2.212 Internal
problem
ID
[9772]Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948 Section
:
Chapter
1,
Additional
non-linear
first
order Problem
number
:
788Date
solved
:
Friday, October 11, 2024 at 11:41:05 AMCAS
classification
:
[_Bernoulli]
Solve
\begin{align*} y^{\prime }&=-\frac {y \left (\ln \left (x -1\right )+\coth \left (x +1\right ) x -\coth \left (x +1\right ) x^{2} y\right )}{x \ln \left (x -1\right )} \end{align*}
2.212.1 Time used: 1.044 (sec)
In canonical form, the ODE is
\begin{align*} y' &= F(x,y)\\ &= \frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )}{x \ln \left (x -1\right )} \end{align*}
This is a Bernoulli ODE.
\[ y' = \left (\frac {-\coth \left (x +1\right ) x -\ln \left (x -1\right )}{x \ln \left (x -1\right )}\right ) + \left (\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )}\right )y^{2} \tag {1} \]
The standard Bernoulli ODE has the form
\[ y' = f_0(x)y+f_1(x)y^n \tag {2} \]
Comparing this to (1)
shows that
\begin{align*} f_0 &=\frac {-\coth \left (x +1\right ) x -\ln \left (x -1\right )}{x \ln \left (x -1\right )}\\ f_1 &=\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )} \end{align*}
The first step is to divide the above equation by \(y^n \) which gives
\[ \frac {y'}{y^n} = f_0(x) y^{1-n} +f_1(x) \tag {3} \]
The next step is use the
substitution \(v = y^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be
easily solved using an integrating factor. Backsubstitution then gives the solution \(y(x)\) which is
what we want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows
that
\begin{align*} f_0(x)&=\frac {-\coth \left (x +1\right ) x -\ln \left (x -1\right )}{x \ln \left (x -1\right )}\\ f_1(x)&=\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )}\\ n &=2 \end{align*}
Dividing both sides of ODE (1) by \(y^n=y^{2}\) gives
\begin{align*} y'\frac {1}{y^{2}} &= \frac {-\coth \left (x +1\right ) x -\ln \left (x -1\right )}{x \ln \left (x -1\right ) y} +\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )} \tag {4} \end{align*}
Let
\begin{align*} v &= y^{1-n} \\ &= \frac {1}{y} \tag {5} \end{align*}
Taking derivative of equation (5) w.r.t \(x\) gives
\begin{align*} v' &= -\frac {1}{y^{2}}y' \tag {6} \end{align*}
Substituting equations (5) and (6) into equation (4) gives
\begin{align*} -v^{\prime }\left (x \right )&= \frac {\left (-\coth \left (x +1\right ) x -\ln \left (x -1\right )\right ) v \left (x \right )}{x \ln \left (x -1\right )}+\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )}\\ v' &= -\frac {\left (-\coth \left (x +1\right ) x -\ln \left (x -1\right )\right ) v}{x \ln \left (x -1\right )}-\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )} \tag {7} \end{align*}
The above now is a linear ODE in \(v \left (x \right )\) which is now solved.
