Internal problem ID [9079]
Internal file name [OUTPUT/8014_Monday_June_06_2022_01_16_56_AM_28316288/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 745.
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
[[_1st_order, `_with_symmetry_[F(x),G(y)]`], [_Abel, `2nd type`, `class C`]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {\left (y \ln \left (x \right )-1\right )^{3}}{\left (-1+y \ln \left (x \right )-y\right ) x}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {y^{3} \ln \left (x \right )^{3}-3 \ln \left (x \right )^{2} y^{2}+3 y \ln \left (x \right )-1}{\left (-1+y \ln \left (x \right )-y \right ) x}\\ 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}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{2} b_{4}+y x 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 (y^{3} \ln \left (x \right )^{3}-3 \ln \left (x \right )^{2} y^{2}+3 y \ln \left (x \right )-1\right ) \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{\left (-1+y \ln \left (x \right )-y \right ) x}-\frac {\left (y^{3} \ln \left (x \right )^{3}-3 \ln \left (x \right )^{2} y^{2}+3 y \ln \left (x \right )-1\right )^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{\left (-1+y \ln \left (x \right )-y \right )^{2} x^{2}}-\left (\frac {\frac {3 y^{3} \ln \left (x \right )^{2}}{x}-\frac {6 \ln \left (x \right ) y^{2}}{x}+\frac {3 y}{x}}{\left (-1+y \ln \left (x \right )-y \right ) x}-\frac {\left (y^{3} \ln \left (x \right )^{3}-3 \ln \left (x \right )^{2} y^{2}+3 y \ln \left (x \right )-1\right ) y}{\left (-1+y \ln \left (x \right )-y \right )^{2} x^{2}}-\frac {y^{3} \ln \left (x \right )^{3}-3 \ln \left (x \right )^{2} y^{2}+3 y \ln \left (x \right )-1}{\left (-1+y \ln \left (x \right )-y \right ) x^{2}}\right ) \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {3 \ln \left (x \right )^{3} y^{2}-6 \ln \left (x \right )^{2} y +3 \ln \left (x \right )}{\left (-1+y \ln \left (x \right )-y \right ) x}-\frac {\left (y^{3} \ln \left (x \right )^{3}-3 \ln \left (x \right )^{2} y^{2}+3 y \ln \left (x \right )-1\right ) \left (-1+\ln \left (x \right )\right )}{\left (-1+y \ln \left (x \right )-y \right )^{2} x}\right ) \left (x^{2} b_{4}+y x 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. \[ \{x, y, \ln \left (x \right )\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}, \ln \left (x \right ) = v_{3}\} \] 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 \[ \{v_{1}, v_{2}, v_{3}\} \] 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_{6}&=0\\ b_{5}&=0\\ -a_{3}&=0\\ 6 a_{3}&=0\\ 3 a_{4}&=0\\ -6 a_{5}&=0\\ -2 a_{5}&=0\\ -a_{5}&=0\\ 3 a_{5}&=0\\ 6 a_{5}&=0\\ -6 a_{6}&=0\\ -3 a_{6}&=0\\ -2 a_{6}&=0\\ 3 a_{6}&=0\\ 12 a_{6}&=0\\ -6 b_{1}&=0\\ -2 b_{1}&=0\\ 2 b_{1}&=0\\ 6 b_{1}&=0\\ -6 b_{2}&=0\\ -2 b_{2}&=0\\ 2 b_{2}&=0\\ 6 b_{2}&=0\\ -6 b_{4}&=0\\ -4 b_{4}&=0\\ -2 b_{4}&=0\\ -b_{4}&=0\\ 2 b_{4}&=0\\ 3 b_{4}&=0\\ 4 b_{4}&=0\\ 6 b_{4}&=0\\ -4 a_{1}+6 a_{3}&=0\\ -4 a_{1}+20 a_{3}&=0\\ a_{1}-15 a_{3}&=0\\ a_{1}-a_{3}&=0\\ 6 a_{1}-15 a_{3}&=0\\ 6 a_{2}+6 b_{6}&=0\\ -6 a_{3}-9 a_{6}&=0\\ -3 a_{3}-4 a_{6}&=0\\ a_{3}-30 a_{6}&=0\\ 3 a_{3}+3 a_{6}&=0\\ 3 a_{3}+9 a_{6}&=0\\ -6 a_{4}-2 b_{2}&=0\\ -6 a_{4}-2 b_{5}&=0\\ -a_{4}-b_{5}&=0\\ -a_{4}+b_{5}&=0\\ 3 a_{4}+b_{5}&=0\\ -15 a_{5}-3 b_{1}&=0\\ -15 a_{5}-b_{3}&=0\\ 6 a_{5}-2 b_{3}&=0\\ -9 a_{1}-4 a_{3}+12 a_{6}&=0\\ -6 a_{1}-9 a_{3}-4 a_{6}&=0\\ -3 a_{1}-4 a_{3}+40 a_{6}&=0\\ 3 a_{1}+a_{3}-2 a_{6}&=0\\ 3 a_{1}+3 a_{3}+a_{6}&=0\\ 3 a_{1}+9 a_{3}+6 a_{6}&=0\\ 