Internal problem ID [11485]
Internal file name [OUTPUT/10468_Thursday_May_18_2023_04_20_21_AM_74297487/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 2.4.1 Cauchy-Euler equations.
Exercises page 120
Problem number: 2.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {x^{\prime \prime }+t^{2} x^{\prime }=0} \] With initial conditions \begin {align*} [x \left (0\right ) = 0, x^{\prime }\left (0\right ) = 1] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end {align*}
Where here \begin {align*} p(t) &=t^{2}\\ q(t) &=0\\ F &=0 \end {align*}
Hence the ode is \begin {align*} x^{\prime \prime }+t^{2} x^{\prime } = 0 \end {align*}
The domain of \(p(t)=t^{2}\) is \[
\{-\infty
This is second order ode with missing dependent variable \(x\). Let \begin {align*} p(t) &= x^{\prime } \end {align*}
Then \begin {align*} p'(t) &= x^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} p^{\prime }\left (t \right )+t^{2} p \left (t \right ) = 0 \end {align*}
Which is now solve for \(p(t)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(t,p)\\ &= f( t) g(p)\\ &= -t^{2} p \end {align*}
Where \(f(t)=-t^{2}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -t^{2} \,d t\\ \int { \frac {1}{p} \,dp} &= \int {-t^{2} \,d t}\\ \ln \left (p \right )&=-\frac {t^{3}}{3}+c_{1}\\ p&={\mathrm e}^{-\frac {t^{3}}{3}+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {t^{3}}{3}} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(p=1\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 1 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 1 \end {align*}
Trying the constant \begin {align*} c_{1} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (t \right )&={\mathrm e}^{-\frac {t^{3}}{3}} \end {align*}
The constant \(c_{1} = 1\) gives valid solution.
Since \(p=x^{\prime }\) then the new first order ode to solve is \begin {align*} x^{\prime } = {\mathrm e}^{-\frac {t^{3}}{3}} \end {align*}
Integrating both sides gives \begin {align*} x &= \int { {\mathrm e}^{-\frac {t^{3}}{3}}\,\mathop {\mathrm {d}t}}\\ &= \frac {3^{\frac {1}{3}} \left (\frac {3 t 3^{\frac {5}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right )}{4 \left (t^{3}\right )^{\frac {1}{6}}}+\frac {3 \,3^{\frac {5}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right )}{t^{2} \left (t^{3}\right )^{\frac {1}{6}}}\right )}{3}+c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(t=0\) and \(x=0\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 0 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 0 \end {align*}
Trying the constant \begin {align*} c_{2} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} x&=\frac {3 \,3^{\frac {1}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right ) t^{3}+12 \,3^{\frac {1}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right )}{4 \left (t^{3}\right )^{\frac {1}{6}} t^{2}} \end {align*}
But this does not satisfy the initial conditions. Hence no solution can be found. The constant
\(c_{2} = 0\) does not give valid solution.
Which is valid for any constant of integration. Therefore keeping the constant in place.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} x &= \frac {3 \,3^{\frac {1}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right ) t^{3}+12 \,3^{\frac {1}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right )}{4 \left (t^{3}\right )^{\frac {1}{6}} t^{2}} \\
\end{align*} Verification of solutions
\[
x = \frac {3 \,3^{\frac {1}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right ) t^{3}+12 \,3^{\frac {1}{6}} {\mathrm e}^{-\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {7}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right )}{4 \left (t^{3}\right )^{\frac {1}{6}} t^{2}}
\] Verified OK. Writing the ode as \begin {align*} x^{\prime \prime }+t^{2} x^{\prime } &= 0 \tag {1} \\ A x^{\prime \prime } + B x^{\prime } + C x &= 0 \tag {2} \end {align*}
Comparing (1) and (2) shows that \begin {align*} A &= 1 \\ B &= t^{2}\tag {3} \\ C &= 0 \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(t) &= x e^{\int \frac {B}{2 A} \,dt} \end {align*}
Then (2) becomes \begin {align*} z''(t) = r z(t)\tag {4} \end {align*}
Where \(r\) is given by \begin {align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end {align*}
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives \begin {align*} r &= \frac {t \left (t^{3}+4\right )}{4}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= t \left (t^{3}+4\right )\\ t &= 4 \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(t) &= \left ( \frac {t \left (t^{3}+4\right )}{4}\right ) z(t)\tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(t)\) then \(x\) is found using the inverse transformation
\begin {align*} x &= z \left (t \right ) e^{-\int \frac {B}{2 A} \,dt} \end {align*}
The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3
cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table
summarizes these cases. Case Allowed pole order for \(r\) Allowed value for \(\mathcal {O}(\infty )\) 1 \(\left \{ 0,1,2,4,6,8,\cdots \right \} \) \(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \) 2
Need to have at least one pole
that is either order \(2\) or odd order
greater than \(2\). Any other pole order
is allowed as long as the above
condition is satisfied. Hence the
following set of pole orders are all
allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\). no condition 3 \(\left \{ 1,2\right \} \) \(\left \{ 2,3,4,5,6,7,\cdots \right \} \) The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore \begin {align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - 4 \\ &= -4 \end {align*}
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole
larger than \(2\) and the order at \(\infty \) is \(-4\) then the necessary conditions for case one are met.
