Internal problem ID [9258]
Internal file name [OUTPUT/8194_Monday_June_06_2022_02_14_35_AM_40385857/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 925.
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]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {y^{2}+2 y x +x^{2}+{\mathrm e}^{2 \left (x -y\right )^{2} \left (x +y\right )^{2}}}{y^{2}+2 y x +x^{2}-{\mathrm e}^{2 \left (x -y\right )^{2} \left (x +y\right )^{2}}}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {y^{2}+2 y x +x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}}{-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}}\\ 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 {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right ) \left (b_{3}-a_{2}\right )}{-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}}-\frac {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right )^{2} a_{3}}{\left (-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right )^{2}}-\left (-\frac {2 y +2 x +\left (-4 \left (y -x \right ) \left (x +y \right )^{2}+4 \left (y -x \right )^{2} \left (x +y \right )\right ) {\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}}{-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}}+\frac {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right ) \left (-2 y -2 x +\left (-4 \left (y -x \right ) \left (x +y \right )^{2}+4 \left (y -x \right )^{2} \left (x +y \right )\right ) {\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right )}{\left (-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {2 y +2 x +\left (4 \left (y -x \right ) \left (x +y \right )^{2}+4 \left (y -x \right )^{2} \left (x +y \right )\right ) {\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}}{-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}}+\frac {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right ) \left (-2 y -2 x +\left (4 \left (y -x \right ) \left (x +y \right )^{2}+4 \left (y -x \right )^{2} \left (x +y \right )\right ) {\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right )}{\left (-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}\right )^{2}}\right ) \left (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} 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. \[ \{x, y, {\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}}, {\mathrm e}^{4 \left (y -x \right )^{2} \left (x +y \right )^{2}}\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}, {\mathrm e}^{2 \left (y -x \right )^{2} \left (x +y \right )^{2}} = v_{3}, {\mathrm e}^{4 \left (y -x \right )^{2} \left (x +y \right )^{2}} = v_{4}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -16 