Internal problem ID [8942]
Internal file name [OUTPUT/7877_Monday_June_06_2022_12_49_42_AM_35821911/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 608.
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.
\[ \boxed {y^{\prime }-\frac {\sqrt {y}}{\sqrt {y}+F \left (\frac {x -y}{\sqrt {y}}\right )}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {\sqrt {y}}{\sqrt {y}+F \left (-\frac {y -x}{\sqrt {y}}\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*}
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 {\sqrt {y}\, \left (b_{3}-a_{2}\right )}{\sqrt {y}+F \left (-\frac {y -x}{\sqrt {y}}\right )}-\frac {y a_{3}}{\left (\sqrt {y}+F \left (-\frac {y -x}{\sqrt {y}}\right )\right )^{2}}+\frac {D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )}{\left (\sqrt {y}+F \left (-\frac {y -x}{\sqrt {y}}\right )\right )^{2}}-\left (\frac {1}{2 \sqrt {y}\, \left (\sqrt {y}+F \left (-\frac {y -x}{\sqrt {y}}\right )\right )}-\frac {\sqrt {y}\, \left (\frac {1}{2 \sqrt {y}}+D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) \left (-\frac {1}{\sqrt {y}}+\frac {y -x}{2 y^{{3}/{2}}}\right )\right )}{\left (\sqrt {y}+F \left (-\frac {y -x}{\sqrt {y}}\right )\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {2 y^{2} a_{2}-2 y^{2} b_{2}-2 y^{2} b_{3}+2 y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) a_{2}-4 y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) b_{2}-y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) b_{3}-2 y F \left (\frac {x -y}{\sqrt {y}}\right )^{2} b_{2}-2 y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x a_{2}+y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x b_{2}-2 y^{2} D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) a_{3}+y^{2} D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) b_{3}-2 y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) a_{1}+y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) b_{1}+2 y^{2} a_{3}+\sqrt {y}\, F \left (\frac {x -y}{\sqrt {y}}\right ) x b_{2}+D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x^{2} b_{2}+D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x y b_{3}+\sqrt {y}\, F \left (\frac {x -y}{\sqrt {y}}\right ) b_{1}+D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x b_{1}}{2 y \left (\sqrt {y}+F \left (\frac {x -y}{\sqrt {y}}\right )\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -2 y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) a_{2}+4 y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) b_{2}+y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) b_{3}-\sqrt {y}\, F \left (\frac {x -y}{\sqrt {y}}\right ) x b_{2}+2 y F \left (\frac {x -y}{\sqrt {y}}\right )^{2} b_{2}-D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x^{2} b_{2}+2 y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x a_{2}-y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x b_{2}-D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x y b_{3}+2 y^{2} D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) a_{3}-y^{2} D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) b_{3}-\sqrt {y}\, F \left (\frac {x -y}{\sqrt {y}}\right ) b_{1}-D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x b_{1}+2 y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) a_{1}-y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) b_{1}-2 y^{2} a_{2}-2 y^{2} a_{3}+2 y^{2} b_{2}+2 y^{2} b_{3} = 0 \end{equation} Simplifying the above gives \begin{equation} \tag{6E} -\sqrt {y}\, \left (2 y^{2} a_{2}-2 y^{2} b_{2}-2 y^{2} b_{3}+2 y^{{3}/{2}} F \left (-\frac {y -x}{\sqrt {y}}\right ) a_{2}-4 y^{{3}/{2}} F \left (-\frac {y -x}{\sqrt {y}}\right ) b_{2}-y^{{3}/{2}} F \left (-\frac {y -x}{\sqrt {y}}\right ) b_{3}-2 y D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) x a_{2}+D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) x y b_{2}-2 y^{2} D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) a_{3}+D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) y^{2} b_{3}-2 y F \left (-\frac {y -x}{\sqrt {y}}\right )^{2} b_{2}-2 y D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) a_{1}+D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) y b_{1}+2 y^{2} a_{3}+\sqrt {y}\, F \left (-\frac {y -x}{\sqrt {y}}\right ) x b_{2}+D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) x^{2} b_{2}+D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) x y b_{3}+\sqrt {y}\, F \left (-\frac {y -x}{\sqrt {y}}\right ) b_{1}+D\left (F \right )\left (-\frac {y -x}{\sqrt {y}}\right ) x b_{1}\right ) = 0 \end{equation} Since the PDE has radicals, simplifying gives \[ -\sqrt {y}\, \left (2 y^{2} a_{2}-2 y^{2} b_{2}-2 y^{2} b_{3}+2 y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) a_{2}-4 y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) b_{2}-y^{{3}/{2}} F \left (\frac {x -y}{\sqrt {y}}\right ) b_{3}-2 y F \left (\frac {x -y}{\sqrt {y}}\right )^{2} b_{2}-2 y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x a_{2}+y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x b_{2}-2 y^{2} D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) a_{3}+y^{2} D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) b_{3}-2 y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) a_{1}+y D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) b_{1}+2 y^{2} a_{3}+\sqrt {y}\, F \left (\frac {x -y}{\sqrt {y}}\right ) x b_{2}+D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x^{2} b_{2}+D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x y b_{3}+\sqrt {y}\, F \left (\frac {x -y}{\sqrt {y}}\right ) b_{1}+D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) x b_{1}\right ) = 0 \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \frac {1}{\sqrt {y}}, \sqrt {y}, y^{{3}/{2}}, F \left (\frac {x -y}{\sqrt {y}}\right ), D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right )\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \frac {1}{\sqrt {y}} = v_{3}, \sqrt {y} = v_{4}, y^{{3}/{2}} = v_{5}, F \left (\frac {x -y}{\sqrt {y}}\right ) = v_{6}, D\left (F \right )\left (\frac {x -y}{\sqrt {y}}\right ) = v_{7}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -v_{4} \left (-2 v_{2} v_{7} v_{1} a_{2}-2 v_{2}^{2} v_{7} a_{3}+v_{7} v_{1}^{2} b_{2}+v_{2} v_{7} v_{1} b_{2}+v_{4} v_{6} v_{1} b_{2}-2 v_{2} v_{6}^{2} b_{2}+v_{7} v_{1} v_{2} b_{3}+v_{2}^{2} v_{7} b_{3}-2 v_{2} v_{7} a_{1}+2 v_{2}^{2} a_{2}+2 v_{5} v_{6} a_{2}+2 v_{2}^{2} a_{3}+v_{7} v_{1} b_{1}+v_{2} v_{7} b_{1}+v_{4} v_{6} b_{1}-2 v_{2}^{2} b_{2}-4 v_{5} v_{6} b_{2}-2 v_{2}^{2} b_{3}-v_{5} v_{6} b_{3}\right ) = 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} -v_{4} b_{2} v_{7} v_{1}^{2}+\left (2 a_{2}-b_{2}-b_{3}\right ) v_{7} v_{1} v_{2} v_{4}-b_{2} v_{6} v_{1} v_{4}^{2}-b_{1} v_{7} v_{1} v_{4}+\left (2 a_{3}-b_{3}\right ) v_{7} v_{2}^{2} v_{4}+\left (-2 a_{2}-2 a_{3}+2 b_{2}+2 b_{3}\right ) v_{2}^{2} v_{4}+2 b_{2} v_{6}^{2} v_{2} v_{4}+\left (2 a_{1}-b_{1}\right ) v_{7} v_{2} v_{4}-b_{1} v_{6} v_{4}^{2}+\left (-2 a_{2}+4 b_{2}+b_{3}\right ) v_{5} v_{6} v_{4} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -b_{1}&=0\\ -b_{2}&=0\\ 2 b_{2}&=0\\ 2 a_{1}-b_{1}&=0\\ 2 a_{3}-b_{3}&=0\\ -2 a_{2}+4 b_{2}+b_{3}&=0\\ 2 a_{2}-b_{2}-b_{3}&=0\\ -2 a_{2}-2 a_{3}+2 b_{2}+2 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=a_{3}\\ a_{3}&=a_{3}\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=2 a_{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 &= x +y \\ \eta &= 2 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } F \left (-\frac {y-x}{\sqrt {y}}\right )+y^{\prime } \sqrt {y}-\sqrt {y}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {y}}{\sqrt {y}+F \left (-\frac {y-x}{\sqrt {y}}\right )} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: 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: 0.093 (sec). Leaf size: 38
dsolve(diff(y(x),x) = y(x)^(1/2)/(y(x)^(1/2)+F((x-y(x))/y(x)^(1/2))),y(x), singsol=all)
\[ \frac {\ln \left (y \left (x \right )\right )}{2}-\left (\int _{}^{\frac {x -y \left (x \right )}{\sqrt {y \left (x \right )}}}\frac {1}{2 F \left (\textit {\_a} \right )-\textit {\_a}}d \textit {\_a} \right )-c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 0.508 (sec). Leaf size: 274
DSolve[y'[x] == Sqrt[y[x]]/(F[(x - y[x])/Sqrt[y[x]]] + Sqrt[y[x]]),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (-\frac {F\left (\frac {x-K[2]}{\sqrt {K[2]}}\right )}{x \sqrt {K[2]}}-\int _1^x-\frac {-\frac {F\left (\frac {K[1]-K[2]}{\sqrt {K[2]}}\right )}{\sqrt {K[2]}}-2 \left (-\frac {K[1]-K[2]}{2 K[2]^{3/2}}-\frac {1}{\sqrt {K[2]}}\right ) \sqrt {K[2]} F'\left (\frac {K[1]-K[2]}{\sqrt {K[2]}}\right )-1}{\left (-2 \sqrt {K[2]} F\left (\frac {K[1]-K[2]}{\sqrt {K[2]}}\right )+K[1]-K[2]\right )^2}dK[1]+\frac {2 F\left (\frac {x-K[2]}{\sqrt {K[2]}}\right )^2+\sqrt {K[2]} F\left (\frac {x-K[2]}{\sqrt {K[2]}}\right )+x}{x \left (-x+K[2]+2 F\left (\frac {x-K[2]}{\sqrt {K[2]}}\right ) \sqrt {K[2]}\right )}\right )dK[2]+\int _1^x\frac {1}{-2 \sqrt {y(x)} F\left (\frac {K[1]-y(x)}{\sqrt {y(x)}}\right )+K[1]-y(x)}dK[1]=c_1,y(x)\right ] \]