2.208 problem 784

Internal problem ID [9768]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, Additional non-linear first order
Problem number : 784
Date solved : Friday, October 11, 2024 at 11:38:21 AM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]

Solve

\begin{align*} y^{\prime }&=\frac {-\sinh \left (x \right )+x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +\ln \left (x \right )+y^{2} \ln \left (x \right )}{\sinh \left (x \right )} \end{align*}

2.208.1 Solved using Lie symmetry for first order ode

Time used: 1.575 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=\frac {x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +y^{2} \ln \left (x \right )-\sinh \left (x \right )+\ln \left (x \right )}{\sinh \left (x \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 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 (x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +y^{2} \ln \left (x \right )-\sinh \left (x \right )+\ln \left (x \right )\right ) \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{\sinh \left (x \right )}-\frac {\left (x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +y^{2} \ln \left (x \right )-\sinh \left (x \right )+\ln \left (x \right )\right )^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{\sinh \left (x \right )^{2}}-\left (\frac {2 \ln \left (x \right ) x +x +2 y +2 y \ln \left (x \right )+\frac {y^{2}}{x}-\cosh \left (x \right )+\frac {1}{x}}{\sinh \left (x \right )}-\frac {\left (x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +y^{2} \ln \left (x \right )-\sinh \left (x \right )+\ln \left (x \right )\right ) \cosh \left (x \right )}{\sinh \left (x \right )^{2}}\right ) \left (x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (2 \ln \left (x \right ) x +2 y \ln \left (x \right )\right ) \left (x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right )}{\sinh \left (x \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.

\[ \{x, y, \cosh \left (x \right ), \cosh \left (2 x \right ), \ln \left (x \right ), \sinh \left (x \right )\} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \{x = v_{1}, y = v_{2}, \cosh \left (x \right ) = v_{3}, \cosh \left (2 x \right ) = v_{4}, \ln \left (x \right ) = v_{5}, \sinh \left (x \right ) = v_{6}\} \]

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

\[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}\} \]

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_{2}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{6}&=0\\ -a_{1}&=0\\ -a_{2}&=0\\ -6 a_{3}&=0\\ -4 a_{3}&=0\\ -2 a_{3}&=0\\ -a_{3}&=0\\ -a_{4}&=0\\ -2 a_{5}&=0\\ -a_{5}&=0\\ -4 a_{6}&=0\\ -2 a_{6}&=0\\ -a_{6}&=0\\ -2 a_{1}-a_{5}&=0\\ -a_{1}-a_{6}&=0\\ -2 a_{2}-a_{3}&=0\\ -a_{2}-2 a_{3}&=0\\ -a_{2}-b_{3}&=0\\ a_{2}+2 a_{3}&=0\\ 2 a_{2}+a_{3}&=0\\ -2 a_{4}-a_{5}&=0\\ -a_{4}-a_{1}&=0\\ a_{4}+a_{1}&=0\\ 2 a_{4}+a_{5}&=0\\ -6 a_{5}-8 a_{6}&=0\\ -4 a_{5}-12 a_{6}&=0\\ -4 a_{5}-4 a_{6}&=0\\ -4 a_{5}-2 a_{6}&=0\\ -2 a_{5}-8 a_{6}&=0\\ -a_{5}-8 a_{6}&=0\\ -a_{5}-2 a_{6}&=0\\ a_{5}+2 a_{1}&=0\\ a_{5}+2 a_{6}&=0\\ a_{6}+a_{1}&=0\\ 2 a_{6}-a_{5}&=0\\ -4 a_{2}+2 a_{3}-2 b_{2}&=0\\ -a_{2}+2 a_{3}+b_{3}&=0\\ -a_{4}-2 a_{5}-a_{6}&=0\\ a_{4}+2 a_{5}+a_{6}&=0\\ -3 a_{2}+2 a_{3}-2 b_{2}+b_{3}&=0\\ -a_{4}-b_{4}+\frac {b_{5}}{2}+\frac {a_{5}}{2}&=0\\ \frac {a_{5}}{2}-a_{6}+\frac {b_{5}}{2}-b_{6}&=0\\ a_{6}-\frac {a_{5}}{2}-\frac {b_{5}}{2}+b_{6}&=0\\ -\frac {b_{3}}{2}+\frac {a_{2}}{2}-\frac {a_{3}}{2}+\frac {b_{2}}{2}&=0\\ \frac {b_{3}}{2}+\frac {a_{3}}{2}-\frac {a_{2}}{2}-\frac {b_{2}}{2}&=0\\ b_{4}-\frac {b_{5}}{2}+a_{4}-\frac {a_{5}}{2}&=0\\ b_{5}-4 a_{4}+2 a_{5}-2 b_{4}&=0\\ -6 a_{4}+a_{5}+4 a_{6}-2 b_{4}+2 b_{6}&=0\\ -2 a_{4}-2 a_{5}+6 a_{6}-b_{5}+2 b_{6}&=0\\ -2 a_{4}+2 a_{5}+b_{5}-2 a_{1}-2 b_{1}&=0\\ -a_{5}+4 a_{6}+2 b_{6}-2 a_{1}-2 b_{1}&=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\\ b_{1}&=b_{6}\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=b_{6}\\ b_{5}&=2 b_{6}\\ 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 &= 0 \\ \eta &= x^{2}+2 x y +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 \(S(R)\). 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}{x^{2}+2 x y +y^{2}+1}} dy \end{align*}

