Internal problem ID [12047]
Internal file name [OUTPUT/10700_Sunday_September_03_2023_12_36_41_PM_98333301/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: 17.1.
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_1", "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {t^{2} y^{\prime \prime }-\left (t^{2}+2 t \right ) y^{\prime }+\left (t +2\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= t \end {align*}
Given one basis solution \(y_{1}\left (t \right )\), then the second basis solution is given by \[ y_{2}\left (t \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d t \right )}}{y_{1}^{2}}d t \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y = f \left (t \right ) \] Looking at the ode to solve shows that \[ p \left (t \right ) = \frac {-t^{2}-2 t}{t^{2}} \] Therefore \begin{align*} y_{2}\left (t \right ) &= t \left (\int \frac {{\mathrm e}^{-\left (\int \frac {-t^{2}-2 t}{t^{2}}d t \right )}}{t^{2}}d t \right ) \\ y_{2}\left (t \right ) &= t \int \frac {{\mathrm e}^{t +2 \ln \left (t \right )}}{t^{2}} , dt \\ y_{2}\left (t \right ) &= t \left (\int {\mathrm e}^{t}d t \right ) \\ y_{2}\left (t \right ) &= t \,{\mathrm e}^{t} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= t c_{1} +c_{2} t \,{\mathrm e}^{t} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= t c_{1} +c_{2} t \,{\mathrm e}^{t} \\ \end{align*}
Verification of solutions
\[ y = t c_{1} +c_{2} t \,{\mathrm e}^{t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {d}{d t}y^{\prime }\right )+\left (-t^{2}-2 t \right ) y^{\prime }+\left (t +2\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-\frac {\left (t +2\right ) y}{t^{2}}+\frac {\left (t +2\right ) y^{\prime }}{t} \\ \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 t}y^{\prime }-\frac {\left (t +2\right ) y^{\prime }}{t}+\frac {\left (t +2\right ) y}{t^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=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}{t}, P_{3}\left (t \right )=\frac {t +2}{t^{2}}\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}}}=2 \\ {} & \circ & t =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {d}{d t}y^{\prime }\right )-t \left (t +2\right ) y^{\prime }+\left (t +2\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\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 y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & t^{m}\cdot y=\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 y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & t^{m}\cdot y^{\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 y^{\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^{2}\cdot \left (\frac {d}{d t}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t^{2}\cdot \left (\frac {d}{d t}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+r \right ) \left (-2+r \right ) t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r -1\right ) \left (k +r -2\right )-a_{k -1} \left (k +r -2\right )\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, 2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -2\right ) \left (a_{k} \left (k +r -1\right )-a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r -1\right ) \left (a_{k +1} \left (k +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 +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}, 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 +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +2}, a_{k +1}=\frac {a_{k}}{k +2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k +2}\right ), a_{k +1}=\frac {a_{k}}{k +1}, b_{k +1}=\frac {b_{k}}{k +2}\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([t^2*diff(y(t),t$2)-(t^2+2*t)*diff(y(t),t)+(t+2)*y(t)=0,t],singsol=all)
\[ y \left (t \right ) = t \left (c_{1} +c_{2} {\mathrm e}^{t}\right ) \]
✓ Solution by Mathematica
Time used: 0.032 (sec). Leaf size: 16
DSolve[t^2*y''[t]-(t^2+2*t)*y'[t]+(t+2)*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to t \left (c_2 e^t+c_1\right ) \]