2.9.4 problem 8
Internal
problem
ID
[18271]
Book
:
Elementary
Differential
Equations.
By
Thornton
C.
Fry.
D
Van
Nostrand.
NY.
First
Edition
(1929)
Section
:
Chapter
VII.
Linear
equations
of
order
higher
than
the
first.
section
63.
Problems
at
page
196
Problem
number
:
8
Date
solved
:
Monday, December 23, 2024 at 09:47:19 PM
CAS
classification
:
[[_2nd_order, _exact, _linear, _nonhomogeneous]]
Solve
\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+y&=\frac {1}{x} \end{align*}
Solved as second order Euler type ode
Time used: 0.298 (sec)
This is second order non-homogeneous ODE. In standard form the ODE is
\[ A y''(x) + B y'(x) + C y(x) = f(x) \]
Where \(A=x^{2}, B=3 x, C=1, f(x)=\frac {1}{x}\). Let the
solution be
\[ y = y_h + y_p \]
Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution
to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from
\[ x^{2} y^{\prime \prime }+3 x y^{\prime }+y = 0 \]
This is Euler second order ODE. Let the
solution be \(y = x^r\), then \(y'=r x^{r-1}\) and \(y''=r(r-1) x^{r-2}\). Substituting these back into the given ODE gives
\[ x^{2}(r(r-1))x^{r-2}+3 x r x^{r-1}+x^{r} = 0 \]
Simplifying gives
\[ r \left (r -1\right )x^{r}+3 r\,x^{r}+x^{r} = 0 \]
Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives
\[ r \left (r -1\right )+3 r+1 = 0 \]
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
\[ y= c_1 y_1 + c_2 y_2 \]
Where \(y_1 = x^{r}\) and \(y_2 = x^{r} \ln \left (x \right )\). Hence
\[ y = \frac {c_1}{x}+\frac {c_2 \ln \left (x \right )}{x} \]
Next, we find the
particular solution to the ODE
\[ x^{2} y^{\prime \prime }+3 x y^{\prime }+y = \frac {1}{x} \]
The particular solution \(y_p\) can be found using either the
method of undetermined coefficients, or the method of variation of parameters. The method
of variation of parameters will be used as it is more general and can be used when the
coefficients of the ODE depend on \(x\) as well. Let
\begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the
two basis solutions (the two linearly independent solutions of the homogeneous
ODE) found earlier when solving the homogeneous ODE as
\begin{align*}
y_1 &= \frac {1}{x} \\
y_2 &= \frac {\ln \left (x \right )}{x} \\
\end{align*}
In the Variation of
parameters \(u_1,u_2\) are found using
\begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\
\tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\
\end{align*}
Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in
front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ \frac {d}{dx}\left (\frac {1}{x}\right ) & \frac {d}{dx}\left (\frac {\ln \left (x \right )}{x}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ -\frac {1}{x^{2}} & -\frac {\ln \left (x \right )}{x^{2}}+\frac {1}{x^{2}} \end {vmatrix} \]
Therefore
\[
W = \left (\frac {1}{x}\right )\left (-\frac {\ln \left (x \right )}{x^{2}}+\frac {1}{x^{2}}\right ) - \left (\frac {\ln \left (x \right )}{x}\right )\left (-\frac {1}{x^{2}}\right )
\]
Which simplifies to
\[
W = \frac {1}{x^{3}}
\]
Which simplifies to
\[
W = \frac {1}{x^{3}}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {\frac {\ln \left (x \right )}{x^{2}}}{\frac {1}{x}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \frac {\ln \left (x \right )}{x}d x
\]
Hence
\[
u_1 = -\frac {\ln \left (x \right )^{2}}{2}
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {\frac {1}{x^{2}}}{\frac {1}{x}}\,dx
\]
Which simplifies to
\[
u_2 = \int \frac {1}{x}d x
\]
Hence
\[
u_2 = \ln \left (x \right )
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = \frac {\ln \left (x \right )^{2}}{2 x}
\]
Therefore the general solution is
\begin{align*} y &= y_h + y_p \\ &= \frac {\ln \left (x \right )^{2}+2 c_2 \ln \left (x \right )+2 c_1}{2 x} \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \frac {\ln \left (x \right )^{2}+2 c_2 \ln \left (x \right )+2 c_1}{2 x} \\
\end{align*}
Solved as second order linear exact ode
Time used: 0.100 (sec)
An ode of the form
\begin{align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end{align*}
is exact if
\begin{align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end{align*}
For the given ode we have
\begin{align*} p(x) &= x^{2}\\ q(x) &= 3 x\\ r(x) &= 1\\ s(x) &= \frac {1}{x} \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 (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end{align*}
