Internal problem ID [9041]
Internal file name [OUTPUT/7976_Monday_June_06_2022_01_04_19_AM_96726188/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 707.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[`y=_G(x,y')`]
\[ \boxed {y^{\prime }-\frac {\left (-\ln \left (-1+y\right )+\ln \left (y+1\right )+2 \ln \left (x \right )\right )^{2} x \left (y+1\right )^{2}}{16}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right )^{2} x \left (y +1\right )^{2}}{16}\\ 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*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 2 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= 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 \[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} 2 x b_{4}+y b_{5}+b_{2}+\frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right )^{2} x \left (y +1\right )^{2} \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{16}-\frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right )^{4} x^{2} \left (y +1\right )^{4} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{256}-\left (-\frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right ) \left (y +1\right )^{2}}{4}+\frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right )^{2} \left (y +1\right )^{2}}{16}\right ) \left (x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right ) x \left (y +1\right )^{2} \left (\frac {1}{y -1}-\frac {1}{y +1}\right )}{8}+\frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right )^{2} x \left (y +1\right )}{8}\right ) \left (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} Simplifying the above 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. \[ \{x, y, \ln \left (x \right ), \ln \left (y -1\right ), \ln \left (y +1\right )\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}, \ln \left (x \right ) = v_{3}, \ln \left (y -1\right ) = v_{4}, \ln \left (y +1\right ) = v_{5}\} \] 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 \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] 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_{3}&=0\\ a_{5}&=0\\ -64 a_{1}&=0\\ -32 a_{1}&=0\\ 16 a_{1}&=0\\ 64 a_{1}&=0\\ 128 a_{1}&=0\\ -160 a_{3}&=0\\ -120 a_{3}&=0\\ -80 a_{3}&=0\\ -48 a_{3}&=0\\ -40 a_{3}&=0\\ -32 a_{3}&=0\\ -30 a_{3}&=0\\ -24 a_{3}&=0\\ -8 a_{3}&=0\\ -5 a_{3}&=0\\ -4 a_{3}&=0\\ 6 a_{3}&=0\\ 8 a_{3}&=0\\ 16 a_{3}&=0\\ 20 