2.16.1 Problem (a)

Internal problem ID [21002]
Book : Ordinary Differential Equations. By Wolfgang Walter. Graduate texts in Mathematics. Springer. NY. QA372.W224 1998
Section : Chapter V. Complex Linear Systems. Excercise XII at page 244
Problem number : (a)
Date solved : Saturday, November 29, 2025 at 01:36:21 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solved as second order linear exact ode

Time used: 0.331 (sec)

Solve

\begin{align*} z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u&=0 \\ \end{align*}

An ode of the form

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

is exact if

\begin{align*} p''(z) - q'(z) + r(z) &= 0 \tag {1} \end{align*}

For the given ode we have

\begin{align*} p(x) &= z^{2}\\ q(x) &= 3 z +1\\ r(x) &= 1\\ s(x) &= 0 \end{align*}

Hence

\begin{align*} p''(x) &= 2\\ q'(x) &= 3 \end{align*}

Therefore (1) becomes

\begin{align*} 2- \left (3\right ) + \left (1\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 (z \right ) u^{\prime }+\left (q \left (z \right )-p^{\prime }\left (z \right )\right ) u\right )' &= s(x) \end{align*}

Integrating gives

\begin{align*} p \left (z \right ) u^{\prime }+\left (q \left (z \right )-p^{\prime }\left (z \right )\right ) u&=\int {s \left (z \right )\, dz} \end{align*}

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

\begin{align*} z^{2} u^{\prime }+\left (z +1\right ) u&=c_1 \end{align*}

We now have a first order ode to solve which is

\begin{align*} z^{2} u^{\prime }+\left (z +1\right ) u = c_1 \end{align*}

Solve In canonical form a linear first order is

\begin{align*} u^{\prime } + q(z)u &= p(z) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(z) &=-\frac {-z -1}{z^{2}}\\ p(z) &=\frac {c_1}{z^{2}} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dz}}\\ &= {\mathrm e}^{\int -\frac {-z -1}{z^{2}}d z}\\ &= z \,{\mathrm e}^{-\frac {1}{z}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \left (\mu \right ) \left (\frac {c_1}{z^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (z \,{\mathrm e}^{-\frac {1}{z}}\right ) \left (\frac {c_1}{z^{2}}\right ) \\ \mathrm {d} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z}\right )\, \mathrm {d} z \\ \end{align*}

Integrating gives

\begin{align*} u z \,{\mathrm e}^{-\frac {1}{z}}&= \int {\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z} \,dz} \\ &=c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right ) + c_2 \end{align*}

Dividing throughout by the integrating factor \(z \,{\mathrm e}^{-\frac {1}{z}}\) gives the final solution

\[ u = \frac {{\mathrm e}^{\frac {1}{z}} \left (c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right )+c_2 \right )}{z} \]

Summary of solutions found

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

Solved as second order integrable as is ode

Time used: 0.122 (sec)

Solve

\begin{align*} z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u&=0 \\ \end{align*}

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

\begin{align*} \int \left (z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u\right )d z &= 0 \\ z^{2} u^{\prime }+\left (z +1\right ) u = c_1 \end{align*}

Which is now solved for \(u\). Solve In canonical form a linear first order is

\begin{align*} u^{\prime } + q(z)u &= p(z) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(z) &=-\frac {-z -1}{z^{2}}\\ p(z) &=\frac {c_1}{z^{2}} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dz}}\\ &= {\mathrm e}^{\int -\frac {-z -1}{z^{2}}d z}\\ &= z \,{\mathrm e}^{-\frac {1}{z}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \left (\mu \right ) \left (\frac {c_1}{z^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (z \,{\mathrm e}^{-\frac {1}{z}}\right ) \left (\frac {c_1}{z^{2}}\right ) \\ \mathrm {d} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z}\right )\, \mathrm {d} z \\ \end{align*}

Integrating gives

\begin{align*} u z \,{\mathrm e}^{-\frac {1}{z}}&= \int {\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z} \,dz} \\ &=c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right ) + c_2 \end{align*}

Dividing throughout by the integrating factor \(z \,{\mathrm e}^{-\frac {1}{z}}\) gives the final solution

\[ u = \frac {{\mathrm e}^{\frac {1}{z}} \left (c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right )+c_2 \right )}{z} \]

Summary of solutions found

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

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

Time used: 0.088 (sec)

Solve

\begin{align*} z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u&=0 \\ \end{align*}

Writing the ode as

\[ z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u = 0 \]

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

\begin{align*} \int \left (z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u\right )d z &= 0 \\ z^{2} u^{\prime }+\left (z +1\right ) u = c_1 \end{align*}

