2.1.51 Problem 51

Internal problem ID [10037]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 51
Date solved : Monday, December 08, 2025 at 07:09:48 PM
CAS classification : [[_2nd_order, _missing_y]]

2.1.51.1 second order ode missing y

0.372 (sec)

\begin{align*} t^{2} y^{\prime \prime }-2 y^{\prime }&=0 \\ \end{align*}

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(t) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(t) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} u^{\prime }\left (t \right ) t^{2}-2 u \left (t \right ) = 0 \end{align*}

Which is now solved for \(u(t)\) as first order ode.

Entering first order ode linear solverIn 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 {2}{t^{2}}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {2}{t^{2}}d t}\\ &= {\mathrm e}^{\frac {2}{t}} \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 \,{\mathrm e}^{\frac {2}{t}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\frac {2}{t}}&= \int {0 \,dt} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {2}{t}}\) gives the final solution

\[ u \left (t \right ) = {\mathrm e}^{-\frac {2}{t}} c_1 \]

In summary, these are the solution found for \(y\)

\begin{align*} u \left (t \right ) &= {\mathrm e}^{-\frac {2}{t}} c_1 \\ \end{align*}

For solution \(u \left (t \right ) = {\mathrm e}^{-\frac {2}{t}} c_1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = {\mathrm e}^{-\frac {2}{t}} c_1 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {{\mathrm e}^{-\frac {2}{t}} c_1\, dt}\\ y &= c_1 \left (t \,{\mathrm e}^{-\frac {2}{t}}-2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right )\right ) + c_2 \end{align*}

\begin{align*} y&= {\mathrm e}^{-\frac {2}{t}} c_1 t -2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right ) c_1 +c_2 \end{align*}

In summary, these are the solution found for \((y)\)

\begin{align*} y &= {\mathrm e}^{-\frac {2}{t}} c_1 t -2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right ) c_1 +c_2 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= {\mathrm e}^{-\frac {2}{t}} c_1 t -2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right ) c_1 +c_2 \\ \end{align*}

2.1.51.2 second order ode non constant coeff transformation on B

0.169 (sec)

\begin{align*} t^{2} y^{\prime \prime }-2 y^{\prime }&=0 \\ \end{align*}

Entering second order ode non constant coeff transformation on \(B\) solverGiven an ode of the form

\begin{align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(t) \end{align*}

This method reduces the order ode the ODE by one by applying the transformation

\begin{align*} y&= B v \end{align*}

This results in

\begin{align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end{align*}

And now the original ode becomes

\begin{align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end{align*}

If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve

\[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \]

By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is

\[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \]

The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).

This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that

\begin{align*} A &= t^{2}\\ B &= -2\\ C &= 0\\ F &= 0 \end{align*}

The above shows that for this ode

\begin{align*} AB''+BB'+CB &= \left (t^{2}\right ) \left (0\right ) + \left (-2\right ) \left (0\right ) + \left (0\right ) \left (-2\right ) \\ &=0 \end{align*}

Hence the ode in \(v\) given in (1) now simplifies to

\begin{align*} -2 t^{2} v'' +\left ( 4\right ) v' & =0 \end{align*}

Now by applying \(v'=u\) the above becomes

\begin{align*} -2 t^{2} u^{\prime }\left (t \right )+4 u \left (t \right ) = 0 \end{align*}

Which is now solved for \(u\). Entering first order ode linear solverIn 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 {2}{t^{2}}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {2}{t^{2}}d t}\\ &= {\mathrm e}^{\frac {2}{t}} \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 \,{\mathrm e}^{\frac {2}{t}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\frac {2}{t}}&= \int {0 \,dt} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {2}{t}}\) gives the final solution

\[ u \left (t \right ) = {\mathrm e}^{-\frac {2}{t}} c_1 \]

The ode for \(v\) now becomes

\[ v^{\prime }\left (t \right ) = {\mathrm e}^{-\frac {2}{t}} c_1 \]

Which is now solved for \(v\). Entering first order ode quadrature solverSince the ode has the form \(v^{\prime }\left (t \right )=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dv} &= \int {{\mathrm e}^{-\frac {2}{t}} c_1\, dt}\\ v \left (t \right ) &= c_1 \left (t \,{\mathrm e}^{-\frac {2}{t}}-2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right )\right ) + c_2 \end{align*}

\begin{align*} v \left (t \right )&= {\mathrm e}^{-\frac {2}{t}} c_1 t -2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right ) c_1 +c_2 \end{align*}