Integrating gives
\begin{align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end{align*}
Substituting the above values for \(p,q,r,s\) gives
\begin{align*} x^{2} y^{\prime }+y x&=\int {\frac {1}{x}\, dx} \end{align*}
We now have a first order ode to solve which is
\begin{align*} x^{2} y^{\prime }+y x = \ln \left (x \right )+c_1 \end{align*}
In canonical form a linear first order is
\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {1}{x}\\ p(x) &=\frac {\ln \left (x \right )+c_1}{x^{2}} \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{x}d x}\\ &= x \end{align*}
The ode becomes
\begin{align*}
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {\ln \left (x \right )+c_1}{x^{2}}\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y x\right ) &= \left (x\right ) \left (\frac {\ln \left (x \right )+c_1}{x^{2}}\right ) \\
\mathrm {d} \left (y x\right ) &= \left (\frac {\ln \left (x \right )+c_1}{x}\right )\, \mathrm {d} x \\
\end{align*}
Integrating gives
\begin{align*} y x&= \int {\frac {\ln \left (x \right )+c_1}{x} \,dx} \\ &=\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right ) + c_2 \end{align*}
Dividing throughout by the integrating factor \(x\) gives the final solution
\[ y = \frac {\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right )+c_2}{x} \]
Will add steps
showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \frac {\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right )+c_2}{x} \\
\end{align*}
Solved as second order integrable as is ode
Time used: 0.185 (sec)
Integrating both sides of the ODE w.r.t \(x\) gives
\begin{align*} \int \left (x^{2} y^{\prime \prime }+3 x y^{\prime }+y\right )d x &= \int \frac {1}{x}d x\\ x^{2} y^{\prime }+y x = \ln \left (x \right ) + c_1 \end{align*}
Which is now solved for \(y\). In canonical form a linear first order is
\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {1}{x}\\ p(x) &=\frac {\ln \left (x \right )+c_1}{x^{2}} \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{x}d x}\\ &= x \end{align*}
The ode becomes
\begin{align*}
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {\ln \left (x \right )+c_1}{x^{2}}\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y x\right ) &= \left (x\right ) \left (\frac {\ln \left (x \right )+c_1}{x^{2}}\right ) \\
\mathrm {d} \left (y x\right ) &= \left (\frac {\ln \left (x \right )+c_1}{x}\right )\, \mathrm {d} x \\
\end{align*}
Integrating gives
\begin{align*} y x&= \int {\frac {\ln \left (x \right )+c_1}{x} \,dx} \\ &=\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right ) + c_2 \end{align*}
Dividing throughout by the integrating factor \(x\) gives the final solution
\[ y = \frac {\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right )+c_2}{x} \]
Will add steps
showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \frac {\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right )+c_2}{x} \\
\end{align*}
Solved as second order integrable as is ode (ABC method)
Time used: 0.066 (sec)
Writing the ode as
\[
x^{2} y^{\prime \prime }+3 x y^{\prime }+y = \frac {1}{x}
\]
Integrating both sides of the ODE w.r.t \(x\) gives
\begin{align*} \int \left (x^{2} y^{\prime \prime }+3 x y^{\prime }+y\right )d x &= \int \frac {1}{x}d x\\ x^{2} y^{\prime }+y x = \ln \left (x \right ) +c_1 \end{align*}
Which is now solved for \(y\). In canonical form a linear first order is
\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {1}{x}\\ p(x) &=\frac {\ln \left (x \right )+c_1}{x^{2}} \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{x}d x}\\ &= x \end{align*}
The ode becomes
\begin{align*}
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {\ln \left (x \right )+c_1}{x^{2}}\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y x\right ) &= \left (x\right ) \left (\frac {\ln \left (x \right )+c_1}{x^{2}}\right ) \\
\mathrm {d} \left (y x\right ) &= \left (\frac {\ln \left (x \right )+c_1}{x}\right )\, \mathrm {d} x \\
\end{align*}
Integrating gives
\begin{align*} y x&= \int {\frac {\ln \left (x \right )+c_1}{x} \,dx} \\ &=\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right ) + c_2 \end{align*}
Dividing throughout by the integrating factor \(x\) gives the final solution
\[ y = \frac {\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right )+c_2}{x} \]
Will add steps
showing solving for IC soon.
