Internal problem ID [8932]
Internal file name [OUTPUT/7867_Monday_June_06_2022_12_48_45_AM_32618646/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 598.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class D`]]
\[ \boxed {y^{\prime }-\frac {y+F \left (\frac {y}{x}\right )}{x -1}=0} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} -\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x +F \left (u \left (x \right )\right )+u \left (x \right ) x +u^{\prime }\left (x \right ) x +u \left (x \right ) = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u +F \left (u \right )}{x \left (x -1\right )} \end {align*}
Where \(f(x)=\frac {1}{\left (x -1\right ) x}\) and \(g(u)=u +F \left (u \right )\). Integrating both sides gives \begin{align*} \frac {1}{u +F \left (u \right )} \,du &= \frac {1}{\left (x -1\right ) x} \,d x \\ \int { \frac {1}{u +F \left (u \right )} \,du} &= \int {\frac {1}{\left (x -1\right ) x} \,d x} \\ \int _{}^{u}\frac {1}{\textit {\_a} +F \left (\textit {\_a} \right )}d \textit {\_a}&=\ln \left (x -1\right )-\ln \left (x \right )+c_{2} \\ \end{align*} Which results in \[ \int _{}^{u}\frac {1}{\textit {\_a} +F \left (\textit {\_a} \right )}d \textit {\_a}=\ln \left (x -1\right )-\ln \left (x \right )+c_{2} \] The solution is \[ \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a} +F \left (\textit {\_a} \right )}d \textit {\_a} -\ln \left (x -1\right )+\ln \left (x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} +F \left (\textit {\_a} \right )}d \textit {\_a} -\ln \left (x -1\right )+\ln \left (x \right )-c_{2} = 0\\ \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} +F \left (\textit {\_a} \right )}d \textit {\_a} -\ln \left (x -1\right )+\ln \left (x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} +F \left (\textit {\_a} \right )}d \textit {\_a} -\ln \left (x -1\right )+\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} +F \left (\textit {\_a} \right )}d \textit {\_a} -\ln \left (x -1\right )+\ln \left (x \right )-c_{2} = 0 \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=\frac {y +F \left (\frac {y}{x}\right )}{x -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 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 +F \left (\frac {y}{x}\right )\right ) \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{x -1}-\frac {\left (y +F \left (\frac {y}{x}\right )\right )^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{\left (x -1\right )^{2}}-\left (-\frac {D\left (F \right )\left (\frac {y}{x}\right ) y}{x^{2} \left (x -1\right )}-\frac {y +F \left (\frac {y}{x}\right )}{\left (x -1\right )^{2}}\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 (1+\frac {D\left (F \right )\left (\frac {y}{x}\right )}{x}\right ) \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right )}{x -1} = 0 \end{equation} Putting the above in normal form gives \[ -\frac {-x^{5} b_{4}+3 x^{4} b_{4}+F \left (\frac {y}{x}\right ) x^{2} y a_{3}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y a_{2}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y b_{3}-D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{2} a_{3}-D\left (F \right )\left (\frac {y}{x}\right ) x y a_{1}+D\left (F \right )\left (\frac {y}{x}\right ) x y a_{2}-D\left (F \right )\left (\frac {y}{x}\right ) x y b_{3}-y b_{5} x^{2}-F \left (\frac {y}{x}\right ) x^{2} y a_{5}+2 F \left (\frac {y}{x}\right ) x^{2} y b_{6}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y a_{4}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y b_{5}+D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{2} a_{5}-D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{2} b_{6}+2 F \left (\frac {y}{x}\right )^{2} x^{2} y a_{6}+2 F \left (\frac {y}{x}\right ) x^{3} y a_{5}-2 F \left (\frac {y}{x}\right ) x^{3} y b_{6}+3 F \left (\frac {y}{x}\right ) x^{2} y^{2} a_{6}-D\left (F \right )\left (\frac {y}{x}\right ) x^{3} y a_{4}+D\left (F \right )\left (\frac {y}{x}\right ) x^{3} y b_{5}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y^{2} a_{5}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y^{2} b_{6}-D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{3} a_{6}+x^{3} b_{1}+x^{3} b_{2}-x^{2} b_{1}+F \left (\frac {y}{x}\right ) x^{3} b_{5}-D\left (F \right )\left (\frac {y}{x}\right ) x^{3} b_{4}+D\left (F \right )\left (\frac {y}{x}\right ) y^{3} a_{6}-2 x^{3} y a_{4}+2 x^{3} y b_{5}-x^{2} y^{2} a_{5}+x^{2} y^{2} b_{6}+F \left (\frac {y}{x}\right )^{2} x^{3} a_{5}+F \left (\frac {y}{x}\right ) x^{4} a_{4}-F \left (\frac {y}{x}\right ) x^{4} b_{5}+D\left (F \right )\left (\frac {y}{x}\right ) x^{4} b_{4}-2 F \left (\frac {y}{x}\right ) x^{3} a_{4}-2 x^{3} b_{4}+x^{4} y a_{4}-x^{4} y b_{5}+x^{3} y^{2} a_{5}-x^{3} y^{2} b_{6}+x^{2} y^{3} a_{6}-b_{2} x^{2}-x^{2} y a_{1}-x^{2} y a_{2}+F \left (\frac {y}{x}\right )^{2} x^{2} a_{3}-F \left (\frac {y}{x}\right ) x^{3} b_{3}+D\left (F \right )\left (\frac {y}{x}\right ) x^{3} b_{2}-F \left (\frac {y}{x}\right ) x^{2} a_{1}-F \left (\frac {y}{x}\right ) x^{2} a_{2}+F \left (\frac {y}{x}\right ) x^{2} b_{3}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} b_{1}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} b_{2}+D\left (F \right )\left (\frac {y}{x}\right ) y^{2} a_{3}-D\left (F \right )\left (\frac {y}{x}\right ) x b_{1}+D\left (F \right )\left (\frac {y}{x}\right ) y a_{1}}{x^{2} \left (x -1\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} x^{5} b_{4}-3 x^{4} b_{4}-F \left (\frac {y}{x}\right ) x^{2} y a_{3}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y a_{2}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y b_{3}+D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{2} a_{3}+D\left (F \right )\left (\frac {y}{x}\right ) x y a_{1}-D\left (F \right )\left (\frac {y}{x}\right ) x y a_{2}+D\left (F \right )\left (\frac {y}{x}\right ) x y b_{3}+y b_{5} x^{2}+F \left (\frac {y}{x}\right ) x^{2} y a_{5}-2 F \left (\frac {y}{x}\right ) x^{2} y b_{6}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y a_{4}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y b_{5}-D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{2} a_{5}+D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{2} b_{6}-2 F \left (\frac {y}{x}\right )^{2} x^{2} y