a_{3}&=0\\ 24 a_{3}&=0\\ 32 a_{3}&=0\\ 40 a_{3}&=0\\ 120 a_{3}&=0\\ 160 a_{3}&=0\\ 240 a_{3}&=0\\ -128 a_{4}&=0\\ -64 a_{4}&=0\\ 64 a_{4}&=0\\ -240 a_{5}&=0\\ -192 a_{5}&=0\\ -160 a_{5}&=0\\ -128 a_{5}&=0\\ -120 a_{5}&=0\\ -96 a_{5}&=0\\ -80 a_{5}&=0\\ -64 a_{5}&=0\\ -48 a_{5}&=0\\ -40 a_{5}&=0\\ -32 a_{5}&=0\\ -30 a_{5}&=0\\ -24 a_{5}&=0\\ -20 a_{5}&=0\\ -16 a_{5}&=0\\ -8 a_{5}&=0\\ -6 a_{5}&=0\\ -5 a_{5}&=0\\ -4 a_{5}&=0\\ -a_{5}&=0\\ 4 a_{5}&=0\\ 5 a_{5}&=0\\ 6 a_{5}&=0\\ 8 a_{5}&=0\\ 16 a_{5}&=0\\ 20 a_{5}&=0\\ 24 a_{5}&=0\\ 30 a_{5}&=0\\ 32 a_{5}&=0\\ 40 a_{5}&=0\\ 48 a_{5}&=0\\ 64 a_{5}&=0\\ 80 a_{5}&=0\\ 96 a_{5}&=0\\ 120 a_{5}&=0\\ 128 a_{5}&=0\\ 160 a_{5}&=0\\ 192 a_{5}&=0\\ 240 a_{5}&=0\\ -480 a_{6}&=0\\ -320 a_{6}&=0\\ -240 a_{6}&=0\\ -128 a_{6}&=0\\ -80 a_{6}&=0\\ -64 a_{6}&=0\\ -48 a_{6}&=0\\ -40 a_{6}&=0\\ -32 a_{6}&=0\\ -16 a_{6}&=0\\ -12 a_{6}&=0\\ -2 a_{6}&=0\\ 8 a_{6}&=0\\ 10 a_{6}&=0\\ 16 a_{6}&=0\\ 32 a_{6}&=0\\ 48 a_{6}&=0\\ 60 a_{6}&=0\\ 64 a_{6}&=0\\ 80 a_{6}&=0\\ 96 a_{6}&=0\\ 160 a_{6}&=0\\ 240 a_{6}&=0\\ 320 a_{6}&=0\\ -256 b_{2}&=0\\ 256 b_{2}&=0\\ -512 b_{4}&=0\\ -128 b_{4}&=0\\ -64 b_{4}&=0\\ -32 b_{4}&=0\\ 32 b_{4}&=0\\ 64 b_{4}&=0\\ 128 b_{4}&=0\\ 256 b_{4}&=0\\ 512 b_{4}&=0\\ -256 b_{5}&=0\\ 256 b_{5}&=0\\ -256 a_{1}+256 a_{6}&=0\\ -128 a_{1}-256 a_{3}&=0\\ -128 a_{1}-64 a_{3}&=0\\ -128 a_{1}+128 a_{6}&=0\\ -64 a_{1}-128 a_{3}&=0\\ -64 a_{1}-32 a_{3}&=0\\ -32 a_{1}+32 a_{6}&=0\\ -16 a_{1}-32 a_{3}&=0\\ 32 a_{1}+16 a_{3}&=0\\ 32 a_{1}+64 a_{3}&=0\\ 64 a_{1}+128 a_{3}&=0\\ 64 a_{1}-64 a_{6}&=0\\ 128 a_{1}+64 a_{3}&=0\\ 128 a_{1}-128 a_{6}&=0\\ 256 a_{1}+128 a_{3}&=0\\ -64 a_{2}-64 b_{1}&=0\\ 64 a_{2}+64 b_{1}&=0\\ 128 a_{2}+128 b_{1}&=0\\ -240 a_{3}-384 a_{6}&=0\\ -192 a_{3}-96 a_{6}&=0\\ -160 a_{3}-256 a_{6}&=0\\ -128 a_{3}-320 a_{6}&=0\\ -128 a_{3}-256 a_{6}&=0\\ -128 a_{3}-64 a_{6}&=0\\ -120 a_{3}-192 a_{6}&=0\\ -96 a_{3}-240 a_{6}&=0\\ -96 a_{3}-48 a_{6}&=0\\ -64 a_{3}-160 a_{6}&=0\\ -64 a_{3}-128 a_{6}&=0\\ -64 a_{3}-32 a_{6}&=0\\ -40 a_{3}-64 a_{6}&=0\\ -32 a_{3}-256 a_{6}&=0\\ -32 a_{3}-80 a_{6}&=0\\ -32 a_{3}-16 a_{6}&=0\\ -24 a_{3}-192 a_{6}&=0\\ -24 a_{3}-60 a_{6}&=0\\ -20 a_{3}-32 a_{6}&=0\\ -16 a_{3}-128 a_{6}&=0\\ -16 a_{3}-32 a_{6}&=0\\ -16 a_{3}-8 a_{6}&=0\\ -8 a_{3}-64 a_{6}&=0\\ -6 a_{3}-48 a_{6}&=0\\ -4 a_{3}-10 a_{6}&=0\\ -a_{3}-8 a_{6}&=0\\ 4 a_{3}+2 a_{6}&=0\\ 4 a_{3}+32 a_{6}&=0\\ 5 a_{3}+8 