Replacing \(v \left (t \right )\) above by \(-\frac {y}{2}\), then the solution becomes

\begin{align*} y(t) &= B v\\ &= -2 \,{\mathrm e}^{-\frac {2}{t}} c_1 t +4 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right ) c_1 -2 c_2 \end{align*}

Summary of solutions found

\begin{align*} y &= -2 \,{\mathrm e}^{-\frac {2}{t}} c_1 t +4 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right ) c_1 -2 c_2 \\ \end{align*}

2.1.51.3 second order kovacic

1.130 (sec)

\begin{align*} t^{2} y^{\prime \prime }-2 y^{\prime }&=0 \\ \end{align*}

Entering kovacic solverWriting the ode as

\begin{align*} t^{2} y^{\prime \prime }-2 y^{\prime } &= 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 &= -2\tag {3} \\ C &= 0 \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 {1+2 t}{t^{4}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 1+2 t\\ t &= t^{4} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= \left ( \frac {1+2 t}{t^{4}}\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 2.18: 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) \\ &= 4 - 1 \\ &= 3 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=t^{4}\). There is a pole at \(t=0\) of order \(4\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(3\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Attempting to find a solution using case \(n=1\).

Looking at higher order poles of order \(2 v \)≥\( 4\) (must be even order for case one).Then for each pole \(c\), \([\sqrt r]_{c}\) is the sum of terms \(\frac {1}{(t-c)^i}\) for \(2 \leq i \leq v\) in the Laurent series expansion of \(\sqrt r\) expanded around each pole \(c\). Hence

\begin{align*} [\sqrt r]_c &= \sum _2^v \frac {a_i}{ (t-c)^i} \tag {1B} \end{align*}

Let \(a\) be the coefficient of the term \(\frac {1}{ (t-c)^v}\) in the above where \(v\) is the pole order divided by 2. Let \(b\) be the coefficient of \(\frac {1}{ (t-c)^{v+1}} \) in \(r\) minus the coefficient of \(\frac {1}{ (t-c)^{v+1}} \) in \([\sqrt r]_c\). Then

\begin{alignat*}{1} \alpha _c^{+} &= \frac {1}{2} \left ( \frac {b}{a} + v \right ) \\ \alpha _c^{-} &= \frac {1}{2} \left (- \frac {b}{a} + v \right ) \end{alignat*}

The partial fraction decomposition of \(r\) is

\[ r = \frac {1}{t^{4}}+\frac {2}{t^{3}} \]

There is pole in \(r\) at \(t= 0\) of order \(4\), hence \(v=2\). Expanding \(\sqrt {r}\) as Laurent series about this pole \(c=0\) gives

\begin{equation} \tag{2B} [\sqrt {r}]_c \approx \frac {1}{t^{2}}+\frac {1}{t}-\frac {1}{2}+\frac {t}{2}-\frac {5 t^{2}}{8}+\frac {7 t^{3}}{8} + \dots \end{equation}

Using eq. (1B), taking the sum up to \(v=2\) the above becomes

\begin{equation} \tag{3V} [\sqrt {r}]_c = \frac {1}{t^{2}} \end{equation}

The above shows that the coefficient of \(\frac {1}{(t-0)^{2}}\) is

\[ a = 1 \]

Now we need to find \(b\). let \(b\) be the coefficient of the term \(\frac {1}{(t-c)^{v+1}}\) in \(r\) minus the coefficient of the same term but in the sum \([\sqrt r]_c \) found in eq. (3B). Here \(c\) is current pole which is \(c=0\). This term becomes \(\frac {1}{t^{3}}\). The coefficient of this term in the sum \([\sqrt r]_c\) is seen to be \(0\) and the coefficient of this term \(r\) is found from the partial fraction decomposition from above to be \(2\). Therefore

\begin{align*} b &= \left (2\right )-(0)\\ &= 2 \end{align*}

Hence

\begin{alignat*}{3} [\sqrt r]_c &= \frac {1}{t^{2}} \\ \alpha _c^{+} &= \frac {1}{2} \left ( \frac {b}{a} + v \right ) &&= \frac {1}{2} \left ( \frac {2}{1} + 2 \right ) &&=2\\ \alpha _c^{-} &= \frac {1}{2} \left (- \frac {b}{a} + v \right ) &&= \frac {1}{2} \left (- \frac {2}{1} + 2 \right )&&=0 \end{alignat*}