9 a_{1}+6 a_{3}-30 a_{6}&=0\\ -6 a_{2}-6 a_{5}-6 b_{6}&=0\\ -6 a_{2}-3 b_{3}-6 b_{6}&=0\\ -2 a_{2}+b_{3}-2 b_{6}&=0\\ 2 a_{2}+2 b_{3}+2 b_{6}&=0\\ 3 a_{2}+2 a_{5}+3 b_{6}&=0\\ 3 a_{2}+6 a_{5}+3 b_{6}&=0\\ -3 a_{4}-2 b_{2}-5 b_{5}&=0\\ -a_{4}+2 b_{2}+b_{5}&=0\\ a_{4}+2 b_{2}+3 b_{5}&=0\\ 3 a_{4}+b_{2}+2 b_{5}&=0\\ 4 a_{4}-2 b_{2}-2 b_{5}&=0\\ 4 a_{4}+2 b_{2}+2 b_{5}&=0\\ -a_{5}+b_{1}+b_{3}&=0\\ 20 a_{5}+2 b_{1}+2 b_{3}&=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}&=0\\ 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} \\ \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 \(S(R)\). Unable to determine \(R\). Terminating
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \ln \left (x \right )^{3} y^{3}-3 y^{2} \ln \left (x \right )^{2}-\ln \left (x \right ) y^{\prime } y x +y^{\prime } x y+3 y \ln \left (x \right )+y^{\prime } x -1=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-\ln \left (x \right )^{3} y^{3}+3 y^{2} \ln \left (x \right )^{2}-3 y \ln \left (x \right )+1}{-y \ln \left (x \right ) x +y x +x} \end {array} \]
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 trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 77
dsolve(diff(y(x),x) = (-1+y(x)*ln(x))^3/(-1+y(x)*ln(x)-y(x))/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {47 \operatorname {RootOf}\left (-27783 \left (\int _{}^{\textit {\_Z}}\frac {1}{2209 \textit {\_a}^{3}-9261 \textit {\_a} +9261}d \textit {\_a} \right )-7 \ln \left (x \right )+3 c_{1} \right )-84}{21+47 \left (-1+\ln \left (x \right )\right ) \operatorname {RootOf}\left (-27783 \left (\int _{}^{\textit {\_Z}}\frac {1}{2209 \textit {\_a}^{3}-9261 \textit {\_a} +9261}d \textit {\_a} \right )-7 \ln \left (x \right )+3 c_{1} \right )-84 \ln \left (x \right )} \]
✓ Solution by Mathematica
Time used: 1.247 (sec). Leaf size: 546
DSolve[y'[x] == (-1 + Log[x]*y[x])^3/(x*(-1 - y[x] + Log[x]*y[x])),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {\log (x) K[1]-K[1]-1}{\log ^3(x) K[1]^3+\log (x) K[1]^3-K[1]^3-3 \log ^2(x) K[1]^2-K[1]^2+3 \log (x) K[1]-1}+\text {RootSum}\left [K[1]^3-\text {$\#$1} K[1]^2-\text {$\#$1}^3\&,\frac {K[1] \log (K[1] \log (x)-\text {$\#$1}-1)-\log (K[1] \log (x)-\text {$\#$1}-1) \text {$\#$1}}{K[1]^2+3 \text {$\#$1}^2}\&\right ]+\frac {\text {RootSum}\left [K[1]^3-\text {$\#$1} K[1]^2-\text {$\#$1}^3\&,\frac {4 \log (x) K[1]^3-4 \log (x) \log (K[1] \log (x)-\text {$\#$1}-1) K[1]^3-12 \log (K[1] \log (x)-\text {$\#$1}-1) K[1]^3+12 K[1]^3+4 \log (K[1] \log (x)-\text {$\#$1}-1) K[1]^2+5 \log (x) \text {$\#$1} K[1]^2-5 \log (x) \log (K[1] \log (x)-\text {$\#$1}-1) \text {$\#$1} K[1]^2+16 \log (K[1] \log (x)-\text {$\#$1}-1) \text {$\#$1} K[1]^2-16 \text {$\#$1} K[1]^2-12 \log (x) \text {$\#$1}^2 K[1]+12 \log (x) \log (K[1] \log (x)-\text {$\#$1}-1) \text {$\#$1}^2 K[1]+5 \log (K[1] \log (x)-\text {$\#$1}-1) \text {$\#$1}^2 K[1]-5 \text {$\#$1}^2 K[1]+5 \log (K[1] \log (x)-\text {$\#$1}-1) \text {$\#$1} K[1]-12 \log (K[1] \log (x)-\text {$\#$1}-1) \text {$\#$1}^2}{28 \log (x) K[1]^3-9 K[1]^3-27 \log (x) \text {$\#$1} K[1]^2-19 \text {$\#$1} K[1]^2-28 K[1]^2+9 \log (x) \text {$\#$1}^2 K[1]+27 \text {$\#$1}^2 K[1]+27 \text {$\#$1} K[1]-9 \text {$\#$1}^2}\&\right ]}{K[1]}\right )dK[1]-y(x) \text {RootSum}\left [-\text {$\#$1}^3-\text {$\#$1} y(x)^2+y(x)^3\&,\frac {y(x) \log (-\text {$\#$1}+y(x) \log (x)-1)-\text {$\#$1} \log (-\text {$\#$1}+y(x) \log (x)-1)}{3 \text {$\#$1}^2+y(x)^2}\&\right ]-\log (x)=c_1,y(x)\right ] \]