Which results in

\begin{align*} S&= \arctan \left (x +y \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 {x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +y^{2} \ln \left (x \right )-\sinh \left (x \right )+\ln \left (x \right )}{\sinh \left (x \right )} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {1}{\left (x +y \right )^{2}+1}\\ S_{y} &= \frac {1}{\left (x +y \right )^{2}+1} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \ln \left (x \right ) \operatorname {csch}\left (x \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} &= \ln \left (R \right ) \operatorname {csch}\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 \(R,S\).

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 {\ln \left (R \right ) \operatorname {csch}\left (R \right )\, dR}\\ S \left (R \right ) &= \int \ln \left (R \right ) \operatorname {csch}\left (R \right )d R + c_2 \end{align*}

\begin{align*} S \left (R \right )&= \int \ln \left (R \right ) \operatorname {csch}\left (R \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*} \arctan \left (x +y\right ) = \int \ln \left (x \right ) \operatorname {csch}\left (x \right )d x +c_2 \end{align*}

Which gives

\begin{align*} y = -x +\tan \left (\int \ln \left (x \right ) \operatorname {csch}\left (x \right )d x +c_2 \right ) \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 \((R,S)\)
\( \frac {dy}{dx} = \frac {x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +y^{2} \ln \left (x \right )-\sinh \left (x \right )+\ln \left (x \right )}{\sinh \left (x \right )}\)		\( \frac {d S}{d R} = \ln \left (R \right ) \operatorname {csch}\left (R \right )\)
	\(\!\begin {aligned} R&= x\\ S&= \arctan \left (x +y \right ) \end {aligned} \)

Figure 792: Slope field plot
\(y^{\prime } = \frac {-\sinh \left (x \right )+x^{2} \ln \left (x \right )+2 y \ln \left (x \right ) x +\ln \left (x \right )+y^{2} \ln \left (x \right )}{\sinh \left (x \right )}\)

2.208.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-\sinh \left (x \right )+x^{2} \ln \left (x \right )+2 y \left (x \right ) \ln \left (x \right ) x +\ln \left (x \right )+y \left (x \right )^{2} \ln \left (x \right )}{\sinh \left (x \right )} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-\sinh \left (x \right )+x^{2} \ln \left (x \right )+2 y \left (x \right ) \ln \left (x \right ) x +\ln \left (x \right )+y \left (x \right )^{2} \ln \left (x \right )}{\sinh \left (x \right )} \end {array} \]

