Internal problem ID [11499]
Internal file name [OUTPUT/10482_Thursday_May_18_2023_04_20_45_AM_67513854/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: 6.
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]]
\[ \boxed {t^{2} x^{\prime \prime }+x^{\prime } t +\left (t^{2}-\frac {1}{4}\right ) x=0} \] Given that one solution of the ode is \begin {align*} x_1 &= \frac {\cos \left (t \right )}{\sqrt {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 \[ p \left (t \right ) = \frac {1}{t} \] Therefore \begin{align*} x_{2}\left (t \right ) &= \frac {\cos \left (t \right ) \left (\int \frac {{\mathrm e}^{-\left (\int \frac {1}{t}d t \right )} t}{\cos \left (t \right )^{2}}d t \right )}{\sqrt {t}} \\ x_{2}\left (t \right ) &= \frac {\cos \left (t \right )}{\sqrt {t}} \int \frac {\frac {1}{t}}{\frac {\cos \left (t \right )^{2}}{t}} , dt \\ x_{2}\left (t \right ) &= \frac {\cos \left (t \right ) \left (\int \sec \left (t \right )^{2}d t \right )}{\sqrt {t}} \\ x_{2}\left (t \right ) &= \frac {\cos \left (t \right ) \tan \left (t \right )}{\sqrt {t}} \\ \end{align*} Hence the solution is \begin{align*} x &= c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right ) \\ &= \frac {\cos \left (t \right ) c_{1}}{\sqrt {t}}+\frac {c_{2} \cos \left (t \right ) \tan \left (t \right )}{\sqrt {t}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} x &= \frac {\cos \left (t \right ) c_{1}}{\sqrt {t}}+\frac {c_{2} \cos \left (t \right ) \tan \left (t \right )}{\sqrt {t}} \\ \end{align*}
Verification of solutions
\[ x = \frac {\cos \left (t \right ) c_{1}}{\sqrt {t}}+\frac {c_{2} \cos \left (t \right ) \tan \left (t \right )}{\sqrt {t}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} x^{\prime \prime }+x^{\prime } t +\left (t^{2}-\frac {1}{4}\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 (4 t^{2}-1\right ) x}{4 t^{2}}-\frac {x^{\prime }}{t} \\ \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 {x^{\prime }}{t}+\frac {\left (4 t^{2}-1\right ) x}{4 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 {1}{t}, P_{3}\left (t \right )=\frac {4 t^{2}-1}{4 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}}}=1 \\ {} & \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}}}=-\frac {1}{4} \\ {} & \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} \\ {} & {} & 4 t^{2} x^{\prime \prime }+4 x^{\prime } t +\left (4 t^{2}-1\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..2 \\ {} & {} & 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 \cdot x^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot x^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +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+2 r \right ) \left (-1+2 r \right ) t^{r}+a_{1} \left (3+2 r \right ) \left (1+2 r \right ) t^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +2 r +1\right ) \left (2 k +2 r -1\right )+4 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 (1+2 r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+2 r \right ) \left (1+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 (4 k^{2}+8 k r +4 r^{2}-1\right )+4 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (4 \left (k +2\right )^{2}+8 \left (k +2\right ) r +4 r^{2}-1\right )+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+8 k r +4 r^{2}+16 k +16 r +15} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+12 k +8} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & \left [x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k -\frac {1}{2}}, a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+12 k +8}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+20 k +24} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +\frac {1}{2}}, a_{k +2}=-\frac {4 a_{k}}{4 k^{2}+20 k +24}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [x=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k +\frac {1}{2}}\right ), a_{2+k}=-\frac {4 a_{k}}{4 k^{2}+12 k +8}, a_{1}=0, b_{2+k}=-\frac {4 b_{k}}{4 k^{2}+20 k +24}, 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: 0.031 (sec). Leaf size: 17
dsolve([t^2*diff(x(t),t$2)+t*diff(x(t),t)+(t^2-1/4)*x(t)=0,cos(t)/sqrt(t)],singsol=all)
\[ x \left (t \right ) = \frac {c_{1} \sin \left (t \right )+c_{2} \cos \left (t \right )}{\sqrt {t}} \]
✓ Solution by Mathematica
Time used: 0.052 (sec). Leaf size: 39
DSolve[t^2*x''[t]+t*x'[t]+(t^2-1/4)*x[t]==0,x[t],t,IncludeSingularSolutions -> True]
\[ x(t)\to \frac {e^{-i t} \left (2 c_1-i c_2 e^{2 i t}\right )}{2 \sqrt {t}} \]