2.1.49 problem 49
Internal
problem
ID
[8437]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
49
Date
solved
:
Thursday, December 12, 2024 at 09:08:02 AM
CAS
classification
:
[[_Emden, _Fowler]]
Solve
\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \end{align*}
Solved as second order Euler type ode
Time used: 0.108 (sec)
This is Euler second order ODE. Let the solution be \(y = t^r\), then \(y'=r t^{r-1}\) and \(y''=r(r-1) t^{r-2}\). Substituting these back
into the given ODE gives
\[ t^{2}(r(r-1))t^{r-2}-3 t r t^{r-1}+5 t^{r} = 0 \]
Simplifying gives
\[ r \left (r -1\right )t^{r}-3 r\,t^{r}+5 t^{r} = 0 \]
Since \(t^{r}\neq 0\) then dividing throughout by \(t^{r}\) gives
\[ r \left (r -1\right )-3 r+5 = 0 \]
Or
\[ r^{2}-4 r +5 = 0 \tag {1} \]
Equation (1) is the characteristic equation. Its roots determine the form of the general
solution. Using the quadratic equation the roots are
\begin{align*} r_1 &= 2-i\\ r_2 &= 2+i \end{align*}
The roots are complex conjugate of each others. Let the roots be
\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}
Where in this case \(\alpha =2\) and \(\beta =-1\). Hence the solution becomes
\begin{align*} y &= c_1 t^{r_1} + c_2 t^{r_2} \\ &= c_1 t^{\alpha + i \beta } + c_2 t^{\alpha - i \beta } \\ &= t^{\alpha } \left ( c_1 t^{i \beta } + c_2 t^{- i \beta }\right ) \\ &= t^{\alpha } \left ( c_1 e^{\ln \left (t^{i \beta }\right )} + c_2 e^{\ln \left (t^{-i \beta }\right )}\right ) \\ &= t^{\alpha } \left ( c_1 e^{i \left (\beta \ln {t}\right )} + c_2 e^{-i \left (\beta \ln {t}\right )}\right ) \end{align*}
Using the values for \(\alpha =2, \beta =-1\), the above becomes
\begin{align*} y&= t^{2} \left ( c_1 e^{-i \ln \left (t \right )} + c_2 e^{i \ln \left (t \right )} \right ) \end{align*}
Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to
change. Applying this to the above result gives
\begin{align*} y&=t^{2}\left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right ) \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= t^{2} \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right ) \\
\end{align*}
Solved as second order ode using change of variable on x method 2
Time used: 0.483 (sec)
In normal form the ode
\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (t \right )&=-\frac {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}} \end{align*}
Applying change of variables \(\tau = g \left (t \right )\) to (2) gives
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}
Where \(\tau \) is the new independent variable, and
\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {5} \end{align*}
Let \(p_{1} = 0\). Eq (4) simplifies to
\begin{align*} \tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )&=0 \end{align*}
This ode is solved resulting in
\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (t \right )d t}d t\\ &= \int {\mathrm e}^{-\int -\frac {3}{t}d t}d t\\ &= \int e^{3 \ln \left (t \right )} \,dt\\ &= \int t^{3}d t\\ &= \frac {t^{4}}{4}\tag {6} \end{align*}
Using (6) to evaluate \(q_{1}\) from (5) gives
\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\\ &= \frac {\frac {5}{t^{2}}}{t^{6}}\\ &= \frac {5}{t^{8}}\tag {7} \end{align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {5 y \left (\tau \right )}{t^{8}}&=0 \\ \end{align*}
But in terms of \(\tau \)
\begin{align*} \frac {5}{t^{8}}&=\frac {5}{16 \tau ^{2}} \end{align*}
Hence the above ode becomes
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {5 y \left (\tau \right )}{16 \tau ^{2}}&=0 \end{align*}
The above ode is now solved for \(y \left (\tau \right )\). Writing the ode as
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {5 y \left (\tau \right )}{16 \tau ^{2}} &= 0 \tag {1} \\ A \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) + B \frac {d}{d \tau }y \left (\tau \right ) + C y \left (\tau \right ) &= 0 \tag {2} \end{align*}
Comparing (1) and (2) shows that
\begin{align*} A &= 1 \\ B &= 0\tag {3} \\ C &= \frac {5}{16 \tau ^{2}} \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(\tau ) &= y \left (\tau \right ) e^{\int \frac {B}{2 A} \,d\tau } \end{align*}
Then (2) becomes
\begin{align*} z''(\tau ) = r z(\tau )\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 {-5}{16 \tau ^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= -5\\ t &= 16 \tau ^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(\tau ) &= \left ( -\frac {5}{16 \tau ^{2}}\right ) z(\tau )\tag {7} \end{align*}
Equation (7) is now solved. After finding \(z(\tau )\) then \(y \left (\tau \right )\) is found using the inverse transformation
\begin{align*} y \left (\tau \right ) &= z \left (\tau \right ) e^{-\int \frac {B}{2 A} \,d\tau } \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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
| | |
Table 2.7: Necessary conditions for each Kovacic case
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) \\ &= 2 - 0 \\ &= 2 \end{align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=16 \tau ^{2}\).
