Internal problem ID [9001]
Internal file name [OUTPUT/7936_Monday_June_06_2022_12_57_57_AM_84589074/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 667.
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_linear_symmetries], [_Abel, `2nd type`, `class C`]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {y^{3} {\mathrm e}^{-2 x b}}{y \,{\mathrm e}^{-x b}+1}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {y^{3} {\mathrm e}^{-2 x b}}{{\mathrm e}^{-x b} y +1}\\ 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 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}+\frac {y^{3} {\mathrm e}^{-2 x b} \left (b_{3}-a_{2}\right )}{{\mathrm e}^{-x b} y +1}-\frac {y^{6} {\mathrm e}^{-4 x b} a_{3}}{\left ({\mathrm e}^{-x b} y +1\right )^{2}}-\left (\frac {y^{4} {\mathrm e}^{-2 x b} b \,{\mathrm e}^{-x b}}{\left ({\mathrm e}^{-x b} y +1\right )^{2}}-\frac {2 y^{3} b \,{\mathrm e}^{-2 x b}}{{\mathrm e}^{-x b} y +1}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {3 y^{2} {\mathrm e}^{-2 x b}}{{\mathrm e}^{-x b} y +1}-\frac {y^{3} {\mathrm e}^{-2 x b} {\mathrm e}^{-x b}}{\left ({\mathrm e}^{-x b} y +1\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \frac {{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} b x \,y^{4} a_{2}+{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} b \,y^{5} a_{3}-y^{6} {\mathrm e}^{-4 x b} a_{3}+{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} b \,y^{4} a_{1}-2 \,{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} x \,y^{3} b_{2}-{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} y^{4} a_{2}-{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} y^{4} b_{3}+2 \,{\mathrm e}^{-2 x b} b x \,y^{3} a_{2}+2 \,{\mathrm e}^{-2 x b} b \,y^{4} a_{3}-2 \,{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} y^{3} b_{1}+2 \,{\mathrm e}^{-2 x b} b \,y^{3} a_{1}+{\mathrm e}^{-2 x b} y^{2} b_{2}-3 \,{\mathrm e}^{-2 x b} x \,y^{2} b_{2}-{\mathrm e}^{-2 x b} y^{3} a_{2}-2 \,{\mathrm e}^{-2 x b} y^{3} b_{3}-3 \,{\mathrm e}^{-2 x b} y^{2} b_{1}+2 \,{\mathrm e}^{-x b} y b_{2}+b_{2}}{\left ({\mathrm e}^{-x b} y +1\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} {\mathrm e}^{-x b} {\mathrm e}^{-2 x b} b x \,y^{4} a_{2}+{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} b \,y^{5} a_{3}-y^{6} {\mathrm e}^{-4 x b} a_{3}+{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} b \,y^{4} a_{1}-2 \,{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} x \,y^{3} b_{2}-{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} y^{4} a_{2}-{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} y^{4} b_{3}+2 \,{\mathrm e}^{-2 x b} b x \,y^{3} a_{2}+2 \,{\mathrm e}^{-2 x b} b \,y^{4} a_{3}-2 \,{\mathrm e}^{-x b} {\mathrm e}^{-2 x b} y^{3} b_{1}+2 \,{\mathrm e}^{-2 x b} b \,y^{3} a_{1}+{\mathrm e}^{-2 x b} y^{2} b_{2}-3 \,{\mathrm e}^{-2 x b} x \,y^{2} b_{2}-{\mathrm e}^{-2 x b} y^{3} a_{2}-2 \,{\mathrm e}^{-2 x b} y^{3} b_{3}-3 \,{\mathrm e}^{-2 x b} y^{2} b_{1}+2 \,{\mathrm e}^{-x b} y b_{2}+b_{2} = 0 \end{equation} Simplifying the above gives \begin{equation} \tag{6E} b x \,y^{4} a_{2} {\mathrm e}^{-3 x b}+b \,y^{5} a_{3} {\mathrm e}^{-3 x b}-y^{6} {\mathrm e}^{-4 x b} a_{3}+b \,y^{4} a_{1} {\mathrm e}^{-3 x b}-2 x \,y^{3} b_{2} {\mathrm e}^{-3 x b}-y^{4} a_{2} {\mathrm e}^{-3 x b}-y^{4} b_{3} {\mathrm e}^{-3 x b}+2 \,{\mathrm e}^{-2 x b} b x \,y^{3} a_{2}+2 \,{\mathrm e}^{-2 x b} b \,y^{4} a_{3}-2 y^{3} b_{1} {\mathrm e}^{-3 x b}+2 \,{\mathrm e}^{-2 x b} b \,y^{3} a_{1}+{\mathrm e}^{-2 x b} y^{2} b_{2}-3 \,{\mathrm e}^{-2 x b} x \,y^{2} b_{2}-{\mathrm e}^{-2 x b} y^{3} a_{2}-2 \,{\mathrm e}^{-2 x b} y^{3} b_{3}-3 \,{\mathrm e}^{-2 x b} y^{2} b_{1}+2 \,{\mathrm e}^{-x b} y b_{2}+b_{2} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y, {\mathrm e}^{-4 x b}, {\mathrm e}^{-3 x b}, {\mathrm e}^{-2 x b}, {\mathrm e}^{-x b}\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}, {\mathrm e}^{-4 x b} = v_{3}, {\mathrm e}^{-3 x b} = v_{4}, {\mathrm e}^{-2 x b} = v_{5}, {\mathrm e}^{-x b} = v_{6}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} b v_{1} v_{2}^{4} a_{2} v_{4}+b v_{2}^{5} a_{3} v_{4}-v_{2}^{6} v_{3} a_{3}+b v_{2}^{4} a_{1} v_{4}+2 v_{5} b v_{1} v_{2}^{3} a_{2}+2 v_{5} b v_{2}^{4} a_{3}+2 v_{5} b v_{2}^{3} a_{1}-v_{2}^{4} a_{2} v_{4}-2 v_{1} v_{2}^{3} b_{2} v_{4}-v_{2}^{4} b_{3} v_{4}-v_{5} v_{2}^{3} a_{2}-2 v_{2}^{3} b_{1} v_{4}-3 v_{5} v_{1} v_{2}^{2} b_{2}-2 v_{5} v_{2}^{3} b_{3}-3 v_{5} v_{2}^{2} b_{1}+v_{5} v_{2}^{2} b_{2}+2 v_{6} v_{2} b_{2}+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} b v_{1} v_{2}^{4} a_{2} v_{4}-2 v_{1} v_{2}^{3} b_{2} v_{4}+2 v_{5} b v_{1} v_{2}^{3} a_{2}-3 v_{5} v_{1} v_{2}^{2} b_{2}-v_{2}^{6} v_{3} a_{3}+b v_{2}^{5} a_{3} v_{4}+\left (b a_{1}-a_{2}-b_{3}\right ) v_{2}^{4} v_{4}+2 v_{5} b v_{2}^{4} a_{3}-2 v_{2}^{3} b_{1} v_{4}+\left (2 b a_{1}-a_{2}-2 b_{3}\right ) v_{2}^{3} v_{5}+\left (-3 b_{1}+b_{2}\right ) v_{2}^{2} v_{5}+2 v_{6} v_{2} b_{2}+b_{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} b_{2}&=0\\ b a_{2}&=0\\ b a_{3}&=0\\ -a_{3}&=0\\ -2 b_{1}&=0\\ -3 b_{2}&=0\\ -2 b_{2}&=0\\ 2 b_{2}&=0\\ 2 b a_{2}&=0\\ 2 b a_{3}&=0\\ -3 b_{1}+b_{2}&=0\\ b a_{1}-a_{2}-b_{3}&=0\\ 2 b a_{1}-a_{2}-2 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=\frac {b_{3}}{b}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \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 &= \frac {1}{b} \\ \eta &= y \\ \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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{3} {\mathrm e}^{-2 x b}+y^{\prime } y \,{\mathrm e}^{-x b}+y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3} {\mathrm e}^{-2 x b}}{y \,{\mathrm e}^{-x b}+1} \end {array} \]
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 differential order: 1; found: 1 linear symmetries. Trying reduction of order 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful`
✓ Solution by Maple
Time used: 0.219 (sec). Leaf size: 95
dsolve(diff(y(x),x) = y(x)^3/(y(x)*exp(-b*x)+1)*exp(-2*b*x),y(x), singsol=all)
\[ -\frac {-2 \ln \left (y \left (x \right ) {\mathrm e}^{-b x}\right ) \sqrt {b \left (4+b \right )}+\left (-2 b x +\ln \left (-b y \left (x \right ) {\mathrm e}^{-b x}+y \left (x \right )^{2} {\mathrm e}^{-2 b x}-b \right )+2 c_{1} \right ) \sqrt {b \left (4+b \right )}+2 b \,\operatorname {arctanh}\left (\frac {-2 y \left (x \right ) {\mathrm e}^{-b x}+b}{\sqrt {b \left (4+b \right )}}\right )}{2 \sqrt {b \left (4+b \right )}} = 0 \]
✓ Solution by Mathematica
Time used: 2.618 (sec). Leaf size: 95
DSolve[y'[x] == y[x]^3/(E^(2*b*x)*(1 + y[x]/E^(b*x))),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {2 \sqrt {\frac {b}{b+4}} \text {arctanh}\left (\frac {\sqrt {\frac {b}{b+4}} \left (2 e^{b x}+y(x)\right )}{y(x)}\right )-\log \left (b e^{b x} \left (e^{b x}+y(x)\right )-y(x)^2\right )+2 \log \left (e^{b x}\right )}{2 b}+\frac {\log (y(x))}{b}=c_1,y(x)\right ] \]