Solved as second order ode using change of variable on x method 2
Time used: 0.495 (sec)
This is second order non-homogeneous ODE. Let the solution be
\[
y = y_h + y_p
\]
Where \(y_h\) is the solution to
the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the
solution to
\[
x^{2} y^{\prime \prime }+3 x y^{\prime }+y = 0
\]
In normal form the ode
\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+y&=0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=\frac {3}{x}\\ q \left (x \right )&=\frac {1}{x^{2}} \end{align*}
Applying change of variables \(\tau = g \left (x \right )\) 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 (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}
Let \(p_{1} = 0\). Eq (4) simplifies to
\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}
This ode is solved resulting in
\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \frac {3}{x}d x}d x\\ &= \int e^{-3 \ln \left (x \right )} \,dx\\ &= \int \frac {1}{x^{3}}d x\\ &= -\frac {1}{2 x^{2}}\tag {6} \end{align*}
Using (6) to evaluate \(q_{1}\) from (5) gives
\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {1}{x^{2}}}{\frac {1}{x^{6}}}\\ &= x^{4}\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 )+x^{4} y \left (\tau \right )&=0 \\ \end{align*}
But in terms of \(\tau \)
\begin{align*} x^{4}&=\frac {1}{4 \tau ^{2}} \end{align*}
Hence the above ode becomes
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {y \left (\tau \right )}{4 \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 {y \left (\tau \right )}{4 \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 {1}{4 \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 {-1}{4 \tau ^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= -1\\ t &= 4 \tau ^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(\tau ) &= \left ( -\frac {1}{4 \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 2.9: 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 \(t=4 \tau ^{2}\).
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
\[
r = -\frac {1}{4 \tau ^{2}}
\]
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 \(b=-{\frac {1}{4}}\). 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 \(\frac {1}{\tau ^{2}}\) 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 {1}{4 \tau ^{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
\[ r=-\frac {1}{4 \tau ^{2}} \]
| | | | |
| 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)}}{\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 {1}{2 \tau } + (-) \left ( 0 \right ) \\ &= \frac {1}{2 \tau }\\ &= \frac {1}{2 \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 {1}{2 \tau }\right ) \left (0\right ) + \left ( \left (-\frac {1}{2 \tau ^{2}}\right ) + \left (\frac {1}{2 \tau }\right )^2 - \left (-\frac {1}{4 \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 {1}{2 \tau }d \tau }\\ &= \sqrt {\tau } \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 \\
&= \sqrt {\tau } \\
\end{align*}
Which simplifies to
\[
y_1 = \sqrt {\tau }
\]
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 \\
&= \sqrt {\tau }\int \frac {1}{\tau } \,d\tau \\
&= \sqrt {\tau }\left (\ln \left (\tau \right )\right ) \\
\end{align*}
Therefore the solution
is
\begin{align*}
y \left (\tau \right ) &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (\sqrt {\tau }\right ) + c_2 \left (\sqrt {\tau }\left (\ln \left (\tau \right )\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
\[
y = c_1 \sqrt {-\frac {1}{2 x^{2}}}+c_2 \sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{2 x^{2}}\right )
\]
Therefore the
homogeneous solution \(y_h\) is
\[
y_h = c_1 \sqrt {-\frac {1}{2 x^{2}}}+c_2 \sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{2 x^{2}}\right )
\]
The particular solution \(y_p\) can be found using either the method of
undetermined coefficients, or the method of variation of parameters. The method of variation
of parameters will be used as it is more general and can be used when the coefficients of the
ODE depend on \(x\) as well. Let
\begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the
two linearly independent solutions of the homogeneous ODE) found earlier when solving the
homogeneous ODE as
\begin{align*}
y_1 &= \sqrt {-\frac {1}{2 x^{2}}} \\
y_2 &= -\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (2\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right ) \\
\end{align*}
In the Variation of parameters \(u_1,u_2\) are found using
\begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\
\tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\
\end{align*}
Where \(W(x)\) is the
Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given
by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} \sqrt {-\frac {1}{2 x^{2}}} & -\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (2\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right ) \\ \frac {d}{dx}\left (\sqrt {-\frac {1}{2 x^{2}}}\right ) & \frac {d}{dx}\left (-\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (2\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} \sqrt {-\frac {1}{2 x^{2}}} & -\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (2\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right ) \\ \frac {1}{2 \sqrt {-\frac {1}{2 x^{2}}}\, x^{3}} & -\frac {\ln \left (2\right )}{2 \sqrt {-\frac {1}{2 x^{2}}}\, x^{3}}+\frac {\ln \left (-\frac {1}{x^{2}}\right )}{2 \sqrt {-\frac {1}{2 x^{2}}}\, x^{3}}-\frac {2 \sqrt {-\frac {1}{2 x^{2}}}}{x} \end {vmatrix} \]
Therefore
\[
W = \left (\sqrt {-\frac {1}{2 x^{2}}}\right )\left (-\frac {\ln \left (2\right )}{2 \sqrt {-\frac {1}{2 x^{2}}}\, x^{3}}+\frac {\ln \left (-\frac {1}{x^{2}}\right )}{2 \sqrt {-\frac {1}{2 x^{2}}}\, x^{3}}-\frac {2 \sqrt {-\frac {1}{2 x^{2}}}}{x}\right ) - \left (-\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (2\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )\right )\left (\frac {1}{2 \sqrt {-\frac {1}{2 x^{2}}}\, x^{3}}\right )
\]
Which simplifies to
\[
W = \frac {1}{x^{3}}
\]
Which simplifies to
\[
W = \frac {1}{x^{3}}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {\frac {-\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (2\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{x}}{\frac {1}{x}}\,dx
\]
Which simplifies to
\[
u_1 = - \int -\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \left (\ln \left (2\right )-\ln \left (-\frac {1}{x^{2}}\right )\right )}{2}d x
\]
Hence
\[
u_1 = \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (2\right ) \ln \left (x \right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (-\frac {1}{x^{2}}\right )^{2}}{8}
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {\frac {\sqrt {-\frac {1}{2 x^{2}}}}{x}}{\frac {1}{x}}\,dx
\]
Which
simplifies to
\[
u_2 = \int \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}}{2}d x
\]
Hence
\[
u_2 = \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right )}{2}
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = \left (\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (2\right ) \ln \left (x \right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (-\frac {1}{x^{2}}\right )^{2}}{8}\right ) \sqrt {-\frac {1}{2 x^{2}}}+\frac {\left (-\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (2\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )\right ) \sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right )}{2}
\]
Which
simplifies to
\[
y_p(x) = -\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (\ln \left (-\frac {1}{x^{2}}\right )+4 \ln \left (x \right )\right )}{8 x}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_1 \sqrt {-\frac {1}{2 x^{2}}}+c_2 \sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{2 x^{2}}\right )\right ) + \left (-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (\ln \left (-\frac {1}{x^{2}}\right )+4 \ln \left (x \right )\right )}{8 x}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= -\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (\ln \left (-\frac {1}{x^{2}}\right )+4 \ln \left (x \right )\right )}{8 x}+c_1 \sqrt {-\frac {1}{2 x^{2}}}+c_2 \sqrt {-\frac {1}{2 x^{2}}}\, \ln \left (-\frac {1}{2 x^{2}}\right ) \\
\end{align*}
Solved as second order ode using change of variable on y method 2
Time used: 0.318 (sec)
This is second order non-homogeneous ODE. In standard form the ODE is
\[
A y''(x) + B y'(x) + C y(x) = f(x)
\]
Where \(A=x^{2}, B=3 x, C=1, f(x)=\frac {1}{x}\). Let the
solution be
\[
y = y_h + y_p
\]
Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular
solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from
\[
x^{2} y^{\prime \prime }+3 x y^{\prime }+y = 0
\]
In normal form the ode
\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+y&=0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=\frac {3}{x}\\ q \left (x \right )&=\frac {1}{x^{2}} \end{align*}
Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where
the dependent variables is \(v \left (x \right )\) and not \(y\).