In canonical form a linear first order is
\begin{align*} v^{\prime }\left (x \right ) + q(x)v \left (x \right ) &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=-\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}\\ p(x) &=-\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )} \end{align*}
The integrating factor \(\mu \) is
\[ \mu = {\mathrm e}^{\int -\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x} \]
Therefore the solution is
\[ v \left (x \right ) = \left (\int -\frac {\coth \left (x +1\right ) x \,{\mathrm e}^{\int -\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x}}{\ln \left (x -1\right )}d x +c_1 \right ) {\mathrm e}^{-\left (\int -\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x \right )} \]
The substitution \(v = y^{1-n}\) is now used to convert
the above solution back to \(y\) which results in
\[
\frac {1}{y} = \left (\int -\frac {\coth \left (x +1\right ) x \,{\mathrm e}^{\int -\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x}}{\ln \left (x -1\right )}d x +c_1 \right ) {\mathrm e}^{-\left (\int -\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x \right )}
\]
Solving for \(y\) from the above solution(s) gives
(after possible removing of solutions that do not verify)
\begin{align*} y = \frac {{\mathrm e}^{\int -\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x}}{\int -\frac {\coth \left (x +1\right ) x \,{\mathrm e}^{\int -\frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x}}{\ln \left (x -1\right )}d x +c_1} \end{align*}
Figure 794: Slope field plot\(y^{\prime } = -\frac {y \left (\ln \left (x -1\right )+\coth \left (x +1\right ) x -\coth \left (x +1\right ) x^{2} y\right )}{x \ln \left (x -1\right )}\) 2.212.2 Time used: 2.297 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )}{x \ln \left (x -1\right )}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}
To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 3 to
use as anstaz gives
\begin{align*}
\tag{1E} \xi &= x^{3} a_{7}+x^{2} y a_{8}+y^{2} x a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\
\tag{2E} \eta &= x^{3} b_{7}+x^{2} y b_{8}+y^{2} x b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\
\end{align*}
Where the unknown coefficients are
\[
\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}, b_{8}, b_{9}, b_{10}\}
\]
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} 3 x^{2} b_{7}+2 x y b_{8}+y^{2} b_{9}+2 x b_{4}+y b_{5}+b_{2}+\frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right ) \left (-3 x^{2} a_{7}+x^{2} b_{8}-2 x y a_{8}+2 x y b_{9}-y^{2} a_{9}+3 y^{2} b_{10}-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{x \ln \left (x -1\right )}-\frac {y^{2} \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )^{2} \left (x^{2} a_{8}+2 x y a_{9}+3 y^{2} a_{10}+x a_{5}+2 y a_{6}+a_{3}\right )}{x^{2} \ln \left (x -1\right )^{2}}-\left (\frac {y \left (\left (1-\coth \left (x +1\right )^{2}\right ) x^{2} y +2 y \coth \left (x +1\right ) x -\left (1-\coth \left (x +1\right )^{2}\right ) x -\coth \left (x +1\right )-\frac {1}{x -1}\right )}{x \ln \left (x -1\right )}-\frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )}{x^{2} \ln \left (x -1\right )}-\frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )}{x \ln \left (x -1\right )^{2} \left (x -1\right )}\right ) \left (x^{3} a_{7}+x^{2} y a_{8}+y^{2} x a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )}{x \ln \left (x -1\right )}+\frac {y \coth \left (x +1\right ) x}{\ln \left (x -1\right )}\right ) \left (x^{3} b_{7}+x^{2} y b_{8}+y^{2} x b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
\text {Expression too large to display}
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} \text {Expression too large to display}
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y, \coth \left (x +1\right ), \ln \left (x -1\right )\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\{x = v_{1}, y = v_{2}, \coth \left (x +1\right ) = v_{3}, \ln \left (x -1\right ) = v_{4}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} \text {Expression too large to display}