Since the order of \(r\) at \(\infty \) is \(3 > 2\) then

\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= 0 \\ \alpha _{\infty }^{-} &= 1 \end{alignat*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is

\[ r=\frac {1+2 t}{t^{4}} \]

pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)
\(0\)	\(4\)	\(\frac {1}{t^{2}}\)	\(2\)	\(0\)

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(3\)	\(0\)	\(0\)	\(1\)

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 \(\alpha _\infty ^{+} = 0\) then

\begin{align*} d &= \alpha _\infty ^{+} - \left ( \alpha _{c_1}^{-} \right ) \\ &= 0 - \left ( 0 \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*}

Substituting the above values in the above results in

\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{t- c_1}\right ) + (+) [\sqrt r]_\infty \\ &= -\frac {1}{t^{2}} + \left ( 0 \right ) \\ &= -\frac {1}{t^{2}}\\ &= -\frac {1}{t^{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 {1}{t^{2}}\right ) \left (0\right ) + \left ( \left (\frac {2}{t^{3}}\right ) + \left (-\frac {1}{t^{2}}\right )^2 - \left (\frac {1+2 t}{t^{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 {1}{t^{2}}d t}\\ &= {\mathrm e}^{\frac {1}{t}} \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 {-2}{t^{2}} \,dt} \\ &= z_1 e^{-\frac {1}{t}} \\ &= z_1 \left ({\mathrm e}^{-\frac {1}{t}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = 1 \]

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 {-2}{t^{2}} \,dt}}{\left (y_1\right )^2} \,dt \\ &= y_1 \int \frac { e^{-\frac {2}{t}}}{\left (y_1\right )^2} \,dt \\ &= y_1 \left (t \,{\mathrm e}^{-\frac {2}{t}}-2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right )\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (1\right ) + c_2 \left (1\left (t \,{\mathrm e}^{-\frac {2}{t}}-2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right )\right )\right ) \\ \end{align*}

Summary of solutions found

\begin{align*} y &= c_1 +c_2 \left (t \,{\mathrm e}^{-\frac {2}{t}}-2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right )\right ) \\ \end{align*}

2.1.51.4 SSolved using second order ode arccos transformation

14.593 (sec)

\begin{align*} t^{2} y^{\prime \prime }-2 y^{\prime }&=0 \\ \end{align*}

Applying change of variable \(t = \arccos \left (\tau \right )\) to the above ode results in the following new ode

\[ 2 \left (\frac {d}{d \tau }y \left (\tau \right )\right ) \sqrt {-\tau ^{2}+1}-\frac {\left (\left (\tau ^{2}-1\right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )+\left (\frac {d}{d \tau }y \left (\tau \right )\right ) \tau \right ) \left (\pi -2 \arcsin \left (\tau \right )\right )^{2}}{4} = 0 \]

Which is now solved for \(y \left (\tau \right )\). Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y \left (\tau \right )\). Let

\begin{align*} u(\tau ) &= \frac {d}{d \tau }y \left (\tau \right ) \end{align*}

Then

\begin{align*} u'(\tau ) &= \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) \end{align*}

Hence the ode becomes

\begin{align*} 2 u \left (\tau \right ) \sqrt {-\tau ^{2}+1}-\frac {\left (\left (\tau ^{2}-1\right ) \left (\frac {d}{d \tau }u \left (\tau \right )\right )+u \left (\tau \right ) \tau \right ) \left (\pi -2 \arcsin \left (\tau \right )\right )^{2}}{4} = 0 \end{align*}

Which is now solved for \(u(\tau )\) as first order ode.

Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} \frac {d}{d \tau }u \left (\tau \right ) + q(\tau )u \left (\tau \right ) &= p(\tau ) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(\tau ) &=-\frac {-\pi ^{2} \tau +4 \pi \arcsin \left (\tau \right ) \tau -4 \arcsin \left (\tau \right )^{2} \tau +8 \sqrt {-\tau ^{2}+1}}{\left (\tau -1\right ) \left (\tau +1\right ) \left (\pi -2 \arcsin \left (\tau \right )\right )^{2}}\\ p(\tau ) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,d\tau }}\\ &= {\mathrm e}^{\int -\frac {-\pi ^{2} \tau +4 \pi \arcsin \left (\tau \right ) \tau -4 \arcsin \left (\tau \right )^{2} \tau +8 \sqrt {-\tau ^{2}+1}}{\left (\tau -1\right ) \left (\tau +1\right ) \left (\pi -2 \arcsin \left (\tau \right )\right )^{2}}d \tau }\\ &= \sqrt {-\tau ^{2}+1}\, {\mathrm e}^{\frac {2}{\arccos \left (\tau \right )}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \left (u \sqrt {-\tau ^{2}+1}\, {\mathrm e}^{\frac {2}{\arccos \left (\tau \right )}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \sqrt {-\tau ^{2}+1}\, {\mathrm e}^{\frac {2}{\arccos \left (\tau \right )}}&= \int {0 \,d\tau } + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(\sqrt {-\tau ^{2}+1}\, {\mathrm e}^{\frac {2}{\arccos \left (\tau \right )}}\) gives the final solution

