Internal problem ID [8946]
Internal file name [OUTPUT/7881_Monday_June_06_2022_12_50_05_AM_96771428/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 612.
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_symmetry_[F(x),G(y)]`]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{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 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 \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right ) {\mathrm e}^{\frac {x^{2}}{4}} \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{2}-\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right )^{2} {\mathrm e}^{\frac {x^{2}}{2}} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{4}-\left (\frac {\left (-\frac {y \,x^{2} {\mathrm e}^{-\frac {x^{2}}{4}}}{2}+y \,{\mathrm e}^{-\frac {x^{2}}{4}}-D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) y \,{\mathrm e}^{-\frac {x^{2}}{4}} x \right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2}+\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right ) x \,{\mathrm e}^{\frac {x^{2}}{4}}}{4}\right ) \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (x \,{\mathrm e}^{-\frac {x^{2}}{4}}+2 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right )}{2} = 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} 2 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x^{3} y a_{4}-2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x^{2} a_{2}-2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x a_{1}-x^{3} y^{2} a_{5}-2 x^{2} y^{3} a_{6}+2 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x^{2} y^{2} a_{5}+2 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x \,y^{3} a_{6}-4 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x y b_{5}-x^{2} y^{2} a_{3}+2 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x^{2} y a_{2}+2 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x \,y^{2} a_{3}+2 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x y a_{1}-2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x^{3} a_{4}-8 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x a_{4}+4 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x b_{5}-4 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} y a_{5}+8 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} y b_{6}-4 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )^{2} {\mathrm e}^{\frac {x^{2}}{2}} x a_{5}-8 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )^{2} {\mathrm e}^{\frac {x^{2}}{2}} y a_{6}+4 b_{2}-6 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x y a_{3}-6 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x^{2} y a_{5}-10 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} x \,y^{2} a_{6}-4 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) b_{1}-2 x^{2} b_{2}-2 x b_{1}-2 y a_{1}-4 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x b_{2}+8 x b_{4}+4 y b_{5}-6 x^{2} y a_{4}-4 x \,y^{2} a_{5}+2 x \,y^{2} b_{6}-4 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) y^{2} b_{6}-4 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) x^{2} b_{4}-4 x y a_{2}-4 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )^{2} {\mathrm e}^{\frac {x^{2}}{2}} a_{3}-4 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} a_{2}+4 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} b_{3}-2 y^{2} a_{3}-4 D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) y b_{3}-2 y^{3} a_{6}-2 x^{3} b_{4} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ), {\mathrm e}^{-\frac {x^{2}}{4}}, {\mathrm e}^{\frac {x^{2}}{2}}, {\mathrm e}^{\frac {x^{2}}{4}}, D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) = v_{3}, {\mathrm e}^{-\frac {x^{2}}{4}} = v_{4}, {\mathrm e}^{\frac {x^{2}}{2}} = v_{5}, {\mathrm e}^{\frac {x^{2}}{4}} = v_{6}, D\left (F \right )\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right ) = v_{7}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 2 v_{7} v_{1}^{3} v_{2} a_{4}-2 v_{3} v_{6} v_{1}^{3} a_{4}-v_{1}^{3} v_{2}^{2} a_{5}+2 v_{7} v_{1}^{2} v_{2}^{2} a_{5}-6 v_{3} v_{6} v_{1}^{2} v_{2} a_{5}-2 v_{1}^{2} v_{2}^{3} a_{6}+2 v_{7} v_{1} v_{2}^{3} a_{6}-10 v_{3} v_{6} v_{1} v_{2}^{2} a_{6}+2 