a_{6}-2 F \left (\frac {y}{x}\right ) x^{3} y a_{5}+2 F \left (\frac {y}{x}\right ) x^{3} y b_{6}-3 F \left (\frac {y}{x}\right ) x^{2} y^{2} a_{6}+D\left (F \right )\left (\frac {y}{x}\right ) x^{3} y a_{4}-D\left (F \right )\left (\frac {y}{x}\right ) x^{3} y b_{5}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y^{2} a_{5}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} y^{2} b_{6}+D\left (F \right )\left (\frac {y}{x}\right ) x \,y^{3} a_{6}-x^{3} b_{1}-x^{3} b_{2}+x^{2} b_{1}-F \left (\frac {y}{x}\right ) x^{3} b_{5}+D\left (F \right )\left (\frac {y}{x}\right ) x^{3} b_{4}-D\left (F \right )\left (\frac {y}{x}\right ) y^{3} a_{6}+2 x^{3} y a_{4}-2 x^{3} y b_{5}+x^{2} y^{2} a_{5}-x^{2} y^{2} b_{6}-F \left (\frac {y}{x}\right )^{2} x^{3} a_{5}-F \left (\frac {y}{x}\right ) x^{4} a_{4}+F \left (\frac {y}{x}\right ) x^{4} b_{5}-D\left (F \right )\left (\frac {y}{x}\right ) x^{4} b_{4}+2 F \left (\frac {y}{x}\right ) x^{3} a_{4}+2 x^{3} b_{4}-x^{4} y a_{4}+x^{4} y b_{5}-x^{3} y^{2} a_{5}+x^{3} y^{2} b_{6}-x^{2} y^{3} a_{6}+b_{2} x^{2}+x^{2} y a_{1}+x^{2} y a_{2}-F \left (\frac {y}{x}\right )^{2} x^{2} a_{3}+F \left (\frac {y}{x}\right ) x^{3} b_{3}-D\left (F \right )\left (\frac {y}{x}\right ) x^{3} b_{2}+F \left (\frac {y}{x}\right ) x^{2} a_{1}+F \left (\frac {y}{x}\right ) x^{2} a_{2}-F \left (\frac {y}{x}\right ) x^{2} b_{3}-D\left (F \right )\left (\frac {y}{x}\right ) x^{2} b_{1}+D\left (F \right )\left (\frac {y}{x}\right ) x^{2} b_{2}-D\left (F \right )\left (\frac {y}{x}\right ) y^{2} a_{3}+D\left (F \right )\left (\frac {y}{x}\right ) x b_{1}-D\left (F \right )\left (\frac {y}{x}\right ) y a_{1} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, F \left (\frac {y}{x}\right ), D\left (F \right )\left (\frac {y}{x}\right )\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, F \left (\frac {y}{x}\right ) = v_{3}, D\left (F \right )\left (\frac {y}{x}\right ) = v_{4}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -v_{1}^{4} v_{2} a_{4}-v_{3} v_{1}^{4} a_{4}+v_{4} v_{1}^{3} v_{2} a_{4}-v_{1}^{3} v_{2}^{2} a_{5}-2 v_{3} v_{1}^{3} v_{2} a_{5}-v_{3}^{2} v_{1}^{3} a_{5}+v_{4} v_{1}^{2} v_{2}^{2} a_{5}-v_{1}^{2} v_{2}^{3} a_{6}-3 v_{3} v_{1}^{2} v_{2}^{2} a_{6}-2 v_{3}^{2} v_{1}^{2} v_{2} a_{6}+v_{4} v_{1} v_{2}^{3} a_{6}+v_{1}^{5} b_{4}-v_{4} v_{1}^{4} b_{4}+v_{1}^{4} v_{2} b_{5}+v_{3} v_{1}^{4} b_{5}-v_{4} v_{1}^{3} v_{2} b_{5}+v_{1}^{3} v_{2}^{2} b_{6}+2 v_{3} v_{1}^{3} v_{2} b_{6}-v_{4} v_{1}^{2} v_{2}^{2} b_{6}+v_{4} v_{1}^{2} v_{2} a_{2}-v_{3} v_{1}^{2} v_{2} a_{3}-v_{3}^{2} v_{1}^{2} a_{3}+v_{4} v_{1} v_{2}^{2} a_{3}+2 v_{1}^{3} v_{2} a_{4}+2 v_{3} v_{1}^{3} a_{4}-v_{4} v_{1}^{2} v_{2} a_{4}+v_{1}^{2} v_{2}^{2} a_{5}+v_{3} v_{1}^{2} v_{2} a_{5}-v_{4} v_{1} v_{2}^{2} a_{5}-v_{4} v_{2}^{3} a_{6}-v_{4} v_{1}^{3} b_{2}+v_{3} v_{1}^{3} b_{3}-v_{4} v_{1}^{2} v_{2} b_{3}-3 v_{1}^{4} b_{4}+v_{4} v_{1}^{3} b_{4}-2 v_{1}^{3} v_{2} b_{5}-v_{3} v_{1}^{3} b_{5}+v_{4} v_{1}^{2} v_{2} b_{5}-v_{1}^{2} v_{2}^{2} b_{6}-2 