There is a pole at \(\tau =0\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is
\(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then
necessary conditions for case two are met. Since pole order is not larger than \(2\) and
the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore
\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is
\[
r = -\frac {5}{16 \tau ^{2}}
\]
For the pole at \(\tau =0\) let \(b\)
be the coefficient of \(\frac {1}{ \tau ^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {5}{16}}\). Hence
\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+\frac {i}{4}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-\frac {i}{4} \end{alignat*}
Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{\tau ^{2}}\) in the Laurent series expansion of \(r\)
at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\)
from
\begin{alignat*}{2} r &= \frac {s}{t} &&= -\frac {5}{16 \tau ^{2}} \end{alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {5}{16}}\). Hence
\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+\frac {i}{4}\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-\frac {i}{4} \end{alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\)
is
\[ r=-\frac {5}{16 \tau ^{2}} \]
| | | | |
| pole \(c\) location |
pole order |
\([\sqrt r]_c\) |
\(\alpha _c^{+}\) |
\(\alpha _c^{-}\) |
| | | | |
| \(0\) | \(2\) | \(0\) | \(\frac {1}{2}+\frac {i}{4}\) | \(\frac {1}{2}-\frac {i}{4}\) |
| | | | |
| | | |
| Order of \(r\) at \(\infty \) |
\([\sqrt r]_\infty \) |
\(\alpha _\infty ^{+}\) |
\(\alpha _\infty ^{-}\) |
| | | |
| \(2\) |
\(0\) | \(\frac {1}{2}+\frac {i}{4}\) | \(\frac {1}{2}-\frac {i}{4}\) |
| | | |
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 \(\alpha _\infty ^{-} = \frac {1}{2}-\frac {i}{4}\) then
\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= \frac {1}{2}-\frac {i}{4} - \left ( \frac {1}{2}-\frac {i}{4} \right ) \\ &= 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)}}{\tau -c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}
The above gives
\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{\tau - c_1}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {\frac {1}{2}-\frac {i}{4}}{\tau } + (-) \left ( 0 \right ) \\ &= \frac {\frac {1}{2}-\frac {i}{4}}{\tau }\\ &= \frac {\frac {1}{2}-\frac {i}{4}}{\tau } \end{align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(\tau )\) of degree
\(d=0\) to solve the ode. The polynomial \(p(\tau )\) 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(\tau ) &= 1\tag {2A} \end{align*}
Substituting the above in eq. (1A) gives
\begin{align*} \left (0\right ) + 2 \left (\frac {\frac {1}{2}-\frac {i}{4}}{\tau }\right ) \left (0\right ) + \left ( \left (\frac {-\frac {1}{2}+\frac {i}{4}}{\tau ^{2}}\right ) + \left (\frac {\frac {1}{2}-\frac {i}{4}}{\tau }\right )^2 - \left (-\frac {5}{16 \tau ^{2}}\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(\tau ) &= p e^{ \int \omega \,d\tau } \\ &= {\mathrm e}^{\int \frac {\frac {1}{2}-\frac {i}{4}}{\tau }d \tau }\\ &= \tau ^{\frac {1}{2}-\frac {i}{4}} \end{align*}
The first solution to the original ode in \(y \left (\tau \right )\) is found from
\[
y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,d\tau }
\]
Since \(B=0\) then the above reduces to
\begin{align*}
y_1 &= z_1 \\
&= \tau ^{\frac {1}{2}-\frac {i}{4}} \\
\end{align*}
Which simplifies to
\[
