}{t \left (t -1\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} x\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & x^{\prime \prime }+\frac {\left (t^{2}-2\right ) x^{\prime }}{t \left (t -1\right )}+\frac {\left (-2+t \right ) x}{t \left (t -1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=\frac {t^{2}-2}{t \left (t -1\right )}, P_{3}\left (t \right )=\frac {-2+t}{t \left (t -1\right )}\right ] \\ {} & \circ & t \cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t \cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=2 \\ {} & \circ & t^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=0 \\ {} & \circ & t =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{\prime \prime } t \left (t -1\right )+\left (t^{2}-2\right ) x^{\prime }+\left (-2+t \right ) x=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} x \\ {} & {} & x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{m}\cdot x\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & t^{m}\cdot x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & t^{m}\cdot x=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{m}\cdot x^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & t^{m}\cdot x^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & t^{m}\cdot x^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{m}\cdot x^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & t^{m}\cdot x^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & t^{m}\cdot x^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (1+r \right ) t^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (2+r \right )+a_{0} \left (1+r \right ) \left (-2+r \right )\right ) t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+a_{k} \left (k +r +1\right ) \left (k +r -2\right )+a_{k -1} \left (k +r \right )\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} \left (1+r \right ) \left (2+r \right )+a_{0} \left (1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+a_{k} \left (k +r +1\right ) \left (k +r -2\right )+a_{k -1} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -a_{k +2} \left (k +2+r \right ) \left (k +3+r \right )+a_{k +1} \left (k +2+r \right ) \left (k +r -1\right )+a_{k} \left (k +r +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}+2 k r a_{k +1}+r^{2} a_{k +1}+k a_{k}+k a_{k +1}+r a_{k}+r a_{k +1}+a_{k}-2 a_{k +1}}{\left (k +2+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}+k a_{k}-k a_{k +1}-2 a_{k +1}}{\left (k +1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k -1}, a_{k +2}=\frac {k^{2} a_{k +1}+k a_{k}-k a_{k +1}-2 a_{k +1}}{\left (k +1\right ) \left (k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}+k a_{k}+k a_{k +1}+a_{k}-2 a_{k +1}}{\left (k +2\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +2}=\frac {k^{2} a_{k +1}+k a_{k}+k a_{k +1}+a_{k}-2 a_{k +1}}{\left (k +2\right ) \left (k +3\right )}, -2 a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [x=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k}\right ), a_{2+k}=\frac {k^{2} a_{k +1}+k a_{k}-k a_{k +1}-2 a_{k +1}}{\left (k +1\right ) \left (2+k \right )}, 0=0, b_{2+k}=\frac {k^{2} b_{k +1}+k b_{k}+k b_{k +1}+b_{k}-2 b_{k +1}}{\left (2+k \right ) \left (3+k \right )}, -2 b_{1}-2 b_{0}=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: 0.016 (sec). Leaf size: 17
dsolve([(t-t^2)*diff(x(t),t$2)+(2-t^2)*diff(x(t),t)+(2-t)*x(t)=0,exp(-t)],singsol=all)
\[ x \left (t \right ) = \frac {c_{2} {\mathrm e}^{-t} t +c_{1}}{t} \]
✓ Solution by Mathematica
Time used: 0.104 (sec). Leaf size: 42
DSolve[(t-t^2)*x''[t]+(2-t^2)*x'[t]+(2-t)*x[t]==0,x[t],t,IncludeSingularSolutions -> True]
\[ x(t)\to \frac {e^{-t} \sqrt {1-t} \left (c_1 e^t-c_2 t\right )}{\sqrt {t-1} t} \]