\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end{align*}
Let the coefficient of \(v \left (x \right )\) above be zero. Hence
\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end{align*}
Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives
\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {3 n}{x^{2}}+\frac {1}{x^{2}}&=0 \tag {5} \end{align*}
Solving (5) for \(n\) gives
\begin{align*} n&=-1 \tag {6} \end{align*}
Substituting this value in (3) gives
\begin{align*} v^{\prime \prime }\left (x \right )+\frac {v^{\prime }\left (x \right )}{x}&=0 \\ v^{\prime \prime }\left (x \right )+\frac {v^{\prime }\left (x \right )}{x}&=0 \tag {7} \\ \end{align*}
Using the substitution
\begin{align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end{align*}
Then (7) becomes
\begin{align*} u^{\prime }\left (x \right )+\frac {u \left (x \right )}{x} = 0 \tag {8} \\ \end{align*}
The above is now solved for \(u \left (x \right )\). In canonical form a linear first order is
\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {1}{x}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{x}d x}\\ &= x \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u x\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} u x&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}
Dividing throughout by the integrating factor \(x\) gives the final solution
\[ u \left (x \right ) = \frac {c_3}{x} \]
Now that \(u \left (x \right )\) is known,
then
\begin{align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_4\\ &= c_3 \ln \left (x \right )+c_4 \end{align*}
Hence
\begin{align*} y&= v \left (x \right ) x^{n}\\ &= \frac {c_3 \ln \left (x \right )+c_4}{x}\\ &= \frac {c_3 \ln \left (x \right )+c_4}{x}\\ \end{align*}
Now the particular solution to this ODE is found
\[
x^{2} y^{\prime \prime }+3 x y^{\prime }+y = \frac {1}{x}
\]
The particular solution \(y_p\) can
be found using either the method of undetermined coefficients, or the method
of variation of parameters. The method of variation of parameters will be used
as it is more general and can be used when the coefficients of the ODE depend
on \(x\) as well. Let
\begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the
two linearly independent solutions of the homogeneous ODE) found earlier when
solving the homogeneous ODE as
\begin{align*}
y_1 &= \frac {1}{x} \\
y_2 &= \frac {\ln \left (x \right )}{x} \\
\end{align*}
In the Variation of parameters \(u_1,u_2\) are found using
\begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\
\tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\
\end{align*}
Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The
Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ \frac {d}{dx}\left (\frac {1}{x}\right ) & \frac {d}{dx}\left (\frac {\ln \left (x \right )}{x}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ -\frac {1}{x^{2}} & -\frac {\ln \left (x \right )}{x^{2}}+\frac {1}{x^{2}} \end {vmatrix} \]
Therefore
\[
W = \left (\frac {1}{x}\right )\left (-\frac {\ln \left (x \right )}{x^{2}}+\frac {1}{x^{2}}\right ) - \left (\frac {\ln \left (x \right )}{x}\right )\left (-\frac {1}{x^{2}}\right )
\]
Which simplifies to
\[
W = \frac {1}{x^{3}}
\]
Which
simplifies to
\[
W = \frac {1}{x^{3}}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {\frac {\ln \left (x \right )}{x^{2}}}{\frac {1}{x}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \frac {\ln \left (x \right )}{x}d x
\]
Hence
\[
u_1 = -\frac {\ln \left (x \right )^{2}}{2}
\]
And Eq. (3)
becomes
\[
u_2 = \int \frac {\frac {1}{x^{2}}}{\frac {1}{x}}\,dx
\]
Which simplifies to
\[
u_2 = \int \frac {1}{x}d x
\]
Hence
\[
u_2 = \ln \left (x \right )
\]
Therefore the particular solution, from equation
(1) is
\[
y_p(x) = \frac {\ln \left (x \right )^{2}}{2 x}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (\frac {c_3 \ln \left (x \right )+c_4}{x}\right ) + \left (\frac {\ln \left (x \right )^{2}}{2 x}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= \frac {c_3 \ln \left (x \right )+c_4}{x}+\frac {\ln \left (x \right )^{2}}{2 x} \\
\end{align*}
Solved as second order ode using Kovacic algorithm
Time used: 0.293 (sec)
Writing the ode as
\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+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 &= x^{2} \\ B &= 3 x\tag {3} \\ C &= 1 \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}
Then (2) becomes
\begin{align*} z''(x) = r z(x)\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 x^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= -1\\ t &= 4 x^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(x) &= \left ( -\frac {1}{4 x^{2}}\right ) z(x)\tag {7} \end{align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation
\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \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 2.10: 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 \(t=4 x^{2}\).