y_1 = \tau ^{\frac {1}{2}-\frac {i}{4}}
\]
The second solution \(y_2\) to the original
ode is found using reduction of order
\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,d\tau }}{y_1^2} \,d\tau \]
Since \(B=0\) then the above becomes
\begin{align*}
y_2 &= y_1 \int \frac {1}{y_1^2} \,d\tau \\
&= \tau ^{\frac {1}{2}-\frac {i}{4}}\int \frac {1}{\tau ^{1-\frac {i}{2}}} \,d\tau \\
&= \tau ^{\frac {1}{2}-\frac {i}{4}}\left (-2 i \tau \,\tau ^{-1+\frac {i}{2}}\right ) \\
\end{align*}
Therefore the solution
is
\begin{align*}
y \left (\tau \right ) &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (\tau ^{\frac {1}{2}-\frac {i}{4}}\right ) + c_2 \left (\tau ^{\frac {1}{2}-\frac {i}{4}}\left (-2 i \tau \,\tau ^{-1+\frac {i}{2}}\right )\right ) \\
\end{align*}
Will add steps showing solving for IC soon.
The above solution is now transformed back to \(y\) using (6) which results in
\[
y = c_1 \left (\frac {t^{4}}{4}\right )^{\frac {1}{2}-\frac {i}{4}}-2 i c_2 \left (\frac {t^{4}}{4}\right )^{\frac {1}{2}+\frac {i}{4}}
\]
Will add steps
showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 \left (\frac {t^{4}}{4}\right )^{\frac {1}{2}-\frac {i}{4}}-2 i c_2 \left (\frac {t^{4}}{4}\right )^{\frac {1}{2}+\frac {i}{4}} \\
\end{align*}
Solved as second order ode using change of variable on x method 1
Time used: 0.176 (sec)
In normal form the ode
\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (t \right )&=-\frac {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}} \end{align*}
Applying change of variables \(\tau = g \left (t \right )\) to (2) results
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end{align*}
Where \(\tau \) is the new independent variable, and
\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\tag {5} \end{align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5)
\begin{align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {5}\, \sqrt {\frac {1}{t^{2}}}}{c}\tag {6} \\ \tau '' &= -\frac {\sqrt {5}}{c \sqrt {\frac {1}{t^{2}}}\, t^{3}} \end{align*}
Substituting the above into (4) results in
\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (t \right )+p \left (t \right ) \tau ^{\prime }\left (t \right )}{{\tau ^{\prime }\left (t \right )}^{2}}\\ &=\frac {-\frac {\sqrt {5}}{c \sqrt {\frac {1}{t^{2}}}\, t^{3}}-\frac {3}{t}\frac {\sqrt {5}\, \sqrt {\frac {1}{t^{2}}}}{c}}{\left (\frac {\sqrt {5}\, \sqrt {\frac {1}{t^{2}}}}{c}\right )^2} \\ &=-\frac {4 c \sqrt {5}}{5} \end{align*}
Therefore ode (3) now becomes
\begin{align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )-\frac {4 c \sqrt {5}\, \left (\frac {d}{d \tau }y \left (\tau \right )\right )}{5}+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end{align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily
solved to give
\begin{align*} y \left (\tau \right ) &= {\mathrm e}^{\frac {2 \sqrt {5}\, c \tau }{5}} \left (c_1 \cos \left (\frac {\sqrt {5}\, c \tau }{5}\right )+c_2 \sin \left (\frac {\sqrt {5}\, c \tau }{5}\right )\right ) \end{align*}
Now from (6)
\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dt \\ &= \frac {\int \sqrt {5}\, \sqrt {\frac {1}{t^{2}}}d t}{c}\\ &= \frac {\sqrt {5}\, \ln \left (t \right )}{c} \end{align*}
Substituting the above into the solution obtained gives
\[
y = t^{2} \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right )
\]
Will add steps showing solving for
IC soon.