Therefore \begin {align*} L &= [1] \end {align*}
Attempting to find a solution using case \(n=1\).
Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = -4\) then \begin {alignat*} {3} v &= \frac {-O_r(\infty )}{2} &&= \frac {4}{2} &&= 2 \end {alignat*}
\([\sqrt r]_\infty \) is the sum of terms involving \(t^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i t^i \\ &= \sum _{i=0}^{2} a_i t^i \tag {8} \end {align*}
Let \(a\) be the coefficient of \(t^v=t^2\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is \[ \sqrt r \approx \frac {t^{2}}{2}+\frac {1}{t}-\frac {1}{t^{4}}+\frac {2}{t^{7}}-\frac {5}{t^{10}}+\frac {14}{t^{13}}-\frac {42}{t^{16}}+\frac {132}{t^{19}} + \dots \tag {9} \] Comparing Eq. (9)
with Eq. (8) shows that \[ a = {\frac {1}{2}} \] From Eq. (9) the sum up to \(v=2\) gives \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{2} a_i t^i \\ &= \frac {t^{2}}{2} \tag {10} \end {align*}
Now we need to find \(b\), where \(b\) be the coefficient of \(t^{v-1} = t^{1}=t\) in \(r\) minus the coefficient of same term but
in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence \[ \left ( [\sqrt r]_\infty \right )^2 = \frac {t^{4}}{4} \] This shows that the coefficient of \(t\) in the
above is \(0\). Now we need to find the coefficient of \(t\) in \(r\). How this is done depends on if \(v=0\) or
not. Since \(v=2\) which is not zero, then starting \(r=\frac {s}{t}\), we do long division and write this
in the form \[ r = Q + \frac {R}{t} \] Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient
of \(t\) in \(r\) will be the coefficient this term in the quotient. Doing long division gives
\begin {align*} r &= \frac {s}{t} \\ &= \frac {t \left (t^{3}+4\right )}{4} \\ &= Q + \frac {R}{4} \\ &= \left (\frac {1}{4} t^{4}+t\right ) + \left ( 0\right ) \\ &= \frac {1}{4} t^{4}+t \end {align*}
We see that the coefficient of the term \(\frac {1}{t}\) in the quotient is \(1\). Now \(b\) can be found.
\begin {align*} b &= \left (1\right )-\left (0\right )\\ &= 1 \end {align*}
Hence \begin {alignat*} {3} [\sqrt r]_\infty &= \frac {t^{2}}{2}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {1}{{\frac {1}{2}}} - 2 \right ) &&= 0\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {1}{{\frac {1}{2}}} - 2 \right ) &&= -2 \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\)
is \[ r=\frac {t \left (t^{3}+4\right )}{4} \]
Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \)
and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative
integer \(d\) from these using \begin {align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end {align*}
Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until
such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{+} = 0\), and since there are no poles, then
\begin {align*} d &= \alpha _\infty ^{+} \\ &= 0 \end {align*}
Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using \begin {align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{t-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}
Substituting the above values in the above results in \begin {align*} \omega &= (+) [\sqrt r]_\infty \\ &= 0 + \left ( \frac {t^{2}}{2} \right ) \\ &= \frac {t^{2}}{2}\\ &= \frac {t^{2}}{2} \end {align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(t)\) of degree
\(d=0\) to solve the ode. The polynomial \(p(t)\) needs to satisfy the equation \begin {align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end {align*}
Let \begin {align*} p(t) &= 1\tag {2A} \end {align*}
Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (\frac {t^{2}}{2}\right ) \left (0\right ) + \left ( \left (t\right ) + \left (\frac {t^{2}}{2}\right )^2 - \left (\frac {t \left (t^{3}+4\right )}{4}\right ) \right ) &= 0\\ 0 = 0 \end {align*}
The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is
\begin {align*} z_1(t) &= p e^{ \int \omega \,dt} \\ &= {\mathrm e}^{\int \frac {t^{2}}{2}d t}\\ &= {\mathrm e}^{\frac {t^{3}}{6}} \end {align*}
The first solution to the original ode in \(x\) is found from \begin{align*}
x_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {t^{2}}{1} \,dt} \\
&= z_1 e^{-\frac {t^{3}}{6}} \\
&= z_1 \left ({\mathrm e}^{-\frac {t^{3}}{6}}\right ) \\
\end{align*} Which simplifies to \[
x_1 = 1
\] The second