v_{7} v_{1}^{2} v_{2} a_{2}-2 v_{3} v_{6} v_{1}^{2} a_{2}-v_{1}^{2} v_{2}^{2} a_{3}+2 v_{7} v_{1} v_{2}^{2} a_{3}-6 v_{3} v_{6} v_{1} v_{2} a_{3}-4 v_{3}^{2} v_{5} v_{1} a_{5}-8 v_{3}^{2} v_{5} v_{2} a_{6}+2 v_{7} v_{1} v_{2} a_{1}-2 v_{3} v_{6} v_{1} a_{1}-4 v_{3}^{2} v_{5} a_{3}-6 v_{1}^{2} v_{2} a_{4}-8 v_{3} v_{6} v_{1} a_{4}-4 v_{1} v_{2}^{2} a_{5}-4 v_{3} v_{6} v_{2} a_{5}-2 v_{2}^{3} a_{6}-2 v_{1}^{3} b_{4}-4 v_{7} v_{1}^{2} b_{4}-4 v_{7} v_{1} v_{2} b_{5}+4 v_{3} v_{6} v_{1} b_{5}+2 v_{1} v_{2}^{2} b_{6}-4 v_{7} v_{2}^{2} b_{6}+8 v_{3} v_{6} v_{2} b_{6}-4 v_{1} v_{2} a_{2}-4 v_{3} v_{6} a_{2}-2 v_{2}^{2} a_{3}-2 v_{1}^{2} b_{2}-4 v_{7} v_{1} b_{2}-4 v_{7} v_{2} b_{3}+4 v_{3} v_{6} b_{3}-2 v_{2} a_{1}-2 v_{1} b_{1}-4 v_{7} b_{1}+8 v_{1} b_{4}+4 v_{2} b_{5}+4 b_{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, v_{7}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (-4 a_{5}+2 b_{6}\right ) v_{1} v_{2}^{2}+\left (-4 a_{2}+4 b_{3}\right ) v_{3} v_{6}+2 v_{7} v_{1}^{3} v_{2} a_{4}-2 v_{3} v_{6} v_{1}^{2} a_{2}+2 v_{7} v_{1}^{2} v_{2}^{2} a_{5}+2 v_{7} v_{1} v_{2}^{3} a_{6}+2 v_{7} v_{1}^{2} v_{2} a_{2}+2 v_{7} v_{1} v_{2}^{2} a_{3}-2 v_{3} v_{6} v_{1}^{3} a_{4}-4 v_{3}^{2} v_{5} v_{1} a_{5}-8 v_{3}^{2} v_{5} v_{2} a_{6}+4 b_{2}-6 v_{3} v_{6} v_{1} v_{2} a_{3}-6 v_{3} v_{6} v_{1}^{2} v_{2} a_{5}-10 v_{3} v_{6} v_{1} v_{2}^{2} a_{6}+\left (2 a_{1}-4 b_{5}\right ) v_{1} v_{2} v_{7}+\left (-2 a_{1}-8 a_{4}+4 b_{5}\right ) v_{1} v_{3} v_{6}+\left (-4 a_{5}+8 b_{6}\right ) v_{2} v_{3} v_{6}-v_{1}^{3} v_{2}^{2} a_{5}-2 v_{1}^{2} v_{2}^{3} a_{6}-v_{1}^{2} v_{2}^{2} a_{3}-4 v_{7} v_{1} b_{2}-6 v_{1}^{2} v_{2} a_{4}-4 v_{7} v_{2}^{2} b_{6}-4 v_{7} v_{1}^{2} b_{4}-4 v_{1} v_{2} a_{2}-4 v_{3}^{2} v_{5} a_{3}-4 v_{7} v_{2} b_{3}+\left (-2 b_{1}+8 b_{4}\right ) v_{1}-4 v_{7} b_{1}-2 v_{1}^{2} b_{2}-2 v_{2}^{2} a_{3}-2 v_{2}^{3} a_{6}-2 v_{1}^{3} b_{4}+\left (-2 a_{1}+4 b_{5}\right ) v_{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -4 a_{2}&=0\\ -2 a_{2}&=0\\ 2 a_{2}&=0\\ -6 a_{3}&=0\\ -4 a_{3}&=0\\ -2 a_{3}&=0\\ -a_{3}&=0\\ 2 a_{3}&=0\\ -6 a_{4}&=0\\ -2 a_{4}&=0\\ 2 a_{4}&=0\\ -6 a_{5}&=0\\ -4 a_{5}&=0\\ -a_{5}&=0\\ 2 a_{5}&=0\\ -10 a_{6}&=0\\ -8 a_{6}&=0\\ -2 a_{6}&=0\\ 2 a_{6}&=0\\ -4 b_{1}&=0\\ -4 b_{2}&=0\\ -2 b_{2}&=0\\ 4 b_{2}&=0\\ -4 b_{3}&=0\\ -4 b_{4}&=0\\ -2 b_{4}&=0\\ -4 b_{6}&=0\\ -2 a_{1}+4 b_{5}&=0\\ 2 a_{1}-4 b_{5}&=0\\ -4 a_{2}+4 b_{3}&=0\\ -4 a_{5}+2 b_{6}&=0\\ -4 a_{5}+8 b_{6}&=0\\ -2 b_{1}+8 b_{4}&=0\\ -2 a_{1}-8 a_{4}+4 b_{5}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=2 b_{5}\\ a_{2}&=0\\ 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}&=b_{5}\\ b_{6}&=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 &= 2 \\ \eta &= y 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2 F \left (y \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying an equivalence to an Abel ODE trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3`[1, 1/2*x*y]
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 38
dsolve(diff(y(x),x) = 1/2*(y(x)*exp(-1/4*x^2)*x+2*F(y(x)*exp(-1/4*x^2)))*exp(1/4*x^2),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \operatorname {RootOf}\left (F \left (\textit {\_Z} \,{\mathrm e}^{-\frac {x^{2}}{4}}\right )\right ) \\ y \left (x \right ) &= \operatorname {RootOf}\left (-x +\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )}d \textit {\_a} +c_{1} \right ) {\mathrm e}^{\frac {x^{2}}{4}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.349 (sec). Leaf size: 199
DSolve[y'[x] == (E^(x^2/4)*(2*F[y[x]/E^(x^2/4)] + (x*y[x])/E^(x^2/4)))/2,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}-\frac {e^{-\frac {x^2}{4}} \left (e^{\frac {x^2}{4}} F\left (e^{-\frac {x^2}{4}} K[2]\right ) \int _1^x\left (\frac {e^{-\frac {1}{4} K[1]^2} K[1]}{2 F\left (e^{-\frac {1}{4} K[1]^2} K[2]\right )}-\frac {e^{-\frac {1}{2} K[1]^2} K[1] K[2] F'\left (e^{-\frac {1}{4} K[1]^2} K[2]\right )}{2 F\left (e^{-\frac {1}{4} K[1]^2} K[2]\right )^2}\right )dK[1]+1\right )}{F\left (e^{-\frac {x^2}{4}} K[2]\right )}dK[2]+\int _1^x\left (\frac {e^{-\frac {1}{4} K[1]^2} K[1] y(x)}{2 F\left (e^{-\frac {1}{4} K[1]^2} y(x)\right )}+1\right )dK[1]=c_1,y(x)\right ] \]