Internal problem ID [14627]
Internal file name [OUTPUT/14307_Thursday_March_28_2024_03_35_06_AM_26965091/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: 61.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {t^{2} y^{\prime \prime }-4 t y^{\prime }+\left (t^{2}+6\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= t^{2} \cos \left (t \right ) \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 \[ p \left (t \right ) = -\frac {4}{t} \] Therefore \begin{align*} y_{2}\left (t \right ) &= t^{2} \cos \left (t \right ) \left (\int \frac {{\mathrm e}^{-\left (\int -\frac {4}{t}d t \right )}}{t^{4} \cos \left (t \right )^{2}}d t \right ) \\ y_{2}\left (t \right ) &= t^{2} \cos \left (t \right ) \int \frac {t^{4}}{t^{4} \cos \left (t \right )^{2}} , dt \\ y_{2}\left (t \right ) &= t^{2} \cos \left (t \right ) \left (\int \sec \left (t \right )^{2}d t \right ) \\ y_{2}\left (t \right ) &= t^{2} \cos \left (t \right ) \tan \left (t \right ) \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= t^{2} \cos \left (t \right ) c_{1} +c_{2} t^{2} \cos \left (t \right ) \tan \left (t \right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= t^{2} \cos \left (t \right ) c_{1} +c_{2} t^{2} \cos \left (t \right ) \tan \left (t \right ) \\ \end{align*}
Verification of solutions
\[ y = t^{2} \cos \left (t \right ) c_{1} +c_{2} t^{2} \cos \left (t \right ) \tan \left (t \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {d}{d t}y^{\prime }\right )-4 t y^{\prime }+\left (t^{2}+6\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}+6\right ) y}{t^{2}}+\frac {4 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 {4 y^{\prime }}{t}+\frac {\left (t^{2}+6\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 {4}{t}, P_{3}\left (t \right )=\frac {t^{2}+6}{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}}}=-4 \\ {} & \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}}}=6 \\ {} & \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 )-4 t y^{\prime }+\left (t^{2}+6\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..2 \\ {} & {} & 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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +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 (-2+r \right ) \left (-3+r \right ) t^{r}+a_{1} \left (-1+r \right ) \left (-2+r \right ) t^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r -2\right ) \left (k +r -3\right )+a_{k -2}\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-2+r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{2, 3\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (-1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r -2\right ) \left (k +r -3\right )+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +r \right ) \left (k +r -1\right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +r \right ) \left (k +r -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +2}, a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +3}, a_{k +2}=-\frac {a_{k}}{\left (k +3\right ) \left (k +2\right )}, 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 +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k +3}\right ), a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )}, a_{1}=0, b_{k +2}=-\frac {b_{k}}{\left (k +3\right ) \left (k +2\right )}, 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: 0.0 (sec). Leaf size: 17
dsolve([t^2*diff(y(t),t$2)-4*t*diff(y(t),t)+(t^2+6)*y(t)=0,t^2*cos(t)],singsol=all)
\[ y \left (t \right ) = t^{2} \left (c_{1} \sin \left (t \right )+c_{2} \cos \left (t \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.053 (sec). Leaf size: 37
DSolve[t^2*y''[t]-4*t*y'[t]+(t^2+6)*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{2} e^{-i t} t^2 \left (2 c_1-i c_2 e^{2 i t}\right ) \]