Which is now solved for \(u\). Solve In canonical form a linear first order is

\begin{align*} u^{\prime } + q(z)u &= p(z) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(z) &=-\frac {-z -1}{z^{2}}\\ p(z) &=\frac {c_1}{z^{2}} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dz}}\\ &= {\mathrm e}^{\int -\frac {-z -1}{z^{2}}d z}\\ &= z \,{\mathrm e}^{-\frac {1}{z}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \left (\mu \right ) \left (\frac {c_1}{z^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (z \,{\mathrm e}^{-\frac {1}{z}}\right ) \left (\frac {c_1}{z^{2}}\right ) \\ \mathrm {d} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z}\right )\, \mathrm {d} z \\ \end{align*}

Integrating gives

\begin{align*} u z \,{\mathrm e}^{-\frac {1}{z}}&= \int {\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z} \,dz} \\ &=c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right ) + c_2 \end{align*}

Dividing throughout by the integrating factor \(z \,{\mathrm e}^{-\frac {1}{z}}\) gives the final solution

\[ u = \frac {{\mathrm e}^{\frac {1}{z}} \left (c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right )+c_2 \right )}{z} \]

Solve In canonical form a linear first order is

\begin{align*} u^{\prime } + q(z)u &= p(z) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(z) &=-\frac {-z -1}{z^{2}}\\ p(z) &=\frac {c_1}{z^{2}} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dz}}\\ &= {\mathrm e}^{\int -\frac {-z -1}{z^{2}}d z}\\ &= z \,{\mathrm e}^{-\frac {1}{z}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}}\left ( \mu u\right ) &= \left (\mu \right ) \left (\frac {c_1}{z^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}z}} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (z \,{\mathrm e}^{-\frac {1}{z}}\right ) \left (\frac {c_1}{z^{2}}\right ) \\ \mathrm {d} \left (u z \,{\mathrm e}^{-\frac {1}{z}}\right ) &= \left (\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z}\right )\, \mathrm {d} z \\ \end{align*}

Integrating gives

\begin{align*} u z \,{\mathrm e}^{-\frac {1}{z}}&= \int {\frac {c_1 \,{\mathrm e}^{-\frac {1}{z}}}{z} \,dz} \\ &=c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right ) + c_2 \end{align*}

Dividing throughout by the integrating factor \(z \,{\mathrm e}^{-\frac {1}{z}}\) gives the final solution

\[ u = \frac {{\mathrm e}^{\frac {1}{z}} \left (c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right )+c_2 \right )}{z} \]

Solved as second order ode using Kovacic algorithm

Time used: 0.321 (sec)

Solve

\begin{align*} z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u&=0 \\ \end{align*}

Writing the ode as

\begin{align*} z^{2} u^{\prime \prime }+\left (3 z +1\right ) u^{\prime }+u &= 0 \tag {1} \\ A u^{\prime \prime } + B u^{\prime } + C u &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= z^{2} \\ B &= 3 z +1\tag {3} \\ C &= 1 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(z) &= u e^{\int \frac {B}{2 A} \,dz} \end{align*}

Then (2) becomes

\begin{align*} z''(z) = r z(z)\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 {-z^{2}+2 z +1}{4 z^{4}}\tag {6} \end{align*}

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

\begin{align*} z''(z) &= \left ( \frac {-z^{2}+2 z +1}{4 z^{4}}\right ) z(z)\tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(z)\) then \(u\) is found using the inverse transformation

\begin{align*} u &= z \left (z \right ) e^{-\int \frac {B}{2 A} \,dz} \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 2.15: 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 - 2 \\ &= 2 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 z^{4}\). There is a pole at \(z=0\) of order \(4\). 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. 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 \(\frac {1}{(z-c)^i}\) 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}{ (z-c)^i} \tag {1B} \end{align*}

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

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

There is pole in \(r\) at \(z= 0\) 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}{2 z^{2}}+\frac {1}{2 z}-\frac {1}{2}+\frac {z}{2}-\frac {3 z^{2}}{4}+\frac {5 z^{3}}{4} + \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}{2 z^{2}} \end{equation}

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

\[ a = {\frac {1}{2}} \]

Now we need to find \(b\). let \(b\) be the coefficient of the term \(\frac {1}{(z-c)^{v+1}}\) 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 \(\frac {1}{z^{3}}\). 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 \(\frac {1}{2}\). Therefore

\begin{align*} b &= \left ({\frac {1}{2}}\right )-(0)\\ &= {\frac {1}{2}} \end{align*}

Hence

\begin{alignat*}{3} [\sqrt r]_c &= \frac {1}{2 z^{2}} \\ \alpha _c^{+} &= \frac {1}{2} \left ( \frac {b}{a} + v \right ) &&= \frac {1}{2} \left ( \frac {{\frac {1}{2}}}{{\frac {1}{2}}} + 2 \right ) &&={\frac {3}{2}}\\ \alpha _c^{-} &= \frac {1}{2} \left (- \frac {b}{a} + v \right ) &&= \frac {1}{2} \left (- \frac {{\frac {1}{2}}}{{\frac {1}{2}}} + 2 \right )&&={\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 \(\frac {1}{z^{2}}\) 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 {-z^{2}+2 z +1}{4 z^{4}} \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

