Internal problem ID [11498]
Internal file name [OUTPUT/10481_Thursday_May_18_2023_04_20_43_AM_83109919/index.tex]
Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag,
NY. 2015.
Section: Chapter 2, Second order linear equations. Section 2.4.3 Reduction of order. Exercises
page 125
Problem number: 5.
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]]
\[ \boxed {x^{\prime \prime }-\frac {\left (t +2\right ) x^{\prime }}{t}+\frac {\left (t +2\right ) x}{t^{2}}=0} \] Given that one solution of the ode is \begin {align*} x_1 &= 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 \[ p \left (t \right ) = -1-\frac {2}{t} \] Therefore \begin{align*} x_{2}\left (t \right ) &= t \left (\int \frac {{\mathrm e}^{-\left (\int \left (-1-\frac {2}{t}\right )d t \right )}}{t^{2}}d t \right ) \\ x_{2}\left (t \right ) &= t \int \frac {{\mathrm e}^{t +2 \ln \left (t \right )}}{t^{2}} , dt \\ x_{2}\left (t \right ) &= t \left (\int {\mathrm e}^{t}d t \right ) \\ x_{2}\left (t \right ) &= t \,{\mathrm e}^{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} t +c_{2} t \,{\mathrm e}^{t} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} x &= c_{1} t +c_{2} t \,{\mathrm e}^{t} \\ \end{align*}
Verification of solutions
\[ x = c_{1} t +c_{2} t \,{\mathrm e}^{t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime \prime }+\frac {\left (-t -2\right ) x^{\prime }}{t}+\frac {\left (t +2\right ) x}{t^{2}}=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 (t +2\right ) x^{\prime }}{t}-\frac {\left (t +2\right ) x}{t^{2}} \\ \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\right ) x^{\prime }}{t}+\frac {\left (t +2\right ) x}{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} x^{\prime \prime }-\left (t +2\right ) t x^{\prime }+x \left (t +2\right )=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 =1..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^{2}\cdot x^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t^{2}\cdot x^{\prime \prime }=\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 [x=\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 [x=\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 [x=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{2+k}\right ), a_{k +1}=\frac {a_{k}}{k +1}, b_{k +1}=\frac {b_{k}}{2+k}\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([diff(x(t),t$2)-(t+2)/t*diff(x(t),t)+(t+2)/t^2*x(t)=0,t],singsol=all)
\[ x \left (t \right ) = t \left (c_{1} +c_{2} {\mathrm e}^{t}\right ) \]
✓ Solution by Mathematica
Time used: 0.034 (sec). Leaf size: 16
DSolve[x''[t]-(t+2)/t*x'[t]+(t+2)/t^2*x[t]==0,x[t],t,IncludeSingularSolutions -> True]
\[ x(t)\to t \left (c_2 e^t+c_1\right ) \]