There is a pole at \(x=0\) 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
\[
r = -\frac {1}{4 x^{2}}
\]
For the pole at \(x=0\) let \(b\)
be the coefficient of \(\frac {1}{ x^{2}}\) 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 \(\frac {1}{x^{2}}\) 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 x^{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
\[ r=-\frac {1}{4 x^{2}} \]
| | | | |
| 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)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}
The above gives
\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {1}{2 x} + (-) \left ( 0 \right ) \\ &= \frac {1}{2 x}\\ &= \frac {1}{2 x} \end{align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree
\(d=0\) to solve the ode. The polynomial \(p(x)\) 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(x) &= 1\tag {2A} \end{align*}
Substituting the above in eq. (1A) gives
\begin{align*} \left (0\right ) + 2 \left (\frac {1}{2 x}\right ) \left (0\right ) + \left ( \left (-\frac {1}{2 x^{2}}\right ) + \left (\frac {1}{2 x}\right )^2 - \left (-\frac {1}{4 x^{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(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {1}{2 x}d x}\\ &= \sqrt {x} \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} \,dx} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {3 x}{x^{2}} \,dx} \\
&= z_1 e^{-\frac {3 \ln \left (x \right )}{2}} \\
&= z_1 \left (\frac {1}{x^{{3}/{2}}}\right ) \\
\end{align*}
Which simplifies to
\[
y_1 = \frac {1}{x}
\]
The second
solution \(y_2\) to the original ode is found using reduction of order
\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \]
Substituting gives
\begin{align*}
y_2 &= y_1 \int \frac { e^{\int -\frac {3 x}{x^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{-3 \ln \left (x \right )}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (\ln \left (x \right )\right ) \\
\end{align*}
Therefore
the solution is
\begin{align*}
y &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (\frac {1}{x}\right ) + c_2 \left (\frac {1}{x}\left (\ln \left (x \right )\right )\right ) \\
\end{align*}
This is second order nonhomogeneous ODE. Let the solution be
\[
y = y_h + y_p
\]
Where \(y_h\) is the solution to
the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the nonhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the
solution to
\[
x^{2} y^{\prime \prime }+3 x y^{\prime }+y = 0
\]
The homogeneous solution is found using the Kovacic algorithm which results in
\[
y_h = \frac {c_1}{x}+\frac {c_2 \ln \left (x \right )}{x}
\]
The particular solution \(y_p\) can be found using either the method of undetermined coefficients,
or the method of variation of parameters. The method of variation of parameters will be
used as it is more general and can be used when the coefficients of the ODE depend
on \(x\) as well. Let
\begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the
two linearly independent solutions of the homogeneous ODE) found earlier when
solving the homogeneous ODE as
\begin{align*}
y_1 &= \frac {1}{x} \\
y_2 &= \frac {\ln \left (x \right )}{x} \\
\end{align*}