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

pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)
\(0\)	\(4\)	\(\frac {1}{2 z^{2}}\)	\(\frac {3}{2}\)	\(\frac {1}{2}\)

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)}}{z-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

The above gives

\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{z- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= -\frac {1}{2 z^{2}}+\frac {1}{2 z} + (-) \left ( 0 \right ) \\ &= -\frac {1}{2 z^{2}}+\frac {1}{2 z}\\ &= \frac {z -1}{2 z^{2}} \end{align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(z)\) of degree \(d=0\) to solve the ode. The polynomial \(p(z)\) 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(z) &= 1\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

\begin{align*} \left (0\right ) + 2 \left (-\frac {1}{2 z^{2}}+\frac {1}{2 z}\right ) \left (0\right ) + \left ( \left (\frac {1}{z^{3}}-\frac {1}{2 z^{2}}\right ) + \left (-\frac {1}{2 z^{2}}+\frac {1}{2 z}\right )^2 - \left (\frac {-z^{2}+2 z +1}{4 z^{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(z) &= p e^{ \int \omega \,dz} \\ &= {\mathrm e}^{\int \left (-\frac {1}{2 z^{2}}+\frac {1}{2 z}\right )d z}\\ &= \sqrt {z}\, {\mathrm e}^{\frac {1}{2 z}} \end{align*}

The first solution to the original ode in \(u\) is found from

\begin{align*} u_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dz} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {3 z +1}{z^{2}} \,dz} \\ &= z_1 e^{\frac {1}{2 z}-\frac {3 \ln \left (z \right )}{2}} \\ &= z_1 \left (\frac {{\mathrm e}^{\frac {1}{2 z}}}{z^{{3}/{2}}}\right ) \\ \end{align*}

Which simplifies to

\[ u_1 = \frac {{\mathrm e}^{\frac {1}{z}}}{z} \]

The second solution \(u_2\) to the original ode is found using reduction of order

\[ u_2 = u_1 \int \frac { e^{\int -\frac {B}{A} \,dz}}{u_1^2} \,dz \]

Substituting gives

\begin{align*} u_2 &= u_1 \int \frac { e^{\int -\frac {3 z +1}{z^{2}} \,dz}}{\left (u_1\right )^2} \,dz \\ &= u_1 \int \frac { e^{\frac {1}{z}-3 \ln \left (z \right )}}{\left (u_1\right )^2} \,dz \\ &= u_1 \left ({\mathrm e}^{-3 \ln \left (\frac {1}{z}\right )-3 \ln \left (z \right )} \operatorname {Ei}_{1}\left (\frac {1}{z}\right )\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} u &= c_1 u_1 + c_2 u_2 \\ &= c_1 \left (\frac {{\mathrm e}^{\frac {1}{z}}}{z}\right ) + c_2 \left (\frac {{\mathrm e}^{\frac {1}{z}}}{z}\left ({\mathrm e}^{-3 \ln \left (\frac {1}{z}\right )-3 \ln \left (z \right )} \operatorname {Ei}_{1}\left (\frac {1}{z}\right )\right )\right ) \\ \end{align*}

Summary of solutions found

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

✓ Maple. Time used: 0.000 (sec). Leaf size: 21

ode:=z^2*diff(diff(u(z),z),z)+(3*z+1)*diff(u(z),z)+u(z) = 0; 
dsolve(ode,u(z), singsol=all);
 

\[ u = \frac {\left (c_1 \,\operatorname {Ei}_{1}\left (\frac {1}{z}\right )+c_2 \right ) {\mathrm e}^{\frac {1}{z}}}{z} \]

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)] 
<- linear_1 successful
 

✓ Mathematica. Time used: 0.051 (sec). Leaf size: 27

ode=z^2*D[u[z],{z,2}]+(3*z+1)*D[u[z],z]+u[z]==0; 
ic={}; 
DSolve[{ode,ic},u[z],z,IncludeSingularSolutions->True]
 

\begin{align*} u(z)&\to \frac {e^{\frac {1}{z}} \left (c_1-c_2 \operatorname {ExpIntegralEi}\left (-\frac {1}{z}\right )\right )}{z} \end{align*}

✗ Sympy

from sympy import * 
z = symbols("z") 
u = Function("u") 
ode = Eq(z**2*Derivative(u(z), (z, 2)) + (3*z + 1)*Derivative(u(z), z) + u(z),0) 
ics = {} 
dsolve(ode,func=u(z),ics=ics)
 

NotImplementedError : The given ODE Derivative(u(z), z) - (-z**2*Derivative(u(z), (z, 2)) - u(z))/(3*z + 1) cannot be solved by the factorable group method