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}, v_{4}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} \text {Expression too large to display}
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} a_{1}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ a_{7}&=0\\ a_{8}&=0\\ a_{9}&=0\\ a_{10}&=0\\ b_{7}&=0\\ -a_{1}&=0\\ -a_{2}&=0\\ -a_{3}&=0\\ 2 a_{3}&=0\\ -a_{4}&=0\\ -a_{6}&=0\\ 2 a_{6}&=0\\ 3 a_{6}&=0\\ 4 a_{6}&=0\\ -a_{7}&=0\\ -a_{8}&=0\\ 2 a_{8}&=0\\ -2 a_{9}&=0\\ -a_{9}&=0\\ -6 a_{10}&=0\\ -5 a_{10}&=0\\ -4 a_{10}&=0\\ -3 a_{10}&=0\\ -a_{10}&=0\\ 3 a_{10}&=0\\ 4 a_{10}&=0\\ 5 a_{10}&=0\\ 6 a_{10}&=0\\ -b_{1}&=0\\ -2 b_{7}&=0\\ 4 b_{7}&=0\\ -a_{1}+a_{2}&=0\\ a_{1}-a_{2}&=0\\ a_{1}-a_{5}&=0\\ -a_{2}+a_{4}&=0\\ a_{2}-a_{4}&=0\\ -a_{3}+a_{5}&=0\\ -a_{4}+a_{7}&=0\\ -a_{4}-b_{5}&=0\\ a_{4}-a_{7}&=0\\ -a_{5}+a_{8}&=0\\ -a_{5}-b_{6}&=0\\ a_{5}-a_{8}&=0\\ 2 a_{5}-2 a_{8}&=0\\ -4 a_{6}-3 a_{10}&=0\\ -3 a_{6}-6 a_{10}&=0\\ -2 a_{6}+2 a_{9}&=0\\ -a_{6}+a_{9}&=0\\ -a_{6}-a_{10}&=0\\ a_{6}-a_{9}&=0\\ a_{6}+a_{10}&=0\\ 2 a_{6}+6 a_{10}&=0\\ -a_{8}+a_{2}&=0\\ -2 a_{9}-2 b_{10}&=0\\ -a_{9}+a_{3}&=0\\ 2 a_{9}+b_{10}&=0\\ b_{1}-2 b_{2}&=0\\ b_{1}-b_{2}&=0\\ b_{2}-b_{4}&=0\\ 2 b_{2}-3 b_{4}&=0\\ b_{4}-b_{7}&=0\\ 3 b_{4}-4 b_{7}&=0\\ 2 b_{7}-2 b_{4}&=0\\ -b_{8}-4 a_{7}&=0\\ 2 b_{8}+2 a_{7}&=0\\ -a_{1}-a_{3}+a_{5}&=0\\ a_{1}+a_{3}-a_{5}&=0\\ -a_{2}+a_{4}+a_{8}&=0\\ a_{2}-2 a_{4}+2 b_{1}&=0\\ -2 a_{3}+a_{5}+b_{6}&=0\\ -2 a_{3}-2 a_{6}+2 a_{9}&=0\\ -a_{3}+a_{5}+a_{9}&=0\\ -a_{3}+a_{5}+4 a_{9}&=0\\ a_{3}+4 a_{6}-4 a_{9}&=0\\ 2 a_{3}-2 a_{5}-2 a_{9}&=0\\ -a_{4}+a_{2}-a_{8}&=0\\ -a_{5}+a_{3}-a_{9}&=0\\ -3 a_{6}+2 a_{9}+2 b_{10}&=0\\ -a_{6}-a_{3}+a_{9}&=0\\ a_{6}-a_{9}+a_{3}&=0\\ 3 a_{7}+2 b_{4}-2 b_{2}&=0\\ -3 a_{9}-2 b_{10}+a_{3}&=0\\ -2 a_{9}-b_{10}+3 a_{6}&=0\\ -a_{1}+a_{2}+a_{5}-a_{8}&=0\\ a_{1}-2 a_{3}+a_{5}+b_{6}&=0\\ a_{4}+b_{5}-2 b_{8}-2 a_{7}&=0\\ -a_{5}+a_{1}-a_{2}+a_{8}&=0\\ -4 a_{6}-a_{3}+3 a_{9}+2 b_{10}&=0\\ -3 a_{7}+2 a_{4}-2 b_{1}+2 b_{2}&=0\\ 4 a_{7}+b_{8}-3 a_{4}-b_{5}&=0\\ -b_{9}+3 a_{4}+b_{5}-2 a_{2}-b_{3}&=0\\ -a_{5}-b_{6}+b_{9}-a_{1}+2 a_{2}+b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ a_{7}&=0\\ a_{8}&=0\\ a_{9}&=0\\ a_{10}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=-b_{9}\\ b_{4}&=0\\ b_{5}&=0\\ b_{6}&=0\\ b_{7}&=0\\ b_{8}&=0\\ b_{9}&=b_{9}\\ b_{10}&=0 \end{align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any
unknown in the RHS) gives
\begin{align*}
\xi &= 0 \\
\eta &= y^{2} x -y \\
\end{align*}
The next step is to determine the canonical coordinates \(R,S\) . The
canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode
become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\) . Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\) . Since
\(\xi =0\) then in this special case
\begin{align*} R = x \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{y^{2} x -y}} dy \end{align*}
Which results in
\begin{align*} S&= -\ln \left (y \right )+\ln \left (x y -1\right ) \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode
given by
\begin{align*} \omega (x,y) &= \frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )}{x \ln \left (x -1\right )} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {y}{x y -1}\\ S_{y} &= \frac {1}{y \left (x y -1\right )} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= \frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}\tag {2A} \end{align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of
\(R,S\) from the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= \frac {\coth \left (R +1\right ) R +\ln \left (R -1\right )}{R \ln \left (R -1\right )} \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts
an ode, no matter how complicated it is, to one that can be solved by integration when the
ode is in the canonical coordiates \(R,S\) .