Summary of solutions found
\begin{align*}
y &= t^{2} \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right ) \\
\end{align*}
Solved as second order ode using change of variable on y method 2
Time used: 0.110 (sec)
In normal form the ode
\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y&=0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (t \right )&=-\frac {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}} \end{align*}
Applying change of variables on the depndent variable \(y = v \left (t \right ) t^{n}\) to (2) gives the following ode where
the dependent variables is \(v \left (t \right )\) and not \(y\).
\begin{align*} v^{\prime \prime }\left (t \right )+\left (\frac {2 n}{t}+p \right ) v^{\prime }\left (t \right )+\left (\frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q \right ) v \left (t \right )&=0 \tag {3} \end{align*}
Let the coefficient of \(v \left (t \right )\) above be zero. Hence
\begin{align*} \frac {n \left (n -1\right )}{t^{2}}+\frac {n p}{t}+q&=0 \tag {4} \end{align*}
Substituting the earlier values found for \(p \left (t \right )\) and \(q \left (t \right )\) into (4) gives
\begin{align*} \frac {n \left (n -1\right )}{t^{2}}-\frac {3 n}{t^{2}}+\frac {5}{t^{2}}&=0 \tag {5} \end{align*}
Solving (5) for \(n\) gives
\begin{align*} n&=2+i \tag {6} \end{align*}
Substituting this value in (3) gives
\begin{align*} v^{\prime \prime }\left (t \right )+\left (\frac {4+2 i}{t}-\frac {3}{t}\right ) v^{\prime }\left (t \right )&=0 \\ v^{\prime \prime }\left (t \right )+\frac {\left (1+2 i\right ) v^{\prime }\left (t \right )}{t}&=0 \tag {7} \\ \end{align*}
Using the substitution
\begin{align*} u \left (t \right ) = v^{\prime }\left (t \right ) \end{align*}
Then (7) becomes
\begin{align*} u^{\prime }\left (t \right )+\frac {\left (1+2 i\right ) u \left (t \right )}{t} = 0 \tag {8} \\ \end{align*}
The above is now solved for \(u \left (t \right )\). In canonical form a linear first order is
\begin{align*} u^{\prime }\left (t \right ) + q(t)u \left (t \right ) &= p(t) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(t) &=\frac {1+2 i}{t}\\ p(t) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int \frac {1+2 i}{t}d t}\\ &= t^{1+2 i} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (u \,t^{1+2 i}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} u \,t^{1+2 i}&= \int {0 \,dt} + c_1 \\ &=c_1 \end{align*}
Dividing throughout by the integrating factor \(t^{1+2 i}\) gives the final solution
\[ u \left (t \right ) = t^{-1-2 i} c_1 \]
Now that \(u \left (t \right )\) is known,
then
\begin{align*} v^{\prime }\left (t \right )&= u \left (t \right )\\ v \left (t \right )&= \int u \left (t \right )d t +c_2\\ &= \frac {i t^{-2 i} c_1}{2}+c_2 \end{align*}
Hence
\begin{align*} y&= v \left (t \right ) t^{n}\\ &= \left (\frac {i t^{-2 i} c_1}{2}+c_2 \right ) t^{2+i}\\ &= \frac {\left (i t^{-2 i} c_1 +2 c_2 \right ) t^{2+i}}{2}\\ \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \left (\frac {i t^{-2 i} c_1}{2}+c_2 \right ) t^{2+i} \\
\end{align*}
Solved as second order ode using Kovacic algorithm
Time used: 0.187 (sec)
Writing the ode as
\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}
Comparing (1) and (2) shows that
\begin{align*} A &= t^{2} \\ B &= -3 t\tag {3} \\ C &= 5 \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(t) &= y 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 {-5}{4 t^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= -5\\ t &= 4 t^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(t) &= \left ( -\frac {5}{4 t^{2}}\right ) z(t)\tag {7} \end{align*}
Equation (7) is now solved. After finding \(z(t)\) then \(y\) is found using the inverse transformation
\begin{align*} y &= 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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
| | |
Table 2.8: Necessary conditions for each Kovacic case
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) \\ &= 2 - 0 \\ &= 2 \end{align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 t^{2}\).