v_{3} v_{1}^{2} v_{2} b_{6}+v_{4} v_{1} v_{2}^{2} b_{6}+v_{1}^{2} v_{2} a_{1}+v_{3} v_{1}^{2} a_{1}+v_{4} v_{1} v_{2} a_{1}+v_{1}^{2} v_{2} a_{2}+v_{3} v_{1}^{2} a_{2}-v_{4} v_{1} v_{2} a_{2}-v_{4} v_{2}^{2} a_{3}-v_{1}^{3} b_{1}-v_{4} v_{1}^{2} b_{1}-v_{1}^{3} b_{2}+v_{4} v_{1}^{2} b_{2}-v_{3} v_{1}^{2} b_{3}+v_{4} v_{1} v_{2} b_{3}+2 v_{1}^{3} b_{4}+v_{2} b_{5} v_{1}^{2}-v_{4} v_{2} a_{1}+v_{1}^{2} b_{1}+v_{4} v_{1} b_{1}+b_{2} v_{1}^{2} = 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} \left (-a_{4}+b_{5}\right ) v_{1}^{4} v_{2}+\left (-a_{4}+b_{5}\right ) v_{1}^{4} v_{3}+\left (-a_{5}+b_{6}\right ) v_{1}^{3} v_{2}^{2}+\left (2 a_{4}-2 b_{5}\right ) v_{1}^{3} v_{2}+\left (2 a_{4}+b_{3}-b_{5}\right ) v_{1}^{3} v_{3}+\left (-b_{2}+b_{4}\right ) v_{1}^{3} v_{4}+\left (a_{5}-b_{6}\right ) v_{1}^{2} v_{2}^{2}+\left (a_{1}+a_{2}+b_{5}\right ) v_{1}^{2} v_{2}+\left (a_{1}+a_{2}-b_{3}\right ) v_{1}^{2} v_{3}+\left (-b_{1}+b_{2}\right ) v_{1}^{2} v_{4}-2 v_{3}^{2} v_{1}^{2} v_{2} a_{6}-3 v_{3} v_{1}^{2} v_{2}^{2} a_{6}+v_{4} v_{1} v_{2}^{3} a_{6}+\left (a_{5}-b_{6}\right ) v_{1}^{2} v_{2}^{2} v_{4}+\left (-a_{3}+a_{5}-2 b_{6}\right ) v_{1}^{2} v_{2} v_{3}+\left (a_{2}-a_{4}-b_{3}+b_{5}\right ) v_{1}^{2} v_{2} v_{4}+\left (a_{3}-a_{5}+b_{6}\right ) v_{1} v_{2}^{2} v_{4}+\left (a_{1}-a_{2}+b_{3}\right ) v_{1} v_{2} v_{4}+\left (-2 a_{5}+2 b_{6}\right ) v_{1}^{3} v_{2} v_{3}+\left (a_{4}-b_{5}\right ) v_{1}^{3} v_{2} v_{4}+v_{1}^{5} b_{4}-3 v_{1}^{4} b_{4}+\left (b_{1}+b_{2}\right ) v_{1}^{2}+\left (-b_{1}-b_{2}+2 b_{4}\right ) v_{1}^{3}-v_{4} v_{2}^{3} a_{6}-v_{3}^{2} v_{1}^{3} a_{5}-v_{4} v_{1}^{4} b_{4}-v_{1}^{2} v_{2}^{3} a_{6}-v_{3}^{2} v_{1}^{2} a_{3}-v_{4} v_{2}^{2} a_{3}+v_{4} v_{1} b_{1}-v_{4} v_{2} a_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} a_{6}&=0\\ b_{1}&=0\\ b_{4}&=0\\ -a_{1}&=0\\ -a_{3}&=0\\ -a_{5}&=0\\ -3 a_{6}&=0\\ -2 a_{6}&=0\\ -a_{6}&=0\\ -3 b_{4}&=0\\ -b_{4}&=0\\ -a_{4}+b_{5}&=0\\ a_{4}-b_{5}&=0\\ 2 a_{4}-2 b_{5}&=0\\ -2 a_{5}+2 b_{6}&=0\\ -a_{5}+b_{6}&=0\\ a_{5}-b_{6}&=0\\ -b_{1}+b_{2}&=0\\ b_{1}+b_{2}&=0\\ -b_{2}+b_{4}&=0\\ a_{1}-a_{2}+b_{3}&=0\\ a_{1}+a_{2}-b_{3}&=0\\ a_{1}+a_{2}+b_{5}&=0\\ -a_{3}+a_{5}-2 b_{6}&=0\\ a_{3}-a_{5}+b_{6}&=0\\ 2 a_{4}+b_{3}-b_{5}&=0\\ -b_{1}-b_{2}+2 b_{4}&=0\\ a_{2}-a_{4}-b_{3}+b_{5}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=-b_{5}\\ a_{3}&=0\\ a_{4}&=b_{5}\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=-b_{5}\\ 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 &= x^{2}-x \\ \eta &= y x -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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{\prime } x +F \left (\frac {y}{x}\right )+y+y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y+F \left (\frac {y}{x}\right )}{1-x} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 29
dsolve(diff(y(x),x) = (y(x)+F(y(x)/x))/(x-1),y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} \right )+\ln \left (x -1\right )-\ln \left (x \right )+c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 0.139 (sec). Leaf size: 37
DSolve[y'[x] == (F[y[x]/x] + y[x])/(-1 + x),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{\frac {y(x)}{x}}\frac {1}{F(K[1])+K[1]}dK[1]=\log (1-x)-\log (x)+c_1,y(x)\right ] \]