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\) , then we only need to integrate \(f(R)\) .
\begin{align*} \int {dS} &= \int {\frac {\coth \left (R +1\right ) R +\ln \left (R -1\right )}{R \ln \left (R -1\right )}\, dR}\\ S \left (R \right ) &= \int \frac {\coth \left (R +1\right ) R +\ln \left (R -1\right )}{R \ln \left (R -1\right )}d R + c_2 \end{align*}
\begin{align*} S \left (R \right )&= \int \frac {\coth \left (R +1\right ) R +\ln \left (R -1\right )}{R \ln \left (R -1\right )}d R +c_2 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} -\ln \left (y\right )+\ln \left (y x -1\right ) = \int \frac {\coth \left (x +1\right ) x +\ln \left (x -1\right )}{x \ln \left (x -1\right )}d x +c_2 \end{align*}
Which gives
\begin{align*} y = -\frac {1}{{\mathrm e}^{\frac {\left (2 c_2 \,{\mathrm e}+2 \left (\int \frac {1}{x}d x \right ) {\mathrm e}+{\mathrm e}^{2} \left (\int \frac {\cosh \left (x \right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}d x \right )+{\mathrm e}^{2} \left (\int \frac {\sinh \left (x \right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}d x \right )+\int \frac {\cosh \left (x \right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}d x -\left (\int \frac {\sinh \left (x \right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}d x \right )\right ) {\mathrm e}^{-1}}{2}}-x} \end{align*}
The following diagram shows solution curves of the original ode and how they transform in
the canonical coordinates space using the mapping shown.
Original ode in \(x,y\) coordinates
Canonical
coordinates
transformation
ODE in canonical coordinates \((R,S)\)
\( \frac {dy}{dx} = \frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )}{x \ln \left (x -1\right )}\)
\( \frac {d S}{d R} = \frac {\coth \left (R +1\right ) R +\ln \left (R -1\right )}{R \ln \left (R -1\right )}\)
\(\!\begin {aligned} R&= x\\ S&= -\ln \left (y \right )+\ln \left (x y -1\right ) \end {aligned} \)
Figure 795: Slope field plot\(y^{\prime } = -\frac {y \left (\ln \left (x -1\right )+\coth \left (x +1\right ) x -\coth \left (x +1\right ) x^{2} y\right )}{x \ln \left (x -1\right )}\) 2.212.3 Time used: 1.311 (sec)
In canonical form the ODE is
\begin{align*} y' &= F(x,y)\\ &= \frac {y \left (\coth \left (x +1\right ) x^{2} y -\coth \left (x +1\right ) x -\ln \left (x -1\right )\right )}{x \ln \left (x -1\right )} \end{align*}
This is a Riccati ODE. Comparing the ODE to solve
\[ y' = \frac {y^{2} x \cosh \left (x \right ) \cosh \left (1\right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}+\frac {y^{2} x \sinh \left (x \right ) \sinh \left (1\right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}-\frac {y \cosh \left (x \right ) \cosh \left (1\right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}-\frac {y \sinh \left (x \right ) \sinh \left (1\right )}{\ln \left (x -1\right ) \left (\sinh \left (x \right ) \cosh \left (1\right )+\cosh \left (x \right ) \sinh \left (1\right )\right )}-\frac {y}{x} \]
With Riccati ODE standard form
\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows that \(f_0(x)=0\) , \(f_1(x)=\frac {-\coth \left (x +1\right ) x -\ln \left (x -1\right )}{x \ln \left (x -1\right )}\) and \(f_2(x)=\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )}\) . Let
\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u \coth \left (x +1\right ) x}{\ln \left (x -1\right )}} \tag {1} \end{align*}
Using the above substitution in the given ODE results (after some simplification)in a second
order ODE to solve for \(u(x)\) which is
\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}
But