\[ u \left (\tau \right ) = \frac {{\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}} c_1}{\sqrt {-\tau ^{2}+1}} \]

In summary, these are the solution found for \(y \left (\tau \right )\)

\begin{align*} u \left (\tau \right ) &= \frac {{\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}} c_1}{\sqrt {-\tau ^{2}+1}} \\ \end{align*}

For solution \(u \left (\tau \right ) = \frac {{\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}} c_1}{\sqrt {-\tau ^{2}+1}}\), since \(u=\frac {d}{d \tau }y \left (\tau \right )\) then we now have a new first order ode to solve which is

\begin{align*} \frac {d}{d \tau }y \left (\tau \right ) = \frac {{\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}} c_1}{\sqrt {-\tau ^{2}+1}} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(\frac {d}{d \tau }y \left (\tau \right )=f(\tau )\), then we only need to integrate \(f(\tau )\).

\begin{align*} \int {dy} &= \int {\frac {{\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}} c_1}{\sqrt {-\tau ^{2}+1}}\, d\tau }\\ y \left (\tau \right ) &= c_1 \left (-\arccos \left (\tau \right ) {\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}}+2 \,\operatorname {Ei}_{1}\left (\frac {2}{\arccos \left (\tau \right )}\right )\right ) + c_2 \end{align*}

\begin{align*} y \left (\tau \right )&= -\arccos \left (\tau \right ) {\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}} c_1 +2 \,\operatorname {Ei}_{1}\left (\frac {2}{\arccos \left (\tau \right )}\right ) c_1 +c_2 \end{align*}

In summary, these are the solution found for \((y \left (\tau \right ))\)

\begin{align*} y \left (\tau \right ) &= -\arccos \left (\tau \right ) {\mathrm e}^{-\frac {2}{\arccos \left (\tau \right )}} c_1 +2 \,\operatorname {Ei}_{1}\left (\frac {2}{\arccos \left (\tau \right )}\right ) c_1 +c_2 \\ \end{align*}

Applying change of variable \(\tau = \cos \left (t \right )\) to the solutions above gives

\begin{align*} y &= -\arccos \left (\cos \left (t \right )\right ) {\mathrm e}^{-\frac {2}{\arccos \left (\cos \left (t \right )\right )}} c_1 +2 \,\operatorname {Ei}_{1}\left (\frac {2}{\arccos \left (\cos \left (t \right )\right )}\right ) c_1 +c_2 \\ \end{align*}

2.1.51.5 ✓ Maple. Time used: 0.002 (sec). Leaf size: 25

ode:=t^2*diff(diff(y(t),t),t)-2*diff(y(t),t) = 0; 
dsolve(ode,y(t), singsol=all);
 

\[ y = {\mathrm e}^{-\frac {2}{t}} c_2 t -2 \,\operatorname {Ei}_{1}\left (\frac {2}{t}\right ) c_2 +c_1 \]

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 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
<- LODE missing y successful
 

2.1.51.6 ✓ Mathematica. Time used: 0.01 (sec). Leaf size: 26

ode=t^2*D[y[t],{t,2}]-2*D[y[t],t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\begin{align*} y(t)&\to \int _1^te^{-\frac {2}{K[1]}} c_1dK[1]+c_2 \end{align*}

2.1.51.7 ✓ Sympy. Time used: 0.479 (sec). Leaf size: 26

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t**2*Derivative(y(t), (t, 2)) - 2*Derivative(y(t), t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = C_{1} - C_{2} t e^{- \frac {2}{t}} - 2 C_{2} \operatorname {Ei}{\left (\frac {2 e^{i \pi }}{t} \right )} \]