1.560 problem 563

Internal problem ID [9542]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, linear first order
Problem number : 563
Date solved : Thursday, October 17, 2024 at 08:45:04 PM
CAS classification : [[_1st_order, _with_linear_symmetries], _dAlembert]

Solve

\begin{align*} \ln \left (y^{\prime }\right )+x y^{\prime }+a y+b&=0 \end{align*}

1.560.1 Solved using Lie symmetry for first order ode

Time used: 1.073 (sec)

Writing the ode as

\begin{align*} y^{\prime }&={\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b}\\ 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 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \left (b_{3}-a_{2}\right )-{\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} a_{3}+\frac {\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) {\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \left (x a_{2}+y a_{3}+a_{1}\right )}{\left (1+\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )\right ) x}-\left (\frac {a \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )}{1+\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )}-a \right ) {\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {{\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} a_{3} x \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a \,x^{2} b_{2}-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a x y b_{3}+{\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} a_{3} x -{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) x b_{3}-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) y a_{3}-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a x b_{1}-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) a_{1}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} x a_{2}-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} x b_{3}-x b_{2} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-x b_{2}}{\left (1+\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )\right ) x} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -{\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} a_{3} x \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a \,x^{2} b_{2}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a x y b_{3}-{\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} a_{3} x +{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) x b_{3}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) y a_{3}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a x b_{1}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) a_{1}-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} x a_{2}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} x b_{3}+x b_{2} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )+x b_{2} = 0 \end{equation}

Simplifying the above gives

\begin{equation} \tag{6E} -{\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} a_{3} x \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a \,x^{2} b_{2}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a x y b_{3}-{\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} a_{3} x +{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) x b_{3}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) y a_{3}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} a x b_{1}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) a_{1}-{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} x a_{2}+{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} x b_{3}+x b_{2} \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )+x b_{2} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \{x, y, {\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b}, {\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b}, {\mathrm e}^{-a y -b}, \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )\} \]

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

\[ \{x = v_{1}, y = v_{2}, {\mathrm e}^{-2 \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-2 a y -2 b} = v_{3}, {\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} = v_{4}, {\mathrm e}^{-a y -b} = v_{5}, \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right ) = v_{6}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} v_{4} a v_{1}^{2} b_{2}+v_{4} a v_{1} v_{2} b_{3}+v_{4} a v_{1} b_{1}-v_{3} a_{3} v_{1} v_{6}+v_{4} v_{6} v_{2} a_{3}+v_{4} v_{6} v_{1} b_{3}+v_{4} v_{6} a_{1}-v_{4} v_{1} a_{2}-v_{3} a_{3} v_{1}+v_{1} b_{2} v_{6}+v_{4} v_{1} b_{3}+v_{1} b_{2} = 0 \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} v_{4} a v_{1}^{2} b_{2}+v_{4} a v_{1} v_{2} b_{3}-v_{3} a_{3} v_{1} v_{6}-v_{3} a_{3} v_{1}+v_{4} v_{6} v_{1} b_{3}+\left (a b_{1}-a_{2}+b_{3}\right ) v_{1} v_{4}+v_{1} b_{2} v_{6}+v_{1} b_{2}+v_{4} v_{6} v_{2} a_{3}+v_{4} v_{6} a_{1} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} a_{1}&=0\\ a_{3}&=0\\ b_{2}&=0\\ b_{3}&=0\\ a b_{2}&=0\\ a b_{3}&=0\\ -a_{3}&=0\\ a b_{1}-a_{2}+b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=a b_{1}\\ a_{3}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ b_{3}&=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 &= a x \\ \eta &= 1 \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= 1 - \left ({\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b}\right ) \left (a x\right ) \\ &= 1-a x \,{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )} {\mathrm e}^{-a y} {\mathrm e}^{-b}\\ \xi &= 0 \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}{1-a x \,{\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )} {\mathrm e}^{-a y} {\mathrm e}^{-b}}} dy \end{align*}

Which results in

\begin{align*} S&= -\frac {\frac {\left (-a -1\right ) \ln \left (a \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-1\right )}{a}+\ln \left (\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )\right )}{a} \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) &= {\mathrm e}^{-\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-a y -b} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {1}{a \left (a \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-1\right ) x}\\ S_{y} &= -\frac {1}{a \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y -b}\right )-1} \end{align*}

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

\begin{align*} \frac {dS}{dR} &= -\frac {1}{a x}\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 {1}{a R} \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 {1}{a R}\, dR}\\ S \left (R \right ) &= -\frac {\ln \left (R \right )}{a} + 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*} \frac {\left (a +1\right ) \ln \left (a \operatorname {LambertW}\left (x \,{\mathrm e}^{-a y-b}\right )-1\right )+a \left (a y+b -\ln \left (x \right )+\operatorname {LambertW}\left (x \,{\mathrm e}^{-a y-b}\right )\right )}{a^{2}} = -\frac {\ln \left (x \right )}{a}+c_2 \end{align*}

1.560.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \ln \left (\frac {d}{d x}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+a y \left (x \right )+b =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )={\mathrm e}^{-\mathit {LambertW}\left (x \,{\mathrm e}^{-a y \left (x \right )-b}\right )-a y \left (x \right )-b} \end {array} \]

1.560.3 Maple trace

`Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
trying dAlembert 
<- dAlembert successful`
 

1.560.4 Maple dsolve solution

Solving time : 0.086 (sec)
Leaf size : 73

dsolve(ln(diff(y(x),x))+x*diff(y(x),x)+y(x)*a+b = 0, 
       y(x),singsol=all)
 

\[ \frac {-a \left ({\left (\frac {\operatorname {LambertW}\left (x \,{\mathrm e}^{-y a -b}\right )}{x}\right )}^{-\frac {1}{a +1}} c_1 -x \right ) \operatorname {LambertW}\left (x \,{\mathrm e}^{-y a -b}\right )-x}{a \operatorname {LambertW}\left (x \,{\mathrm e}^{-y a -b}\right )} = 0 \]

1.560.5 Mathematica DSolve solution

Solving time : 0.128 (sec)
Leaf size : 59

DSolve[{b + Log[D[y[x],x]] + a*y[x] + x*D[y[x],x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [a \left (\frac {(a+1) \log \left (1-a W\left (x e^{-a y(x)-b}\right )\right )}{a^2}+\frac {W\left (x e^{-a y(x)-b}\right )}{a}\right )+a y(x)=c_1,y(x)\right ] \]