There is a pole at \(t=0\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is
\(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then
necessary conditions for case two are met. Since pole order is not larger than \(2\) and
the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore
\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is
\[
r = -\frac {5}{4 t^{2}}
\]
For the pole at \(t=0\) let \(b\)
be the coefficient of \(\frac {1}{ t^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {5}{4}}\). Hence
\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+i\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-i \end{alignat*}
Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{t^{2}}\) in the Laurent series expansion of \(r\)
at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\)
from
\begin{alignat*}{2} r &= \frac {s}{t} &&= -\frac {5}{4 t^{2}} \end{alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {5}{4}}\). Hence
\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+i\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-i \end{alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\)
is
\[ r=-\frac {5}{4 t^{2}} \]
| | | | |
| pole \(c\) location |
pole order |
\([\sqrt r]_c\) |
\(\alpha _c^{+}\) |
\(\alpha _c^{-}\) |
| | | | |
| \(0\) | \(2\) | \(0\) | \(\frac {1}{2}+i\) | \(\frac {1}{2}-i\) |
| | | | |
| | | |
| Order of \(r\) at \(\infty \) |
\([\sqrt r]_\infty \) |
\(\alpha _\infty ^{+}\) |
\(\alpha _\infty ^{-}\) |
| | | |
| \(2\) |
\(0\) | \(\frac {1}{2}+i\) | \(\frac {1}{2}-i\) |
| | | |
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 \(\alpha _\infty ^{-} = \frac {1}{2}-i\) then
\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= \frac {1}{2}-i - \left ( \frac {1}{2}-i \right ) \\ &= 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*}
The above gives
\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{t- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {\frac {1}{2}-i}{t} + (-) \left ( 0 \right ) \\ &= \frac {\frac {1}{2}-i}{t}\\ &= \frac {\frac {1}{2}-i}{t} \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 {\frac {1}{2}-i}{t}\right ) \left (0\right ) + \left ( \left (\frac {-\frac {1}{2}+i}{t^{2}}\right ) + \left (\frac {\frac {1}{2}-i}{t}\right )^2 - \left (-\frac {5}{4 t^{2}}\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 {\frac {1}{2}-i}{t}d t}\\ &= t^{\frac {1}{2}-i} \end{align*}
The first solution to the original ode in \(y\) is found from
\begin{align*}
y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {-3 t}{t^{2}} \,dt} \\
&= z_1 e^{\frac {3 \ln \left (t \right )}{2}} \\
&= z_1 \left (t^{{3}/{2}}\right ) \\
\end{align*}
Which simplifies to
\[
y_1 = t^{2-i}
\]
The second
solution \(y_2\) to the original ode is found using reduction of order
\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \]
Substituting gives
\begin{align*}
y_2 &= y_1 \int \frac { e^{\int -\frac {-3 t}{t^{2}} \,dt}}{\left (y_1\right )^2} \,dt \\
&= y_1 \int \frac { e^{3 \ln \left (t \right )}}{\left (y_1\right )^2} \,dt \\
&= y_1 \left (-\frac {i t^{4} t^{-4+2 i}}{2}\right ) \\
\end{align*}
Therefore
the solution is
\begin{align*}
y &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (t^{2-i}\right ) + c_2 \left (t^{2-i}\left (-\frac {i t^{4} t^{-4+2 i}}{2}\right )\right ) \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 \,t^{2-i}-\frac {i c_2 \,t^{2+i}}{2} \\
\end{align*}
Solved as second order ode adjoint method
Time used: 0.879 (sec)
In normal form the ode
\begin{align*} t^{2} y^{\prime \prime }-3 t y^{\prime }+5 y = 0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y&=r \left (t \right ) \tag {2} \end{align*}
Where
\begin{align*} p \left (t \right )&=-\frac {3}{t}\\ q \left (t \right )&=\frac {5}{t^{2}}\\ r \left (t \right )&=0 \end{align*}
The Lagrange adjoint ode is given by
\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (-\frac {3 \xi \left (t \right )}{t}\right )' + \left (\frac {5 \xi \left (t \right )}{t^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (t \right )+\frac {2 \xi \left (t \right )}{t^{2}}+\frac {3 \xi ^{\prime }\left (t \right )}{t}&= 0 \end{align*}
Which is solved for \(\xi (t)\). This is Euler second order ODE. Let the solution be \(\xi = t^r\), then \(\xi '=r t^{r-1}\) and \(\xi ''=r(r-1) t^{r-2}\).