v_{3} v_{1}^{6} a_{2}-32 v_{3} v_{1}^{5} v_{2} a_{2}+32 v_{3} v_{1}^{3} v_{2}^{3} a_{2}+16 v_{3} v_{1}^{2} v_{2}^{4} a_{2}-16 v_{3} v_{1}^{5} v_{2} a_{3}-32 v_{3} v_{1}^{4} v_{2}^{2} a_{3}+32 v_{3} v_{1}^{2} v_{2}^{4} a_{3}+16 v_{3} v_{1} v_{2}^{5} a_{3}+16 v_{3} v_{1}^{5} v_{2} b_{2}+32 v_{3} v_{1}^{4} v_{2}^{2} b_{2}-32 v_{3} v_{1}^{2} v_{2}^{4} b_{2}-16 v_{3} v_{1} v_{2}^{5} b_{2}+16 v_{3} v_{1}^{4} v_{2}^{2} b_{3}+32 v_{3} v_{1}^{3} v_{2}^{3} b_{3}-32 v_{3} v_{1} v_{2}^{5} b_{3}-16 v_{3} v_{2}^{6} b_{3}-16 v_{3} v_{1}^{5} a_{1}-32 v_{3} v_{1}^{4} v_{2} a_{1}+32 v_{3} v_{1}^{2} v_{2}^{3} a_{1}+16 v_{3} v_{1} v_{2}^{4} a_{1}+16 v_{3} v_{1}^{4} v_{2} b_{1}+32 v_{3} v_{1}^{3} v_{2}^{2} b_{1}-32 v_{3} v_{1} v_{2}^{4} b_{1}-16 v_{3} v_{2}^{5} b_{1}-v_{1}^{4} a_{2}-4 v_{1}^{3} v_{2} a_{2}-6 v_{1}^{2} v_{2}^{2} a_{2}-4 v_{1} v_{2}^{3} a_{2}-v_{2}^{4} a_{2}-v_{1}^{4} a_{3}-4 v_{1}^{3} v_{2} a_{3}-6 v_{1}^{2} v_{2}^{2} a_{3}-4 v_{1} v_{2}^{3} a_{3}-v_{2}^{4} a_{3}+v_{1}^{4} b_{2}+4 v_{1}^{3} v_{2} b_{2}+6 v_{1}^{2} v_{2}^{2} b_{2}+4 v_{1} v_{2}^{3} b_{2}+v_{2}^{4} b_{2}+v_{1}^{4} b_{3}+4 v_{1}^{3} v_{2} b_{3}+6 v_{1}^{2} v_{2}^{2} b_{3}+4 v_{1} v_{2}^{3} b_{3}+v_{2}^{4} b_{3}+4 v_{3} v_{1}^{2} a_{2}+4 v_{3} v_{1} v_{2} a_{2}-2 v_{3} v_{1}^{2} a_{3}+2 v_{3} v_{2}^{2} a_{3}+2 v_{3} v_{1}^{2} b_{2}-2 v_{3} v_{2}^{2} b_{2}+4 v_{3} v_{1} v_{2} b_{3}+4 v_{3} v_{2}^{2} b_{3}+4 v_{3} v_{1} a_{1}+4 v_{3} v_{2} a_{1}+4 v_{3} v_{1} b_{1}+4 v_{3} v_{2} b_{1}+v_{4} a_{2}-v_{4} a_{3}+v_{4} b_{2}-v_{4} b_{3} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -16 v_{3} v_{1}^{6} a_{2}+\left (-32 a_{2}-16 a_{3}+16 b_{2}\right ) v_{1}^{5} v_{2} v_{3}-16 v_{3} v_{1}^{5} a_{1}+\left (-32 a_{3}+32 b_{2}+16 b_{3}\right ) v_{1}^{4} v_{2}^{2} v_{3}+\left (-32 a_{1}+16 b_{1}\right ) v_{1}^{4} v_{2} v_{3}+\left (-a_{2}-a_{3}+b_{2}+b_{3}\right ) v_{1}^{4}+\left (32 a_{2}+32 b_{3}\right ) v_{1}^{3} v_{2}^{3} v_{3}+32 v_{3} v_{1}^{3} v_{2}^{2} b_{1}+\left (-4 a_{2}-4 a_{3}+4 b_{2}+4 b_{3}\right ) v_{1}^{3} v_{2}+\left (16 a_{2}+32 a_{3}-32 b_{2}\right ) v_{1}^{2} v_{2}^{4} v_{3}+32 v_{3} v_{1}^{2} v_{2}^{3} a_{1}+\left (-6 a_{2}-6 a_{3}+6 b_{2}+6 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (4 a_{2}-2 a_{3}+2 b_{2}\right ) v_{1}^{2} v_{3}+\left (16 a_{3}-16 b_{2}-32 b_{3}\right ) v_{1} v_{2}^{5} v_{3}+\left (16 a_{1}-32 b_{1}\right ) v_{1} v_{2}^{4} v_{3}+\left (-4 a_{2}-4 a_{3}+4 b_{2}+4 b_{3}\right ) v_{1} v_{2}^{3}+\left (4 a_{2}+4 b_{3}\right ) v_{1} v_{2} v_{3}+\left (4 a_{1}+4 b_{1}\right ) v_{1} v_{3}-16 v_{3} v_{2}^{6} b_{3}-16 v_{3} v_{2}^{5} b_{1}+\left (-a_{2}-a_{3}+b_{2}+b_{3}\right ) v_{2}^{4}+\left (2 a_{3}-2 b_{2}+4 b_{3}\right ) v_{2}^{2} v_{3}+\left (4 a_{1}+4 b_{1}\right ) v_{2} v_{3}+\left (a_{2}-a_{3}+b_{2}-b_{3}\right ) v_{4} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -16 a_{1}&=0\\ 32 a_{1}&=0\\ -16 a_{2}&=0\\ -16 b_{1}&=0\\ 32 b_{1}&=0\\ -16 b_{3}&=0\\ -32 a_{1}+16 b_{1}&=0\\ 4 a_{1}+4 b_{1}&=0\\ 16 a_{1}-32 b_{1}&=0\\ 4 a_{2}+4 b_{3}&=0\\ 32 a_{2}+32 b_{3}&=0\\ -32 a_{2}-16 a_{3}+16 b_{2}&=0\\ 4 a_{2}-2 a_{3}+2 b_{2}&=0\\ 16 a_{2}+32 a_{3}-32 b_{2}&=0\\ -32 a_{3}+32 b_{2}+16 b_{3}&=0\\ 2 a_{3}-2 b_{2}+4 b_{3}&=0\\ 16 a_{3}-16 b_{2}-32 b_{3}&=0\\ -6 a_{2}-6 a_{3}+6 b_{2}+6 b_{3}&=0\\ -4 a_{2}-4 a_{3}+4 b_{2}+4 b_{3}&=0\\ -a_{2}-a_{3}+b_{2}+b_{3}&=0\\ a_{2}-a_{3}+b_{2}-b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=b_{2}\\ b_{1}&=0\\ b_{2}&=b_{2}\\ 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 &= y \\ \eta &= x \\ \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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{2} y^{\prime }-2 y^{\prime } x y-x^{2} y^{\prime }+{\mathrm e}^{2 \left (y-x \right )^{2} \left (x +y\right )^{2}} y^{\prime }+y^{2}+2 y x +x^{2}+{\mathrm e}^{2 \left (y-x \right )^{2} \left (x +y\right )^{2}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y^{2}-2 y x -x^{2}-{\mathrm e}^{2 \left (y-x \right )^{2} \left (x +y\right )^{2}}}{-y^{2}-2 y x -x^{2}+{\mathrm e}^{2 \left (y-x \right )^{2} \left (x +y\right )^{2}}} \end {array} \]
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: 0.109 (sec). Leaf size: 38
dsolve(diff(y(x),x) = (y(x)^2+2*x*y(x)+x^2+exp(2*(x-y(x))^2*(x+y(x))^2))/(y(x)^2+2*x*y(x)+x^2-exp(2*(x-y(x))^2*(x+y(x))^2)),y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\operatorname {RootOf}\left (-\textit {\_Z} +\int _{}^{{\mathrm e}^{2 \textit {\_Z}}-2 x \,{\mathrm e}^{\textit {\_Z}}}\frac {1}{{\mathrm e}^{2 \textit {\_a}^{2}}+\textit {\_a}}d \textit {\_a} +c_{1} \right )}-x \]
✓ Solution by Mathematica
Time used: 12.399 (sec). Leaf size: 228
DSolve[y'[x] == (E^(2*(x - y[x])^2*(x + y[x])^2) + x^2 + 2*x*y[x] + y[x]^2)/(-E^(2*(x - y[x])^2*(x + y[x])^2) + x^2 + 2*x*y[x] + y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (-\frac {2 K[2]}{-x^2+e^{2 (x-K[2])^2 (x+K[2])^2}+K[2]^2}-\int _1^x\left (\frac {2 K[1] \left (-2 K[2]-e^{2 (K[1]-K[2])^2 (K[1]+K[2])^2} \left (4 (K[1]-K[2])^2 (K[1]+K[2])-4 (K[1]-K[2]) (K[1]+K[2])^2\right )\right )}{\left (K[1]^2-e^{2 (K[1]-K[2])^2 (K[1]+K[2])^2}-K[2]^2\right )^2}-\frac {1}{(K[1]+K[2])^2}\right )dK[1]+\frac {1}{x+K[2]}\right )dK[2]+\int _1^x\left (\frac {1}{K[1]+y(x)}-\frac {2 K[1]}{K[1]^2-e^{2 (K[1]-y(x))^2 (K[1]+y(x))^2}-y(x)^2}\right )dK[1]=c_1,y(x)\right ] \]