Internal problem ID [9246]
Internal file name [OUTPUT/8182_Monday_June_06_2022_02_06_42_AM_18241665/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 913.
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
[[_Abel, `2nd type`, `class C`]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }+\frac {-y^{3}-y+2 y^{2} \ln \left (x \right )-\ln \left (x \right )^{2} y^{3}-1+3 y \ln \left (x \right )-3 y^{2} \ln \left (x \right )^{2}+\ln \left (x \right )^{3} y^{3}}{y x}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {-y^{3}-y +2 y^{2} \ln \left (x \right )-\ln \left (x \right )^{2} y^{3}-1+3 y \ln \left (x \right )-3 \ln \left (x \right )^{2} y^{2}+y^{3} \ln \left (x \right )^{3}}{y 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}-y +2 y^{2} \ln \left (x \right )-\ln \left (x \right )^{2} y^{3}-1+3 y \ln \left (x \right )-3 \ln \left (x \right )^{2} y^{2}+y^{3} \ln \left (x \right )^{3}\right ) \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{y x}-\frac {\left (-y^{3}-y +2 y^{2} \ln \left (x \right )-\ln \left (x \right )^{2} y^{3}-1+3 y \ln \left (x \right )-3 \ln \left (x \right )^{2} y^{2}+y^{3} \ln \left (x \right )^{3}\right )^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{y^{2} x^{2}}-\left (-\frac {\frac {2 y^{2}}{x}-\frac {2 \ln \left (x \right ) y^{3}}{x}+\frac {3 y}{x}-\frac {6 \ln \left (x \right ) y^{2}}{x}+\frac {3 y^{3} \ln \left (x \right )^{2}}{x}}{y x}+\frac {-y^{3}-y +2 y^{2} \ln \left (x \right )-\ln \left (x \right )^{2} y^{3}-1+3 y \ln \left (x \right )-3 \ln \left (x \right )^{2} y^{2}+y^{3} \ln \left (x \right )^{3}}{y \,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 y^{2}-1+4 y \ln \left (x \right )-3 \ln \left (x \right )^{2} y^{2}+3 \ln \left (x \right )-6 \ln \left (x \right )^{2} y +3 y^{2} \ln \left (x \right )^{3}}{y x}+\frac {-y^{3}-y +2 y^{2} \ln \left (x \right )-\ln \left (x \right )^{2} y^{3}-1+3 y \ln \left (x \right )-3 \ln \left (x \right )^{2} y^{2}+y^{3} \ln \left (x \right )^{3}}{y^{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_{1}&=0\\ b_{2}&=0\\ b_{4}&=0\\ -15 a_{3}&=0\\ -a_{3}&=0\\ 6 a_{3}&=0\\ 20 a_{3}&=0\\ -4 a_{4}&=0\\ -2 a_{4}&=0\\ -15 a_{5}&=0\\ -10 a_{5}&=0\\ -3 a_{5}&=0\\ -2 a_{5}&=0\\ -a_{5}&=0\\ 2 a_{5}&=0\\ 4 a_{5}&=0\\ 6 a_{5}&=0\\ -4 a_{6}&=0\\ -2 a_{6}&=0\\ 4 a_{6}&=0\\ 6 a_{6}&=0\\ -3 b_{2}&=0\\ 2 b_{2}&=0\\ -b_{3}&=0\\ -3 b_{4}&=0\\ -2 b_{4}&=0\\ 2 b_{4}&=0\\ 4 a_{1}-4 a_{6}&=0\\ -2 a_{2}-2 b_{6}&=0\\ -10 a_{3}-30 a_{6}&=0\\ -2 a_{3}-9 a_{6}&=0\\ -2 a_{3}-8 a_{6}&=0\\ -a_{3}-20 a_{6}&=0\\ -a_{3}+a_{6}&=0\\ a_{3}-2 a_{6}&=0\\ 2 a_{3}+4 a_{6}&=0\\ 2 a_{3}+7 a_{6}&=0\\ 2 a_{3}+12 a_{6}&=0\\ 3 a_{3}+40 a_{6}&=0\\ -3 a_{4}-2 b_{2}&=0\\ -a_{4}-b_{5}&=0\\ -a_{4}+2 b_{5}&=0\\ a_{4}+b_{5}&=0\\ 2 a_{4}-b_{5}&=0\\ -20 a_{5}-2 b_{1}&=0\\ -15 a_{5}-3 b_{1}&=0\\ -2 a_{5}+2 b_{3}&=0\\ -a_{5}+b_{1}&=0\\ 20 a_{5}+2 b_{1}&=0\\ 20 a_{5}+b_{3}&=0\\ -8 a_{1}+a_{3}+20 a_{6}&=0\\ -3 a_{1}+10 a_{3}+12 a_{6}&=0\\ -2 a_{1}-2 a_{3}+5 a_{6}&=0\\ -a_{1}+20 a_{3}+40 a_{6}&=0\\ a_{1}-2 a_{3}-2 a_{6}&=0\\ 2 a_{1}+2 a_{3}-a_{6}&=0\\ 3 a_{1}-20 a_{3}-30 a_{6}&=0\\ 4 a_{1}-3 a_{3}-40 a_{6}&=0\\ -6 a_{2}+4 a_{5}-6 b_{6}&=0\\ 2 a_{4}-2 b_{2}+b_{5}&=0\\ 2 a_{4}+b_{2}+b_{5}&=0\\ 3 a_{4}+2 b_{2}-3 b_{5}&=0\\ 10 a_{5}+2 b_{1}-3 b_{3}&=0\\ 2 a_{2}+a_{5}-2 b_{1}+2 b_{6}&=0\\ 3 a_{2}-6 a_{5}-b_{3}+3 b_{6}&=0\\ 3 