2.1.50 problem 50

Internal problem ID [8438]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 50
Date solved : Thursday, December 12, 2024 at 09:08:04 AM
CAS classification : [[_2nd_order, _missing_y]]

Solve

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

Solved as second order linear exact ode

Time used: 0.095 (sec)

An ode of the form

\begin{align*} p \left (t \right ) y^{\prime \prime }+q \left (t \right ) y^{\prime }+r \left (t \right ) y&=s \left (t \right ) \end{align*}

is exact if

\begin{align*} p''(t) - q'(t) + r(t) &= 0 \tag {1} \end{align*}

For the given ode we have

\begin{align*} p(x) &= t\\ q(x) &= 1\\ r(x) &= 0\\ s(x) &= 0 \end{align*}

Hence

\begin{align*} p''(x) &= 0\\ q'(x) &= 0 \end{align*}

Therefore (1) becomes

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

Hence the ode is exact. Since we now know the ode is exact, it can be written as

\begin{align*} \left (p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y\right )' &= s(x) \end{align*}

Integrating gives

\begin{align*} p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y&=\int {s \left (t \right )\, dt} \end{align*}

Substituting the above values for \(p,q,r,s\) gives

\begin{align*} t y^{\prime }&=c_1 \end{align*}

We now have a first order ode to solve which is

\begin{align*} t y^{\prime } = c_1 \end{align*}

Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {\frac {c_1}{t}\, dt}\\ y &= c_1 \ln \left (t \right ) + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \ln \left (t \right )+c_2 \\ \end{align*}

Solved as second order missing y ode

Time used: 0.058 (sec)

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

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

Then

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

Hence the ode becomes

\begin{align*} t p^{\prime }\left (t \right )+p \left (t \right ) = 0 \end{align*}

Which is now solve for \(p(t)\) as first order ode. In canonical form a linear first order is

\begin{align*} p^{\prime }\left (t \right ) + q(t)p \left (t \right ) &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=\frac {1}{t}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int \frac {1}{t}d t}\\ &= t \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (p t\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} p t&= \int {0 \,dt} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(t\) gives the final solution

\[ p \left (t \right ) = \frac {c_1}{t} \]

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = \frac {c_1}{t} \end{align*}

Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {\frac {c_1}{t}\, dt}\\ y &= c_1 \ln \left (t \right ) + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \ln \left (t \right )+c_2 \\ \end{align*}

Solved as second order integrable as is ode

Time used: 0.036 (sec)

Integrating both sides of the ODE w.r.t \(t\) gives

\begin{align*} \int \left (t y^{\prime \prime }+y^{\prime }\right )d t &= 0 \\ t y^{\prime } = c_1 \end{align*}

Which is now solved for \(y\). Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {\frac {c_1}{t}\, dt}\\ y &= c_1 \ln \left (t \right ) + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \ln \left (t \right )+c_2 \\ \end{align*}

Solved as second order integrable as is ode (ABC method)

Time used: 0.026 (sec)

Writing the ode as

\[ t y^{\prime \prime }+y^{\prime } = 0 \]

Integrating both sides of the ODE w.r.t \(t\) gives

\begin{align*} \int \left (t y^{\prime \prime }+y^{\prime }\right )d t &= 0 \\ t y^{\prime } = c_1 \end{align*}

Which is now solved for \(y\). Since the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {\frac {c_1}{t}\, dt}\\ y &= c_1 \ln \left (t \right ) + c_2 \end{align*}

Will add steps showing solving for IC soon.

Solved as second order ode using non constant coeff transformation on B method

Time used: 0.062 (sec)

Given 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\\ B &= 1\\ C &= 0\\ F &= 0 \end{align*}

The above shows that for this ode

\begin{align*} AB''+BB'+CB &= \left (t\right ) \left (0\right ) + \left (1\right ) \left (0\right ) + \left (0\right ) \left (1\right ) \\ &=0 \end{align*}

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

\begin{align*} t v'' +\left ( 1\right ) v' & =0 \end{align*}

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

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

Which is now solved for \(u\). 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}{t}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int \frac {1}{t}d t}\\ &= 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 t\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u t&= \int {0 \,dt} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(t\) gives the final solution

\[ u \left (t \right ) = \frac {c_1}{t} \]

The ode for \(v\) now becomes

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

Which is now solved for \(v\). Since 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 {\frac {c_1}{t}\, dt}\\ v \left (t \right ) &= c_1 \ln \left (t \right ) + c_2 \end{align*}

Replacing \(v \left (t \right )\) above by \(y\), then the solution becomes

\begin{align*} y(t) &= B v\\ &= c_1 \ln \left (t \right )+c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \ln \left (t \right )+c_2 \\ \end{align*}

Solved as second order ode using Kovacic algorithm

Time used: 0.210 (sec)

Writing the ode as

\begin{align*} t y^{\prime \prime }+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 \\ B &= 1\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}{4 t^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

\begin{align*} z''(t) &= \left ( -\frac {1}{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 2.9: 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