Substituting these back into the given ODE gives
\[ t^{2}(r(r-1))t^{r-2}+3 t r t^{r-1}+2 t^{r} = 0 \]
Simplifying gives
\[ r \left (r -1\right )t^{r}+3 r\,t^{r}+2 t^{r} = 0 \]
Since \(t^{r}\neq 0\) then dividing
throughout by \(t^{r}\) gives
\[ r \left (r -1\right )+3 r+2 = 0 \]
Or
\[ r^{2}+2 r +2 = 0 \tag {1} \]
Equation (1) is the characteristic equation. Its roots
determine the form of the general solution. Using the quadratic equation the roots are
\begin{align*} r_1 &= -1-i\\ r_2 &= -1+i \end{align*}
The roots are complex conjugate of each others. Let the roots be
\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}
Where in this case \(\alpha =-1\) and \(\beta =-1\). Hence the solution becomes
\begin{align*} \xi &= c_1 t^{r_1} + c_2 t^{r_2} \\ &= c_1 t^{\alpha + i \beta } + c_2 t^{\alpha - i \beta } \\ &= t^{\alpha } \left ( c_1 t^{i \beta } + c_2 t^{- i \beta }\right ) \\ &= t^{\alpha } \left ( c_1 e^{\ln \left (t^{i \beta }\right )} + c_2 e^{\ln \left (t^{-i \beta }\right )}\right ) \\ &= t^{\alpha } \left ( c_1 e^{i \left (\beta \ln {t}\right )} + c_2 e^{-i \left (\beta \ln {t}\right )}\right ) \end{align*}
Using the values for \(\alpha =-1, \beta =-1\), the above becomes
\begin{align*} \xi &= t^{-1} \left ( c_1 e^{-i \ln \left (t \right )} + c_2 e^{i \ln \left (t \right )} \right ) \end{align*}
Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to
change. Applying this to the above result gives
\begin{align*} \xi &=\frac {1}{t}\left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right ) \end{align*}
Will add steps showing solving for IC soon.
The original ode now reduces to first order ode
\begin{align*} \xi \left (t \right ) y^{\prime }-y \xi ^{\prime }\left (t \right )+\xi \left (t \right ) p \left (t \right ) y&=\int \xi \left (t \right ) r \left (t \right )d t\\ y^{\prime }+y \left (p \left (t \right )-\frac {\xi ^{\prime }\left (t \right )}{\xi \left (t \right )}\right )&=\frac {\int \xi \left (t \right ) r \left (t \right )d t}{\xi \left (t \right )} \end{align*}
Or
\begin{align*} y^{\prime }+y \left (-\frac {3}{t}-\frac {\left (-\frac {c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )}{t^{2}}+\frac {-\frac {c_1 \sin \left (\ln \left (t \right )\right )}{t}+\frac {c_2 \cos \left (\ln \left (t \right )\right )}{t}}{t}\right ) t}{c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )}\right )&=0 \end{align*}
Which is now a first order ode. This is now solved for \(y\). In canonical form a linear first order
is
\begin{align*} y^{\prime } + q(t)y &= p(t) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(t) &=-\frac {\left (2 c_1 +c_2 \right ) \cos \left (\ln \left (t \right )\right )-\sin \left (\ln \left (t \right )\right ) \left (c_1 -2 c_2 \right )}{t \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right )}\\ p(t) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {\left (2 c_1 +c_2 \right ) \cos \left (\ln \left (t \right )\right )-\sin \left (\ln \left (t \right )\right ) \left (c_1 -2 c_2 \right )}{t \left (c_1 \cos \left (\ln \left (t \right )\right )+c_2 \sin \left (\ln \left (t \right )\right )\right )}d t}\\ &= {\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (y \,{\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} y \,{\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )}&= \int {0 \,dt} + c_3 \\ &=c_3 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{\frac {\ln \left (\tan \left (\ln \left (t \right )\right )^{2}+1\right )}{2}-\ln \left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right )-2 \ln \left (t \right )}\) gives the final solution
\[ y = \frac {\left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right ) t^{2} c_3}{\sqrt {\tan \left (\ln \left (t \right )\right )^{2}+1}} \]
Hence, the solution
found using Lagrange adjoint equation method is
\begin{align*}