a_{6}&=0\\ 8 a_{3}+64 a_{6}&=0\\ 16 a_{3}+40 a_{6}&=0\\ 24 a_{3}+12 a_{6}&=0\\ 24 a_{3}+192 a_{6}&=0\\ 30 a_{3}+48 a_{6}&=0\\ 32 a_{3}+16 a_{6}&=0\\ 32 a_{3}+64 a_{6}&=0\\ 32 a_{3}+80 a_{6}&=0\\ 32 a_{3}+256 a_{6}&=0\\ 40 a_{3}+64 a_{6}&=0\\ 48 a_{3}+384 a_{6}&=0\\ 64 a_{3}+32 a_{6}&=0\\ 64 a_{3}+128 a_{6}&=0\\ 80 a_{3}+128 a_{6}&=0\\ 96 a_{3}+48 a_{6}&=0\\ 96 a_{3}+240 a_{6}&=0\\ 120 a_{3}+192 a_{6}&=0\\ 128 a_{3}+64 a_{6}&=0\\ 128 a_{3}+320 a_{6}&=0\\ 160 a_{3}+256 a_{6}&=0\\ 192 a_{3}+480 a_{6}&=0\\ 256 a_{3}+128 a_{6}&=0\\ -384 a_{4}-128 b_{2}&=0\\ -256 a_{4}+128 b_{5}&=0\\ -192 a_{4}-64 b_{2}&=0\\ -192 a_{4}-64 b_{5}&=0\\ -128 a_{4}+64 b_{5}&=0\\ -96 a_{4}-32 b_{2}&=0\\ -64 a_{4}-64 b_{2}&=0\\ -48 a_{4}-16 b_{5}&=0\\ 64 a_{4}+64 b_{2}&=0\\ 96 a_{4}+32 b_{2}&=0\\ 96 a_{4}+32 b_{5}&=0\\ 128 a_{4}+128 b_{2}&=0\\ 128 a_{4}-64 b_{5}&=0\\ 192 a_{4}+64 b_{2}&=0\\ 192 a_{4}+64 b_{5}&=0\\ 384 a_{4}+128 b_{2}&=0\\ -128 b_{2}+128 b_{5}&=0\\ -64 b_{2}-128 b_{5}&=0\\ -32 b_{2}+32 b_{5}&=0\\ 64 b_{2}-64 b_{5}&=0\\ 64 b_{2}+128 b_{5}&=0\\ 128 b_{2}-128 b_{5}&=0\\ 128 b_{2}+256 b_{5}&=0\\ -256 a_{2}-128 b_{1}+128 b_{6}&=0\\ -256 a_{2}+128 b_{3}+256 b_{6}&=0\\ -128 a_{2}-256 a_{5}+128 b_{6}&=0\\ -128 a_{2}-128 b_{1}+64 b_{3}&=0\\ -128 a_{2}+64 b_{3}+128 b_{6}&=0\\ -64 a_{2}-128 a_{5}+64 b_{6}&=0\\ -64 a_{2}-64 b_{1}+32 b_{3}&=0\\ -64 a_{2}-32 b_{1}+32 b_{6}&=0\\ 32 a_{2}+32 b_{1}-16 b_{3}&=0\\ 64 a_{2}+128 a_{5}-64 b_{6}&=0\\ 128 a_{2}+64 b_{1}-64 b_{6}&=0\\ 128 a_{2}+128 b_{1}-64 b_{3}&=0\\ 128 a_{2}-64 b_{3}-128 b_{6}&=0\\ 256 a_{2}+128 b_{1}-128 b_{6}&=0\\ -192 a_{4}-128 b_{2}+64 b_{5}&=0\\ -128 a_{4}-128 b_{2}-64 b_{5}&=0\\ -96 a_{4}-64 b_{2}+32 b_{5}&=0\\ 48 a_{4}+32 b_{2}-16 b_{5}&=0\\ 128 a_{4}+128 b_{2}+64 b_{5}&=0\\ 192 a_{4}+128 b_{2}-64 b_{5}&=0\\ 256 a_{4}+256 b_{2}+128 b_{5}&=0\\ -256 a_{2}-128 a_{5}-128 b_{1}+128 b_{6}&=0\\ -128 a_{2}-256 a_{5}-64 b_{3}+128 b_{6}&=0\\ -128 a_{2}-64 a_{5}-128 b_{1}-64 b_{3}&=0\\ -128 a_{2}-64 a_{5}-64 b_{1}+64 b_{6}&=0\\ -32 a_{2}-64 a_{5}-16 b_{3}+32 b_{6}&=0\\ 64 a_{2}+32 a_{5}+32 b_{1}-32 b_{6}&=0\\ 64 a_{2}+128 a_{5}+32 b_{3}-64 b_{6}&=0\\ 128 a_{2}+64 a_{5}+128 b_{1}+64 b_{3}&=0\\ 128 a_{2}+256 a_{5}+64 b_{3}-128 b_{6}&=0\\ 256 a_{2}+128 a_{5}+128 b_{1}-128 b_{6}&=0\\ 256 a_{2}+128 a_{5}+256 b_{1}+128 b_{3}&=0\\ -256 a_{5}+128 b_{1}-128 b_{3}+128 b_{6}&=0\\ -128 a_{5}-64 b_{1}-128 b_{3}-64 b_{6}&=0\\ -128 