2.208.3 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 Riccati 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (2*ln(x)^2*x^2-x*cosh(x)*ln(x)+sinh(x))*(diff(y(x), x))/(sinh(x)*ln(x) 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
      -> Trying changes of variables to rationalize or make the ODE simpler 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
            trying a symmetry of the form [xi=0, eta=F(x)] 
            trying 2nd order exact linear 
            trying symmetries linear in x and y(x) 
            trying to convert to a linear ODE with constant coefficients 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
            trying a symmetry of the form [xi=0, eta=F(x)] 
            trying 2nd order exact linear 
            trying symmetries linear in x and y(x) 
            trying to convert to a linear ODE with constant coefficients 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
            trying a symmetry of the form [xi=0, eta=F(x)] 
            trying 2nd order exact linear 
            trying symmetries linear in x and y(x) 
            trying to convert to a linear ODE with constant coefficients 
            -> trying with_periodic_functions in the coefficients 
      <- unable to find a useful change of variables 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
         trying 2nd order, integrating factor of the form mu(x,y) 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
         -> Trying changes of variables to rationalize or make the ODE simpler 
            trying a symmetry of the form [xi=0, eta=F(x)] 
            checking if the LODE is missing y 
            -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
            -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
               trying a symmetry of the form [xi=0, eta=F(x)] 
               trying 2nd order exact linear 
               trying symmetries linear in x and y(x) 
               trying to convert to a linear ODE with constant coefficients 
            trying a symmetry of the form [xi=0, eta=F(x)] 
            checking if the LODE is missing y 
            -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
            -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
               trying a symmetry of the form [xi=0, eta=F(x)] 
               trying 2nd order exact linear 
               trying symmetries linear in x and y(x) 
               trying to convert to a linear ODE with constant coefficients 
            trying a symmetry of the form [xi=0, eta=F(x)] 
            checking if the LODE is missing y 
            -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
            -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
               trying a symmetry of the form [xi=0, eta=F(x)] 
               trying 2nd order exact linear 
               trying symmetries linear in x and y(x) 
               trying to convert to a linear ODE with constant coefficients 
         <- unable to find a useful change of variables 
            trying a symmetry of the form [xi=0, eta=F(x)] 
         trying to convert to an ODE of Bessel type 
   -> Trying a change of variables to reduce to Bernoulli 
   -> Calling odsolve with the ODE`, diff(y(x), x)-(ln(x)*y(x)^2/sinh(x)+y(x)+2*x^2*ln(x)*y(x)/sinh(x)+x^2*(x^2*ln(x)/sinh(x)-1+ln(x 
      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 Riccati 
      trying Riccati sub-methods: 
         trying Riccati_symmetries 
      trying inverse_Riccati 
      trying 1st order ODE linearizable_by_differentiation 
   -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
   -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
   -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
   <- symmetry pattern of the form [F(x),G(x)*y+H(x)] successful 
   <- Riccati with symmetry pattern of the form [F(x),G(x)*y+H(x)] successful`
 

2.208.4 Maple dsolve solution

Solving time : 0.006 (sec)
Leaf size : 22

dsolve(diff(y(x),x) = (-sinh(x)+x^2*ln(x)+2*y(x)*ln(x)*x+ln(x)+y(x)^2*ln(x))/sinh(x), 
       y(x),singsol=all)
 

\[ y = -x -\tan \left (c_{1} -\left (\int \ln \left (x \right ) \operatorname {csch}\left (x \right )d x \right )\right ) \]

2.208.5 Mathematica DSolve solution

Solving time : 84.501 (sec)
Leaf size : 27

DSolve[{D[y[x],x] == Csch[x]*(Log[x] + x^2*Log[x] - Sinh[x] + 2*x*Log[x]*y[x] + Log[x]*y[x]^2),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to -x+\tan \left (\int _1^x\text {csch}(K[5]) \log (K[5])dK[5]+c_1\right ) \]