\begin{align*} f_2' &=\frac {\left (1-\coth \left (x +1\right )^{2}\right ) x}{\ln \left (x -1\right )}+\frac {\coth \left (x +1\right )}{\ln \left (x -1\right )}-\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )^{2} \left (x -1\right )}\\ f_1 f_2 &=\frac {\left (-\coth \left (x +1\right ) x -\ln \left (x -1\right )\right ) \coth \left (x +1\right )}{\ln \left (x -1\right )^{2}}\\ f_2^2 f_0 &=0 \end{align*}
Substituting the above terms back in equation (2) gives
\begin{align*} \frac {\coth \left (x +1\right ) x u^{\prime \prime }\left (x \right )}{\ln \left (x -1\right )}-\left (\frac {\left (1-\coth \left (x +1\right )^{2}\right ) x}{\ln \left (x -1\right )}+\frac {\coth \left (x +1\right )}{\ln \left (x -1\right )}-\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )^{2} \left (x -1\right )}+\frac {\left (-\coth \left (x +1\right ) x -\ln \left (x -1\right )\right ) \coth \left (x +1\right )}{\ln \left (x -1\right )^{2}}\right ) u^{\prime }\left (x \right ) = 0 \end{align*}
This is second order ode with missing dependent variable \(u\) . Let
\begin{align*} p(x) &= \frac {d u}{d x} \end{align*}
Then
\begin{align*} p'(x) &= \frac {d^{2}u}{d x^{2}} \end{align*}
Hence the ode becomes
\begin{align*} \frac {\coth \left (x +1\right ) x \left (\frac {d}{d x}p \left (x \right )\right )}{\ln \left (x -1\right )}-\left (\frac {\left (1-\coth \left (x +1\right )^{2}\right ) x}{\ln \left (x -1\right )}+\frac {\coth \left (x +1\right )}{\ln \left (x -1\right )}-\frac {\coth \left (x +1\right ) x}{\ln \left (x -1\right )^{2} \left (x -1\right )}+\frac {\left (-\coth \left (x +1\right ) x -\ln \left (x -1\right )\right ) \coth \left (x +1\right )}{\ln \left (x -1\right )^{2}}\right ) p \left (x \right ) = 0 \end{align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form a linear first order is
\begin{align*} \frac {d}{d x}p \left (x \right ) + q(x)p \left (x \right ) &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {\ln \left (x -1\right ) \coth \left (x +1\right )^{2} x -\coth \left (x +1\right )^{2} \ln \left (x -1\right )+\coth \left (x +1\right )^{2} x -x \ln \left (x -1\right )-\coth \left (x +1\right )^{2}+\ln \left (x -1\right )+\coth \left (x +1\right )}{\ln \left (x -1\right ) \coth \left (x +1\right ) \left (x -1\right )}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\[ \mu = {\mathrm e}^{\int \frac {\ln \left (x -1\right ) \coth \left (x +1\right )^{2} x -\coth \left (x +1\right )^{2} \ln \left (x -1\right )+\coth \left (x +1\right )^{2} x -x \ln \left (x -1\right )-\coth \left (x +1\right )^{2}+\ln \left (x -1\right )+\coth \left (x +1\right )}{\ln \left (x -1\right ) \coth \left (x +1\right ) \left (x -1\right )}d x} \]
Therefore the solution is
\[ p \left (x \right ) = c_1 \,{\mathrm e}^{-\left (\int \frac {\ln \left (x -1\right ) \coth \left (x +1\right )^{2} x -\coth \left (x +1\right )^{2} \ln \left (x -1\right )+\coth \left (x +1\right )^{2} x -x \ln \left (x -1\right )-\coth \left (x +1\right )^{2}+\ln \left (x -1\right )+\coth \left (x +1\right )}{\ln \left (x -1\right ) \coth \left (x +1\right ) \left (x -1\right )}d x \right )} \]
For solution (1) found earlier, since \(p=\frac {d u}{d x}\)
then we now have a new first order ode to solve which is
\begin{align*} \frac {d u}{d x} = c_1 \,{\mathrm e}^{-\left (\int \frac {\ln \left (x -1\right ) \coth \left (x +1\right )^{2} x -\coth \left (x +1\right )^{2} \ln \left (x -1\right )+\coth \left (x +1\right )^{2} x -x \ln \left (x -1\right )-\coth \left (x +1\right )^{2}+\ln \left (x -1\right )+\coth \left (x +1\right )}{\ln \left (x -1\right ) \coth \left (x +1\right ) \left (x -1\right )}d x \right )} \end{align*}
Since the ode has the form \(\frac {d u}{d x}=f(x)\) , then we only need to integrate \(f(x)\) .
