Internal problem ID [14629]
Internal file name [OUTPUT/14309_Thursday_March_28_2024_03_35_07_AM_62176316/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.4, page 163
Problem number: 62.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_bessel_ode", "second_order_change_of_variable_on_y_method_1"
Maple gives the following as the ode type
[_Lienard]
\[ \boxed {t y^{\prime \prime }+2 y^{\prime }+y t=0} \] Given that one solution of the ode is \begin {align*} y_1 &= \frac {\cos \left (t \right )}{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 \[ p \left (t \right ) = \frac {2}{t} \] Therefore \begin{align*} y_{2}\left (t \right ) &= \frac {\cos \left (t \right ) \left (\int \frac {{\mathrm e}^{-\left (\int \frac {2}{t}d t \right )} t^{2}}{\cos \left (t \right )^{2}}d t \right )}{t} \\ y_{2}\left (t \right ) &= \frac {\cos \left (t \right )}{t} \int \frac {\frac {1}{t^{2}}}{\frac {\cos \left (t \right )^{2}}{t^{2}}} , dt \\ y_{2}\left (t \right ) &= \frac {\cos \left (t \right ) \left (\int \sec \left (t \right )^{2}d t \right )}{t} \\ y_{2}\left (t \right ) &= \frac {\cos \left (t \right ) \tan \left (t \right )}{t} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= \frac {\cos \left (t \right ) c_{1}}{t}+\frac {c_{2} \cos \left (t \right ) \tan \left (t \right )}{t} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\cos \left (t \right ) c_{1}}{t}+\frac {c_{2} \cos \left (t \right ) \tan \left (t \right )}{t} \\ \end{align*}
Verification of solutions
\[ y = \frac {\cos \left (t \right ) c_{1}}{t}+\frac {c_{2} \cos \left (t \right ) \tan \left (t \right )}{t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t \left (\frac {d}{d t}y^{\prime }\right )+2 y^{\prime }+y t =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 {2 y^{\prime }}{t}-y \\ \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 {2 y^{\prime }}{t}+y=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 {2}{t}, P_{3}\left (t \right )=1\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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & t \left (\frac {d}{d t}y^{\prime }\right )+2 y^{\prime }+y t =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 \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & t \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot \left (\frac {d}{d t}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & t \cdot \left (\frac {d}{d t}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) \left (k +r \right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+r \right ) t^{-1+r}+a_{1} \left (1+r \right ) \left (2+r \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 -1}\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 )=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 -1}=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}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k -1}, a_{k +2}=-\frac {a_{k}}{\left (k +1\right ) \left (k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +3\right )}, 2 a_{1}=0\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}\right ), a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )}, 0=0, b_{k +2}=-\frac {b_{k}}{\left (k +3\right ) \left (k +2\right )}, 2 b_{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 Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 17
dsolve([t*diff(y(t),t$2)+2*diff(y(t),t)+t*y(t)=0,1/t*cos(t)],singsol=all)
\[ y \left (t \right ) = \frac {c_{1} \sin \left (t \right )+c_{2} \cos \left (t \right )}{t} \]
✓ Solution by Mathematica
Time used: 0.023 (sec). Leaf size: 37
DSolve[t*y''[t]+2*y'[t]+t*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {2 c_1 e^{-i t}-i c_2 e^{i t}}{2 t} \]