a_{5}+64 b_{1}-64 b_{3}+64 b_{6}&=0\\ 64 a_{5}-32 b_{1}+32 b_{3}-32 b_{6}&=0\\ 128 a_{5}+64 b_{1}+128 b_{3}+64 b_{6}&=0\\ 256 a_{5}-128 b_{1}+128 b_{3}-128 b_{6}&=0\\ 256 a_{5}+128 b_{1}+256 b_{3}+128 b_{6}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=b_{6}\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=-b_{6}\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=0\\ b_{6}&=b_{6} \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 &= x \\ \eta &= y^{2}-1 \\ \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 \(S(R)\). Therefore \begin {align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {y^{2}-1}{x}\\ &= \frac {y^{2}-1}{x} \end {align*}
This is easily solved to give \begin {align*} y = -\tanh \left (\ln \left (x \right )+c_{1} \right ) \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= -\ln \left (x \right )-\operatorname {arctanh}\left (y \right ) \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{x} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= \ln \left (x \right ) \end {align*}
Where the constant of integration is set to zero as we just need one solution. 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 {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right )^{2} x \left (y +1\right )^{2}}{16} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= -\frac {1}{x}\\ R_{y} &= \frac {1}{y^{2}-1}\\ S_{x} &= \frac {1}{x}\\ S_{y} &= 0 \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= \frac {16 y -16}{x^{2} \left (y +1\right ) \ln \left (y -1\right )^{2}-4 \left (y +1\right ) x^{2} \left (\frac {\ln \left (y +1\right )}{2}+\ln \left (x \right )\right ) \ln \left (y -1\right )+x^{2} \left (y +1\right ) \ln \left (y +1\right )^{2}+4 \left (y +1\right ) x^{2} \ln \left (x \right ) \ln \left (y +1\right )+4 x^{2} \left (y +1\right ) \ln \left (x \right )^{2}-16 y +16}\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 {16 \,{\mathrm e}^{-S \left (R \right )}}{\left (4 i \pi R -\pi ^{2}+4 R^{2}\right ) {\mathrm e}^{-S \left (R \right )-2 R}+16 \,{\mathrm e}^{-S \left (R \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 \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \int -\frac {16 \,{\mathrm e}^{2 R}}{4 i \pi R -\pi ^{2}+4 R^{2}+16 \,{\mathrm e}^{2 R}}d R +c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} \ln \left (x \right ) = \int _{}^{-\ln \left (x \right )-\operatorname {arctanh}\left (y\right )}-\frac {16 \,{\mathrm e}^{2 \textit {\_a}}}{4 i \pi \textit {\_a} -\pi ^{2}+4 \textit {\_a}^{2}+16 \,{\mathrm e}^{2 \textit {\_a}}}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} \ln \left (x \right ) = \int _{}^{-\ln \left (x \right )-\operatorname {arctanh}\left (y\right )}-\frac {16 \,{\mathrm e}^{2 \textit {\_a}}}{4 i \pi \textit {\_a} -\pi ^{2}+4 \textit {\_a}^{2}+16 \,{\mathrm e}^{2 \textit {\_a}}}d \textit {\_a} +c_{1} \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 \((R,S)\) |
| \( \frac {dy}{dx} = \frac {\left (\ln \left (y -1\right )-\ln \left (y +1\right )-2 \ln \left (x \right )\right )^{2} x \left (y +1\right )^{2}}{16}\) |
|
\( \frac {d S}{d R} = -\frac {16 \,{\mathrm e}^{-S \left (R \right )}}{\left (4 i \pi R -\pi ^{2}+4 R^{2}\right ) {\mathrm e}^{-S \left (R \right )-2 R}+16 \,{\mathrm e}^{-S \left (R \right )}}\) |
| |
\(\!\begin {aligned} R&= -\ln \left (x \right )-\operatorname {arctanh}\left (y \right )\\ S&= \ln \left (x \right ) \end {aligned} \) |
|
Summary
The solution(s) found are the following \begin{align*} \tag{1} \ln \left (x \right ) &= \int _{}^{-\ln \left (x \right )-\operatorname {arctanh}\left (y\right )}-\frac {16 \,{\mathrm e}^{2 \textit {\_a}}}{4 i \pi \textit {\_a} -\pi ^{2}+4 \textit {\_a}^{2}+16 \,{\mathrm e}^{2 \textit {\_a}}}d \textit {\_a} +c_{1} \\ \end{align*}
Verification of solutions
\[ \ln \left (x \right ) = \int _{}^{-\ln \left (x \right )-\operatorname {arctanh}\left (y\right )}-\frac {16 \,{\mathrm e}^{2 \textit {\_a}}}{4 i \pi \textit {\_a} -\pi ^{2}+4 \textit {\_a}^{2}+16 \,{\mathrm e}^{2 \textit {\_a}}}d \textit {\_a} +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {\left (-\ln \left (-1+y\right )+\ln \left (y+1\right )+2 \ln \left (x \right )\right )^{2} x \left (y+1\right )^{2}}{16}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (-\ln \left (-1+y\right )+\ln \left (y+1\right )+2 \ln \left (x \right )\right )^{2} x \left (y+1\right )^{2}}{16} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying inverse_Riccati trying an equivalence to an Abel ODE differential order: 1; trying a linearization to 2nd order --- trying a change of variables {x -> y(x), y(x) -> x} differential order: 1; trying a linearization to 2nd order trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3`[x, y^2-1]
✓ Solution by Maple
Time used: 0.437 (sec). Leaf size: 182