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

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 \(b=-{\frac {1}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \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 {1}{4 t^{2}} \end{alignat*}

Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {1}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \end{alignat*}

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

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

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

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(2\)	\(0\)	\(\frac {1}{2}\)	\(\frac {1}{2}\)

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 \(\alpha _\infty ^{-} = {\frac {1}{2}}\) then

\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= {\frac {1}{2}} - \left ( {\frac {1}{2}} \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 {1}{2 t} + (-) \left ( 0 \right ) \\ &= \frac {1}{2 t}\\ &= \frac {1}{2 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 {1}{2 t}\right ) \left (0\right ) + \left ( \left (-\frac {1}{2 t^{2}}\right ) + \left (\frac {1}{2 t}\right )^2 - \left (-\frac {1}{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 {1}{2 t}d t}\\ &= \sqrt {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 {1}{t} \,dt} \\ &= z_1 e^{-\frac {\ln \left (t \right )}{2}} \\ &= z_1 \left (\frac {1}{\sqrt {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 {1}{t} \,dt}}{\left (y_1\right )^2} \,dt \\ &= y_1 \int \frac { e^{-\ln \left (t \right )}}{\left (y_1\right )^2} \,dt \\ &= y_1 \left (\ln \left (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 (\ln \left (t \right )\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 +c_2 \ln \left (t \right ) \\ \end{align*}

Solved as second order ode adjoint method

Time used: 0.347 (sec)

In normal form the ode

\begin{align*} t y^{\prime \prime }+y^{\prime } = 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 {1}{t}\\ q \left (t \right )&=0\\ 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 {\xi \left (t \right )}{t}\right )' + \left (0\right ) &= 0\\ \xi ^{\prime \prime }\left (t \right )+\frac {\xi \left (t \right )}{t^{2}}-\frac {\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

\[ t^{2}(r(r-1))t^{r-2}-t r t^{r-1}+t^{r} = 0 \]

Simplifying gives

\[ r \left (r -1\right )t^{r}-r\,t^{r}+t^{r} = 0 \]

Since \(t^{r}\neq 0\) then dividing throughout by \(t^{r}\) gives

\[ r \left (r -1\right )-r+1 = 0 \]

Or

\[ r^{2}-2 r +1 = 0 \tag {1} \]

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\\ r_2 &= 1 \end{align*}

Since the roots are equal, then the general solution is

\[ \xi = c_1 \xi _1 + c_2 \xi _2 \]

Where \(\xi _1 = t^{r}\) and \(\xi _2 = t^{r} \ln \left (t \right )\). Hence

\[ \xi = c_1 t +c_2 t \ln \left (t \right ) \]

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 {1}{t}-\frac {c_1 +c_2 \ln \left (t \right )+c_2}{c_1 t +c_2 t \ln \left (t \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 {c_2}{t \left (c_1 +c_2 \ln \left (t \right )\right )}\\ p(t) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int -\frac {c_2}{t \left (c_1 +c_2 \ln \left (t \right )\right )}d t}\\ &= \frac {1}{c_1 +c_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 (\frac {y}{c_1 +c_2 \ln \left (t \right )}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} \frac {y}{c_1 +c_2 \ln \left (t \right )}&= \int {0 \,dt} + c_3 \\ &=c_3 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{c_1 +c_2 \ln \left (t \right )}\) gives the final solution

\[ y = \left (c_1 +c_2 \ln \left (t \right )\right ) c_3 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \left (c_1 +c_2 \ln \left (t \right )\right ) c_3 \\ \end{align*}

The constants can be merged to give

\[ y = c_1 +c_2 \ln \left (t \right ) \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 +c_2 \ln \left (t \right ) \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\frac {d}{d t}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 {\frac {d}{d t}y \left (t \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 {\frac {d}{d t}y \left (t \right )}{t}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & t \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\frac {d}{d t}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} \\ {} & {} & t \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 )+\frac {\frac {d}{d s}y \left (s \right )}{t}=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {\frac {d^{2}}{d s^{2}}y \left (s \right )}{t}=0 \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d s^{2}}y \left (s \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (s \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (s \right )\hspace {3pt}\textrm {by}\hspace {3pt} s \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (s \right )=s \\ \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 {C2} s +\mathit {C1} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} s =\ln \left (t \right ) \\ {} & {} & y \left (t \right )=\mathit {C2} \ln \left (t \right )+\mathit {C1} \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 : 0.002 (sec)
Leaf size : 10

dsolve(t*diff(diff(y(t),t),t)+diff(y(t),t) = 0, 
       y(t),singsol=all)
 

\[ y = c_{2} \ln \left (t \right )+c_{1} \]

Mathematica DSolve solution

Solving time : 0.01 (sec)
Leaf size : 13

DSolve[{t*D[y[t],{t,2}]+D[y[t],t]==0,{}}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to c_1 \log (t)+c_2 \]