\begin{align*} \int {du} &= \int {c_1 \,{\mathrm e}^{-\left (\int \frac {\ln \left (x -1\right ) \coth \left (x +1\right )^{2} x -\coth \left (x +1\right )^{2} \ln \left (x -1\right )+\coth \left (x +1\right )^{2} x -x \ln \left (x -1\right )-\coth \left (x +1\right )^{2}+\ln \left (x -1\right )+\coth \left (x +1\right )}{\ln \left (x -1\right ) \coth \left (x +1\right ) \left (x -1\right )}d x \right )}\, dx}\\ u &= \int c_1 \,{\mathrm e}^{-\left (\int \frac {\ln \left (x -1\right ) \coth \left (x +1\right )^{2} x -\coth \left (x +1\right )^{2} \ln \left (x -1\right )+\coth \left (x +1\right )^{2} x -x \ln \left (x -1\right )-\coth \left (x +1\right )^{2}+\ln \left (x -1\right )+\coth \left (x +1\right )}{\ln \left (x -1\right ) \coth \left (x +1\right ) \left (x -1\right )}d x \right )}d x + c_2 \end{align*}
\begin{align*} u&= \int c_1 \,{\mathrm e}^{-\left (\int \frac {\ln \left (x -1\right ) \coth \left (x +1\right )^{2} x -\coth \left (x +1\right )^{2} \ln \left (x -1\right )+\coth \left (x +1\right )^{2} x -x \ln \left (x -1\right )-\coth \left (x +1\right )^{2}+\ln \left (x -1\right )+\coth \left (x +1\right )}{\ln \left (x -1\right ) \coth \left (x +1\right ) \left (x -1\right )}d x \right )}d x +c_2 \end{align*}
Will add steps showing solving for IC soon.
Taking derivative gives
\[
u^{\prime }\left (x \right ) = c_1 \,{\mathrm e}^{-\left (\int \frac {\operatorname {sech}\left (x +1\right ) \operatorname {csch}\left (x +1\right ) \left (x -1\right ) \ln \left (x -1\right )+1+\left (x -1\right ) \coth \left (x +1\right )}{\ln \left (x -1\right ) \left (x -1\right )}d x \right )}
\]
Doing change of constants, the solution becomes
\[
y = -\frac {c_3 \,{\mathrm e}^{-\left (\int \frac {\operatorname {sech}\left (x +1\right ) \operatorname {csch}\left (x +1\right ) \left (x -1\right ) \ln \left (x -1\right )+1+\left (x -1\right ) \coth \left (x +1\right )}{\ln \left (x -1\right ) \left (x -1\right )}d x \right )} \ln \left (x -1\right )}{\coth \left (x +1\right ) x \left (c_3 \left (\int {\mathrm e}^{-\left (\int \frac {\operatorname {sech}\left (x +1\right ) \operatorname {csch}\left (x +1\right ) \left (x -1\right ) \ln \left (x -1\right )+1+\left (x -1\right ) \coth \left (x +1\right )}{\ln \left (x -1\right ) \left (x -1\right )}d x \right )}d x \right )+1\right )}
\]
Figure 796: Slope field plot\(y^{\prime } = -\frac {y \left (\ln \left (x -1\right )+\coth \left (x +1\right ) x -\coth \left (x +1\right ) x^{2} y\right )}{x \ln \left (x -1\right )}\) 2.212.4 \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right ) \left (\ln \left (-1+x \right )+\coth \left (x +1\right ) x -\coth \left (x +1\right ) x^{2} y \left (x \right )\right )}{x \ln \left (-1+x \right )} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right ) \left (\ln \left (-1+x \right )+\coth \left (x +1\right ) x -\coth \left (x +1\right ) x^{2} y \left (x \right )\right )}{x \ln \left (-1+x \right )} \end {array} \]