solution \(x_2\) to the original ode is found using reduction of order \[ x_2 = x_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{x_1^2} \,dt \] Substituting gives \begin{align*}
x_2 &= x_1 \int \frac { e^{\int -\frac {t^{2}}{1} \,dt}}{\left (x_1\right )^2} \,dt \\
&= x_1 \int \frac { e^{-\frac {t^{3}}{3}}}{\left (x_1\right )^2} \,dt \\
&= x_1 \left (\int {\mathrm e}^{-\frac {t^{3}}{3}}d t\right ) \\
\end{align*} Therefore
the solution is
\begin{align*}
x &= c_{1} x_1 + c_{2} x_2 \\
&= c_{1} \left (1\right ) + c_{2} \left (1\left (\int {\mathrm e}^{-\frac {t^{3}}{3}}d t\right )\right ) \\
\end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} x = c_{1} +c_{2} \left (\int {\mathrm e}^{-\frac {t^{3}}{3}}d t \right ) \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(x = 0\) and \(t = 0\) in the above gives
\begin {align*} 0 = \munderset {t \rightarrow 0}{\operatorname {lim}}\left (c_{1} +c_{2} \left (\int {\mathrm e}^{-\frac {t^{3}}{3}}d t \right )\right )\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} x^{\prime } = c_{2} {\mathrm e}^{-\frac {t^{3}}{3}} \end {align*}
substituting \(x^{\prime } = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = c_{2}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=-\left (\int _{}^{0}{\mathrm e}^{-\frac {\textit {\_a}^{3}}{3}}d \textit {\_a} \right )\\ c_{2}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} x = -\left (\int _{}^{0}{\mathrm e}^{-\frac {\textit {\_a}^{3}}{3}}d \textit {\_a} \right )+\int {\mathrm e}^{-\frac {t^{3}}{3}}d t \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} x &= -\left (\int _{}^{0}{\mathrm e}^{-\frac {\textit {\_a}^{3}}{3}}d \textit {\_a} \right )+\int {\mathrm e}^{-\frac {t^{3}}{3}}d t \\
\end{align*} Verification of solutions
\[
x = -\left (\int _{}^{0}{\mathrm e}^{-\frac {\textit {\_a}^{3}}{3}}d \textit {\_a} \right )+\int {\mathrm e}^{-\frac {t^{3}}{3}}d t
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }+t^{2} x^{\prime }=0, x \left (0\right )=0, x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} x \\ {} & {} & x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{2}\cdot x^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t^{2}\cdot x^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} k \,t^{k +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & t^{2}\cdot x^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1\right ) t^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{\prime \prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) t^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & x^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) t^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{2}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )+a_{k -1} \left (k -1\right )\right ) t^{k}\right )=0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 2 a_{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}+a_{k -1} \left (k -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+3 k +5\right ) a_{k +3}+a_{k} k =0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [x=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +3}=-\frac {a_{k} k}{k^{2}+5 k +6}, 2 a_{2}=0\right ] \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 1.297 (sec). Leaf size: 63
\[
x \left (t \right ) = \frac {{\mathrm e}^{-\frac {t^{3}}{3}} \sqrt {t}\, \left (4 \,3^{\frac {5}{6}} \left (t^{3}\right )^{\frac {1}{6}}+9 \,{\mathrm e}^{\frac {t^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, \frac {t^{3}}{3}\right )\right ) 3^{\frac {1}{6}} \left (\left \{\begin {array}{cc} \frac {1}{1-i \sqrt {3}} & t <0 \\ \frac {1}{2} & 0\le t \end {array}\right .\right )}{6}
\]
✓ Solution by Mathematica
Time used: 0.153 (sec). Leaf size: 43
\[
x(t)\to \frac {t^2 \operatorname {Gamma}\left (\frac {1}{3}\right )-\left (t^3\right )^{2/3} \Gamma \left (\frac {1}{3},\frac {t^3}{3}\right )}{3^{2/3} t^2}
\]
11.9.2 Solving as second order ode missing y ode
11.9.3 Solving using Kovacic algorithm
Order of \(r\) at \(\infty \)
\([\sqrt r]_\infty \)
\(\alpha _\infty ^{+}\)
\(\alpha _\infty ^{-}\) \(-4\)
\(\frac {t^{2}}{2}\) \(0\) \(-2\)
11.9.4 Maple step by step solution
`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
<- LODE missing y successful`
dsolve([diff(x(t),t$2)+t^2*diff(x(t),t)=0,x(0) = 0, D(x)(0) = 1],x(t), singsol=all)
DSolve[{x''[t]+t^2*x'[t]==0,{x[0]==0,x'[0]==1}},x[t],t,IncludeSingularSolutions -> True]