In the Variation of parameters \(u_1,u_2\) are found using
\begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\
\tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\
\end{align*}
Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The
Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ \frac {d}{dx}\left (\frac {1}{x}\right ) & \frac {d}{dx}\left (\frac {\ln \left (x \right )}{x}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ -\frac {1}{x^{2}} & -\frac {\ln \left (x \right )}{x^{2}}+\frac {1}{x^{2}} \end {vmatrix} \]
Therefore
\[
W = \left (\frac {1}{x}\right )\left (-\frac {\ln \left (x \right )}{x^{2}}+\frac {1}{x^{2}}\right ) - \left (\frac {\ln \left (x \right )}{x}\right )\left (-\frac {1}{x^{2}}\right )
\]
Which simplifies to
\[
W = \frac {1}{x^{3}}
\]
Which
simplifies to
\[
W = \frac {1}{x^{3}}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {\frac {\ln \left (x \right )}{x^{2}}}{\frac {1}{x}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \frac {\ln \left (x \right )}{x}d x
\]
Hence
\[
u_1 = -\frac {\ln \left (x \right )^{2}}{2}
\]
And Eq. (3)
becomes
\[
u_2 = \int \frac {\frac {1}{x^{2}}}{\frac {1}{x}}\,dx
\]
Which simplifies to
\[
u_2 = \int \frac {1}{x}d x
\]
Hence
\[
u_2 = \ln \left (x \right )
\]
Therefore the particular solution, from equation
(1) is
\[
y_p(x) = \frac {\ln \left (x \right )^{2}}{2 x}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (\frac {c_1}{x}+\frac {c_2 \ln \left (x \right )}{x}\right ) + \left (\frac {\ln \left (x \right )^{2}}{2 x}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= \frac {c_1}{x}+\frac {c_2 \ln \left (x \right )}{x}+\frac {\ln \left (x \right )^{2}}{2 x} \\
\end{align*}
Solved as second order ode adjoint method
Time used: 0.344 (sec)
In normal form the ode
\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+y = \frac {1}{x} \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=\frac {3}{x}\\ q \left (x \right )&=\frac {1}{x^{2}}\\ r \left (x \right )&=\frac {1}{x^{3}} \end{align*}
The Lagrange adjoint ode is given by
\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {3 \xi \left (x \right )}{x}\right )' + \left (\frac {\xi \left (x \right )}{x^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )+\frac {4 \xi \left (x \right )}{x^{2}}-\frac {3 \xi ^{\prime }\left (x \right )}{x}&= 0 \end{align*}
Which is solved for \(\xi (x)\). This is Euler second order ODE. Let the solution be \(\xi = x^r\), then \(\xi '=r x^{r-1}\) and \(\xi ''=r(r-1) x^{r-2}\).
Substituting these back into the given ODE gives
\[ x^{2}(r(r-1))x^{r-2}-3 x r x^{r-1}+4 x^{r} = 0 \]
Simplifying gives
\[ r \left (r -1\right )x^{r}-3 r\,x^{r}+4 x^{r} = 0 \]
Since \(x^{r}\neq 0\) then dividing
throughout by \(x^{r}\) gives
\[ r \left (r -1\right )-3 r+4 = 0 \]
Or
\[ r^{2}-4 r +4 = 0 \tag {1} \]
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\\ r_2 &= 2 \end{align*}
Since the roots are equal, then the general solution is
\[ \xi = c_1 \xi _1 + c_2 \xi _2 \]
Where \(\xi _1 = x^{r}\) and \(\xi _2 = x^{r} \ln \left (x \right )\). Hence
\[ \xi = c_1 \,x^{2}+c_2 \,x^{2} \ln \left (x \right ) \]
Will add steps
showing solving for IC soon.