a_{2}-a_{5}+b_{3}+3 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}&=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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \ln \left (x \right )^{3} y^{3}-\ln \left (x \right )^{2} y^{3}-3 y^{2} \ln \left (x \right )^{2}+y^{\prime } x y-y^{3}+2 y^{2} \ln \left (x \right )+3 y \ln \left (x \right )-y-1=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-\ln \left (x \right )^{3} y^{3}+\ln \left (x \right )^{2} y^{3}+3 y^{2} \ln \left (x \right )^{2}+y^{3}-2 y^{2} \ln \left (x \right )-3 y \ln \left (x \right )+y+1}{y x} \end {array} \]
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: 43
dsolve(diff(y(x),x) = -(-y(x)^3-y(x)+2*y(x)^2*ln(x)-ln(x)^2*y(x)^3-1+3*y(x)*ln(x)-3*ln(x)^2*y(x)^2+ln(x)^3*y(x)^3)/y(x)/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {9}{9 \ln \left (x \right )+56 \operatorname {RootOf}\left (-81 \left (\int _{}^{\textit {\_Z}}\frac {1}{3136 \textit {\_a}^{3}-27 \textit {\_a} +27}d \textit {\_a} \right )-\ln \left (x \right )+3 c_{1} \right )-3} \]
✓ Solution by Mathematica
Time used: 0.529 (sec). Leaf size: 716
DSolve[y'[x] == (1 + y[x] - 3*Log[x]*y[x] - 2*Log[x]*y[x]^2 + 3*Log[x]^2*y[x]^2 + y[x]^3 + Log[x]^2*y[x]^3 - Log[x]^3*y[x]^3)/(x*y[x]),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (2 \text {RootSum}\left [\text {$\#$1}^3 K[1]^3-\text {$\#$1}^2 K[1]^3-2 K[1]^3-3 \text {$\#$1}^2 K[1]^2+2 \text {$\#$1} K[1]^2+3 \text {$\#$1} K[1]-K[1]-1\&,\frac {\log (\log (x)-\text {$\#$1})}{3 \text {$\#$1}^2 K[1]^2-2 \text {$\#$1} K[1]^2-6 \text {$\#$1} K[1]+2 K[1]+3}\&\right ] K[1]-\frac {K[1]}{\log ^3(x) K[1]^3-\log ^2(x) K[1]^3-2 K[1]^3-3 \log ^2(x) K[1]^2+2 \log (x) K[1]^2+3 \log (x) K[1]-K[1]-1}+\frac {\text {RootSum}\left [\text {$\#$1}^3 K[1]^3-\text {$\#$1}^2 K[1]^3-2 K[1]^3-3 \text {$\#$1}^2 K[1]^2+2 \text {$\#$1} K[1]^2+3 \text {$\#$1} K[1]-K[1]-1\&,\frac {-2 \log (x) \log (\log (x)-\text {$\#$1}) \text {$\#$1}^2 K[1]^3+38 \log (\log (x)-\text {$\#$1}) \text {$\#$1}^2 K[1]^3+12 \log (x) \log (\log (x)-\text {$\#$1}) K[1]^3+4 \log (\log (x)-\text {$\#$1}) K[1]^3-36 \log (x) \log (\log (x)-\text {$\#$1}) \text {$\#$1} K[1]^3-12 \log (\log (x)-\text {$\#$1}) \text {$\#$1} K[1]^3+2 \log (\log (x)-\text {$\#$1}) \text {$\#$1}^2 K[1]^2+\text {$\#$1}^2 K[1]^2+36 \log (x) \log (\log (x)-\text {$\#$1}) K[1]^2+4 \log (x) \log (\log (x)-\text {$\#$1}) \text {$\#$1} K[1]^2-40 \log (\log (x)-\text {$\#$1}) \text {$\#$1} K[1]^2+18 \text {$\#$1} K[1]^2-6 K[1]^2-2 \log (x) \log (\log (x)-\text {$\#$1}) K[1]+2 \log (\log (x)-\text {$\#$1}) K[1]-4 \log (\log (x)-\text {$\#$1}) \text {$\#$1} K[1]-2 \text {$\#$1} K[1]-18 K[1]+2 \log (\log (x)-\text {$\#$1})+1}{\log (x) \text {$\#$1}^2 K[1]^3-19 \text {$\#$1}^2 K[1]^3+110 \log (x) K[1]^3+18 \log (x) \text {$\#$1} K[1]^3-110 \text {$\#$1} K[1]^3-2 K[1]^3-\text {$\#$1}^2 K[1]^2-18 \log (x) K[1]^2-2 \log (x) \text {$\#$1} K[1]^2+20 \text {$\#$1} K[1]^2+\log (x) K[1]+2 \text {$\#$1} K[1]-K[1]-1}\&\right ]}{K[1]}\right )dK[1]+y(x)^2 \left (-\text {RootSum}\left [\text {$\#$1}^3 y(x)^3-\text {$\#$1}^2 y(x)^3-3 \text {$\#$1}^2 y(x)^2+2 \text {$\#$1} y(x)^2+3 \text {$\#$1} y(x)-2 y(x)^3-y(x)-1\&,\frac {\log (\log (x)-\text {$\#$1})}{3 \text {$\#$1}^2 y(x)^2-2 \text {$\#$1} y(x)^2-6 \text {$\#$1} y(x)+2 y(x)+3}\&\right ]\right )-\log (x)=c_1,y(x)\right ] \]