dsolve(diff(y(x),x) = 1/16*(-ln(-1+y(x))+ln(y(x)+1)+2*ln(x))^2*x*(y(x)+1)^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= {\mathrm e}^{\operatorname {RootOf}\left (x^{2} {\mathrm e}^{\textit {\_Z}} \textit {\_Z}^{2}-2 x^{2} {\mathrm e}^{\textit {\_Z}} \ln \left (\frac {{\mathrm e}^{\textit {\_Z}}-2}{x^{2}}\right ) \textit {\_Z} +4 \ln \left (x \right )^{2} x^{2} {\mathrm e}^{\textit {\_Z}}-4 \ln \left (x \right ) \ln \left ({\mathrm e}^{\textit {\_Z}}-2\right ) x^{2} {\mathrm e}^{\textit {\_Z}}+\ln \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2} x^{2} {\mathrm e}^{\textit {\_Z}}-16 \,{\mathrm e}^{\textit {\_Z}}+32\right )}-1 \\ \int _{\textit {\_b}}^{y \left (x \right )}\frac {1}{4 \left (\textit {\_a} +1\right ) \left (\frac {x^{2} \left (\textit {\_a} +1\right ) \ln \left (\textit {\_a} -1\right )^{2}}{4}-\left (\ln \left (x \right )+\frac {\ln \left (\textit {\_a} +1\right )}{2}\right ) \left (\textit {\_a} +1\right ) x^{2} \ln \left (\textit {\_a} -1\right )+\frac {x^{2} \left (\textit {\_a} +1\right ) \ln \left (\textit {\_a} +1\right )^{2}}{4}+x^{2} \left (\textit {\_a} +1\right ) \ln \left (x \right ) \ln \left (\textit {\_a} +1\right )+x^{2} \left (\textit {\_a} +1\right ) \ln \left (x \right )^{2}-4 \textit {\_a} +4\right )}d \textit {\_a} -\frac {\ln \left (x \right )}{16}-c_{1} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 4.678 (sec). Leaf size: 1391
DSolve[y'[x] == (x*(2*Log[x] - Log[-1 + y[x]] + Log[1 + y[x]])^2*(1 + y[x])^2)/16,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x-\frac {K[1] (2 \log (K[1])-\log (y(x)-1)+\log (y(x)+1))^2 (y(x)+1)}{4 \log ^2(K[1]) K[1]^2+\log ^2(y(x)-1) K[1]^2+\log ^2(y(x)+1) K[1]^2-4 \log (K[1]) \log (y(x)-1) K[1]^2+4 \log (K[1]) \log (y(x)+1) K[1]^2-2 \log (y(x)-1) \log (y(x)+1) K[1]^2+4 \log ^2(K[1]) y(x) K[1]^2+\log ^2(y(x)-1) y(x) K[1]^2+\log ^2(y(x)+1) y(x) K[1]^2-4 \log (K[1]) \log (y(x)-1) y(x) K[1]^2+4 \log (K[1]) \log (y(x)+1) y(x) K[1]^2-2 \log (y(x)-1) \log (y(x)+1) y(x) K[1]^2-16 y(x)+16}dK[1]+\int _1^{y(x)}\left (\frac {-4 \log ^2(x) x^2-\log ^2(K[2]-1) x^2-\log ^2(K[2]+1) x^2+4 \log (x) \log (K[2]-1) x^2-4 \log (x) \log (K[2]+1) x^2+2 \log (K[2]-1) \log (K[2]+1) x^2+16}{2 \left (4 \log ^2(x) x^2+\log ^2(K[2]-1) x^2+\log ^2(K[2]+1) x^2-4 \log (x) \log (K[2]-1) x^2+4 \log (x) \log (K[2]+1) x^2-2 \log (K[2]-1) \log (K[2]+1) x^2+K[2] \left (4 \log ^2(x) x^2+\log ^2(K[2]-1) x^2+\log ^2(K[2]+1) x^2-4 \log (x) \log (K[2]-1) x^2+4 \log (x) \log (K[2]+1) x^2-2 \log (K[2]-1) \log (K[2]+1) x^2-16\right )+16\right )}-\int _1^x\left (-\frac {K[1] (2 \log (K[1])-\log (K[2]-1)+\log (K[2]+1))^2}{4 K[2] \log ^2(K[1]) K[1]^2+4 \log ^2(K[1]) K[1]^2+K[2] \log ^2(K[2]-1) K[1]^2+\log ^2(K[2]-1) K[1]^2+K[2] \log ^2(K[2]+1) K[1]^2+\log ^2(K[2]+1) K[1]^2-4 K[2] \log (K[1]) \log (K[2]-1) K[1]^2-4 \log (K[1]) \log (K[2]-1) K[1]^2+4 K[2] \log (K[1]) \log (K[2]+1) K[1]^2+4 \log (K[1]) \log (K[2]+1) K[1]^2-2 K[2] \log (K[2]-1) \log (K[2]+1) K[1]^2-2 \log (K[2]-1) \log (K[2]+1) K[1]^2-16 K[2]+16}+\frac {K[1] (K[2]+1) \left (4 \log ^2(K[1]) K[1]^2+\log ^2(K[2]-1) K[1]^2+\log ^2(K[2]+1) K[1]^2-\frac {4 K[2] \log (K[1]) K[1]^2}{K[2]-1}-\frac {4 \log (K[1]) K[1]^2}{K[2]-1}+\frac {4 K[2] \log (K[1]) K[1]^2}{K[2]+1}+\frac {4 \log (K[1]) K[1]^2}{K[2]+1}+\frac {2 K[2] \log (K[2]-1) K[1]^2}{K[2]-1}-4 \log (K[1]) \log (K[2]-1) K[1]^2+\frac {2 \log (K[2]-1) K[1]^2}{K[2]-1}-\frac {2 K[2] \log (K[2]-1) K[1]^2}{K[2]+1}-\frac {2 \log (K[2]-1) K[1]^2}{K[2]+1}-\frac {2 K[2] \log (K[2]+1) K[1]^2}{K[2]-1}+4 \log (K[1]) \log (K[2]+1) K[1]^2-2 \log (K[2]-1) \log (K[2]+1) K[1]^2-\frac {2 \log (K[2]+1) K[1]^2}{K[2]-1}+\frac {2 K[2] \log (K[2]+1) K[1]^2}{K[2]+1}+\frac {2 \log (K[2]+1) K[1]^2}{K[2]+1}-16\right ) (2 \log (K[1])-\log (K[2]-1)+\log (K[2]+1))^2}{\left (4 K[2] \log ^2(K[1]) K[1]^2+4 \log ^2(K[1]) K[1]^2+K[2] \log ^2(K[2]-1) K[1]^2+\log ^2(K[2]-1) K[1]^2+K[2] \log ^2(K[2]+1) K[1]^2+\log ^2(K[2]+1) K[1]^2-4 K[2] \log (K[1]) \log (K[2]-1) K[1]^2-4 \log (K[1]) \log (K[2]-1) K[1]^2+4 K[2] \log (K[1]) \log (K[2]+1) K[1]^2+4 \log (K[1]) \log (K[2]+1) K[1]^2-2 K[2] \log (K[2]-1) \log (K[2]+1) K[1]^2-2 \log (K[2]-1) \log (K[2]+1) K[1]^2-16 K[2]+16\right )^2}-\frac {2 K[1] (K[2]+1) \left (\frac {1}{K[2]+1}-\frac {1}{K[2]-1}\right ) (2 \log (K[1])-\log (K[2]-1)+\log (K[2]+1))}{4 K[2] \log ^2(K[1]) K[1]^2+4 \log ^2(K[1]) K[1]^2+K[2] \log ^2(K[2]-1) K[1]^2+\log ^2(K[2]-1) K[1]^2+K[2] \log ^2(K[2]+1) K[1]^2+\log ^2(K[2]+1) K[1]^2-4 K[2] \log (K[1]) \log (K[2]-1) K[1]^2-4 \log (K[1]) \log (K[2]-1) K[1]^2+4 K[2] \log (K[1]) \log (K[2]+1) K[1]^2+4 \log (K[1]) \log (K[2]+1) K[1]^2-2 K[2] \log (K[2]-1) \log (K[2]+1) K[1]^2-2 \log (K[2]-1) \log (K[2]+1) K[1]^2-16 K[2]+16}\right )dK[1]+\frac {1}{2 (K[2]+1)}\right )dK[2]=c_1,y(x)\right ] \]