y &= \frac {\left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right ) t^{2} c_3}{\sqrt {\tan \left (\ln \left (t \right )\right )^{2}+1}} \\
\end{align*}
The constants can be merged to give
\[
y = \frac {\left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right ) t^{2}}{\sqrt {\tan \left (\ln \left (t \right )\right )^{2}+1}}
\]
Will
add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \frac {\left (c_2 \tan \left (\ln \left (t \right )\right )+c_1 \right ) t^{2}}{\sqrt {\tan \left (\ln \left (t \right )\right )^{2}+1}} \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{2}-3 t \left (\frac {d}{d t}y \left (t \right )\right )+5 y \left (t \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=-\frac {5 y \left (t \right )}{t^{2}}+\frac {3 \left (\frac {d}{d t}y \left (t \right )\right )}{t} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \right )\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^{2}}{d t^{2}}y \left (t \right )-\frac {3 \left (\frac {d}{d t}y \left (t \right )\right )}{t}+\frac {5 y \left (t \right )}{t^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{2}-3 t \left (\frac {d}{d t}y \left (t \right )\right )+5 y \left (t \right )=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & s =\ln \left (t \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=\left (\frac {d}{d s}y \left (s \right )\right ) \left (\frac {d}{d t}s \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=\frac {\frac {d}{d s}y \left (s \right )}{t} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=\left (\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) \left (\frac {d}{d t}s \left (t \right )\right )^{2}+\left (\frac {d^{2}}{d t^{2}}s \left (t \right )\right ) \left (\frac {d}{d s}y \left (s \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=\frac {\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & \left (\frac {\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}}\right ) t^{2}-3 \frac {d}{d s}y \left (s \right )+5 y \left (s \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d s^{2}}y \left (s \right )-4 \frac {d}{d s}y \left (s \right )+5 y \left (s \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-4 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {4\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (2-\mathrm {I}, 2+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (s \right )={\mathrm e}^{2 s} \cos \left (s \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (s \right )={\mathrm e}^{2 s} \sin \left (s \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (s \right )=\mathit {C1} y_{1}\left (s \right )+\mathit {C2} y_{2}\left (s \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (s \right )=\mathit {C1} \,{\mathrm e}^{2 s} \cos \left (s \right )+\mathit {C2} \,{\mathrm e}^{2 s} \sin \left (s \right ) \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} s =\ln \left (t \right ) \\ {} & {} & y \left (t \right )=\mathit {C1} \,t^{2} \cos \left (\ln \left (t \right )\right )+\mathit {C2} \,t^{2} \sin \left (\ln \left (t \right )\right ) \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=t^{2} \left (\mathit {C1} \cos \left (\ln \left (t \right )\right )+\mathit {C2} \sin \left (\ln \left (t \right )\right )\right ) \end {array} \]
Maple trace
`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
<- LODE of Euler type successful`
Maple dsolve solution
Solving time : 0.003 (sec)
Leaf size : 19
dsolve(diff(diff(y(t),t),t)*t^2-3*t*diff(y(t),t)+5*y(t) = 0,
y(t),singsol=all)
\[
y = t^{2} \left (c_{1} \sin \left (\ln \left (t \right )\right )+c_{2} \cos \left (\ln \left (t \right )\right )\right )
\]
Mathematica DSolve solution
Solving time : 0.027 (sec)
Leaf size : 22
DSolve[{t^2*D[y[t],{t,2}]-3*t*D[y[t],t]+5*y[t]==0,{}},
y[t],t,IncludeSingularSolutions->True]
\[
y(t)\to t^2 (c_2 \cos (\log (t))+c_1 \sin (\log (t)))
\]