2.212.5 ` Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
<- Bernoulli successful `
2.212.6 Solving time : 0.006
dsolve ( diff ( y ( x ), x ) = -y(x)*( ln ( x -1)+ coth ( x +1)* x - coth ( x +1)* x ^2* y ( x ))/ x / ln ( x -1), y(x),singsol=all)
\[
y = \frac {{\mathrm e}^{-\left (\int \frac {\ln \left (x -1\right )+\coth \left (x +1\right ) x}{x \ln \left (x -1\right )}d x \right )}}{-\left (\int \frac {x \,{\mathrm e}^{-\left (\int \frac {\ln \left (x -1\right )+\coth \left (x +1\right ) x}{x \ln \left (x -1\right )}d x \right )} \coth \left (x +1\right )}{\ln \left (x -1\right )}d x \right )+c_{1}}
\]
2.212.7 Solving time : 36.6
DSolve [{ D [ y [ x ], x ] == -((y[x]*(x* Coth [1 + x] + Log [-1 + x] - x^2* Coth [1 + x]*y[x]))/(x* Log [-1 + x])),{}},
y[x],x,IncludeSingularSolutions-> True ]
\begin{align*}
y(x)\to \frac {\exp \left (\int _1^x-\frac {\cosh (K[1]) \left (\left (1+e^2\right ) K[1]+\left (-1+e^2\right ) \log (K[1]-1)\right )+\left (\left (-1+e^2\right ) K[1]+\left (1+e^2\right ) \log (K[1]-1)\right ) \sinh (K[1])}{K[1] \log (K[1]-1) \left (\left (-1+e^2\right ) \cosh (K[1])+\left (1+e^2\right ) \sinh (K[1])\right )}dK[1]\right )}{-\int _1^x\frac {\exp \left (\int _1^{K[2]}-\frac {\cosh (K[1]) \left (\left (1+e^2\right ) K[1]+\left (-1+e^2\right ) \log (K[1]-1)\right )+\left (\left (-1+e^2\right ) K[1]+\left (1+e^2\right ) \log (K[1]-1)\right ) \sinh (K[1])}{K[1] \log (K[1]-1) \left (\left (-1+e^2\right ) \cosh (K[1])+\left (1+e^2\right ) \sinh (K[1])\right )}dK[1]\right ) K[2] \left (\left (1+e^2\right ) \cosh (K[2])+\left (-1+e^2\right ) \sinh (K[2])\right )}{\log (K[2]-1) \left (\left (-1+e^2\right ) \cosh (K[2])+\left (1+e^2\right ) \sinh (K[2])\right )}dK[2]+c_1} \\
y(x)\to 0 \\
y(x)\to -\frac {\exp \left (\int _1^x-\frac {\cosh (K[1]) \left (\left (1+e^2\right ) K[1]+\left (-1+e^2\right ) \log (K[1]-1)\right )+\left (\left (-1+e^2\right ) K[1]+\left (1+e^2\right ) \log (K[1]-1)\right ) \sinh (K[1])}{K[1] \log (K[1]-1) \left (\left (-1+e^2\right ) \cosh (K[1])+\left (1+e^2\right ) \sinh (K[1])\right )}dK[1]\right )}{\int _1^x\frac {\exp \left (\int _1^{K[2]}-\frac {\cosh (K[1]) \left (\left (1+e^2\right ) K[1]+\left (-1+e^2\right ) \log (K[1]-1)\right )+\left (\left (-1+e^2\right ) K[1]+\left (1+e^2\right ) \log (K[1]-1)\right ) \sinh (K[1])}{K[1] \log (K[1]-1) \left (\left (-1+e^2\right ) \cosh (K[1])+\left (1+e^2\right ) \sinh (K[1])\right )}dK[1]\right ) K[2] \left (\left (1+e^2\right ) \cosh (K[2])+\left (-1+e^2\right ) \sinh (K[2])\right )}{\log (K[2]-1) \left (\left (-1+e^2\right ) \cosh (K[2])+\left (1+e^2\right ) \sinh (K[2])\right )}dK[2]} \\
\end{align*}