The original ode now reduces to first order ode
\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}
Or
\begin{align*} y^{\prime }+y \left (\frac {3}{x}-\frac {2 c_1 x +2 c_2 x \ln \left (x \right )+c_2 x}{c_1 \,x^{2}+c_2 \,x^{2} \ln \left (x \right )}\right )&=\frac {\frac {c_2 \ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right )}{c_1 \,x^{2}+c_2 \,x^{2} \ln \left (x \right )} \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(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=-\frac {-c_2 \ln \left (x \right )-c_1 +c_2}{x \left (c_2 \ln \left (x \right )+c_1 \right )}\\ p(x) &=\frac {\ln \left (x \right ) \left (c_2 \ln \left (x \right )+2 c_1 \right )}{2 x^{2} \left (c_2 \ln \left (x \right )+c_1 \right )} \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {-c_2 \ln \left (x \right )-c_1 +c_2}{x \left (c_2 \ln \left (x \right )+c_1 \right )}d x}\\ &= \frac {x}{c_2 \ln \left (x \right )+c_1} \end{align*}
The ode becomes
\begin{align*}
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {\ln \left (x \right ) \left (c_2 \ln \left (x \right )+2 c_1 \right )}{2 x^{2} \left (c_2 \ln \left (x \right )+c_1 \right )}\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y x}{c_2 \ln \left (x \right )+c_1}\right ) &= \left (\frac {x}{c_2 \ln \left (x \right )+c_1}\right ) \left (\frac {\ln \left (x \right ) \left (c_2 \ln \left (x \right )+2 c_1 \right )}{2 x^{2} \left (c_2 \ln \left (x \right )+c_1 \right )}\right ) \\
\mathrm {d} \left (\frac {y x}{c_2 \ln \left (x \right )+c_1}\right ) &= \left (\frac {\ln \left (x \right ) \left (c_2 \ln \left (x \right )+2 c_1 \right )}{2 x \left (c_2 \ln \left (x \right )+c_1 \right )^{2}}\right )\, \mathrm {d} x \\
\end{align*}
Integrating gives
\begin{align*} \frac {y x}{c_2 \ln \left (x \right )+c_1}&= \int {\frac {\ln \left (x \right ) \left (c_2 \ln \left (x \right )+2 c_1 \right )}{2 x \left (c_2 \ln \left (x \right )+c_1 \right )^{2}} \,dx} \\ &=\frac {\ln \left (x \right )}{2 c_2}+\frac {c_1^{2}}{2 c_2^{2} \left (c_2 \ln \left (x \right )+c_1 \right )} + c_3 \end{align*}
Dividing throughout by the integrating factor \(\frac {x}{c_2 \ln \left (x \right )+c_1}\) gives the final solution
\[ y = \frac {\ln \left (x \right )^{2} c_2^{2}+c_2 \left (2 c_2^{2} c_3 +c_1 \right ) \ln \left (x \right )+c_1 \left (2 c_2^{2} c_3 +c_1 \right )}{2 x \,c_2^{2}} \]
Hence, the solution
found using Lagrange adjoint equation method is
\begin{align*}
y &= \frac {\ln \left (x \right )^{2} c_2^{2}+c_2 \left (2 c_2^{2} c_3 +c_1 \right ) \ln \left (x \right )+c_1 \left (2 c_2^{2} c_3 +c_1 \right )}{2 x \,c_2^{2}} \\
\end{align*}
The constants can be merged to give
\[
y = \frac {\ln \left (x \right )^{2} c_2^{2}+c_2 \left (2 c_2^{2}+c_1 \right ) \ln \left (x \right )+c_1 \left (2 c_2^{2}+c_1 \right )}{2 x \,c_2^{2}}
\]
Will
add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \frac {\ln \left (x \right )^{2} c_2^{2}+c_2 \left (2 c_2^{2}+c_1 \right ) \ln \left (x \right )+c_1 \left (2 c_2^{2}+c_1 \right )}{2 x \,c_2^{2}} \\
\end{align*}
Maple step by step solution
Maple trace
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
<- high order exact linear fully integrable successful`
Maple dsolve solution
Solving time : 0.003 (sec)
Leaf size : 20
dsolve(x^2*diff(diff(y(x),x),x)+3*x*diff(y(x),x)+y(x) = 1/x,
y(x),singsol=all)
\[
y = \frac {\frac {\ln \left (x \right )^{2}}{2}+c_1 \ln \left (x \right )+c_2}{x}
\]
Mathematica DSolve solution
Solving time : 0.024 (sec)
Leaf size : 27
DSolve[{x^2*D[y[x],{x,2}]+3*x*D[y[x],x]+y[x]==1/x,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to \frac {\log ^2(x)+2 c_2 \log (x)+2 c_1}{2 x}
\]