Internal problem ID [8423]
Internal file name [OUTPUT/7356_Sunday_June_05_2022_10_53_35_PM_57045371/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 86.
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.
\[ \boxed {y^{\prime }-\frac {y-x f \left (x^{2}+a y^{2}\right )}{x +a y f \left (x^{2}+a y^{2}\right )}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {y -x f \left (a \,y^{2}+x^{2}\right )}{x +a y f \left (a \,y^{2}+x^{2}\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 {\left (y -x f \left (a \,y^{2}+x^{2}\right )\right ) \left (b_{3}-a_{2}\right )}{x +a y f \left (a \,y^{2}+x^{2}\right )}-\frac {{\left (y -x f \left (a \,y^{2}+x^{2}\right )\right )}^{2} a_{3}}{{\left (x +a y f \left (a \,y^{2}+x^{2}\right )\right )}^{2}}-\left (\frac {-f \left (a \,y^{2}+x^{2}\right )-2 x^{2} D\left (f \right )\left (a \,y^{2}+x^{2}\right )}{x +a y f \left (a \,y^{2}+x^{2}\right )}-\frac {\left (y -x f \left (a \,y^{2}+x^{2}\right )\right ) \left (1+2 a y D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x \right )}{{\left (x +a y f \left (a \,y^{2}+x^{2}\right )\right )}^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {1-2 a y D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x}{x +a y f \left (a \,y^{2}+x^{2}\right )}-\frac {\left (y -x f \left (a \,y^{2}+x^{2}\right )\right ) \left (a f \left (a \,y^{2}+x^{2}\right )+2 a^{2} y^{2} D\left (f \right )\left (a \,y^{2}+x^{2}\right )\right )}{{\left (x +a y f \left (a \,y^{2}+x^{2}\right )\right )}^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \frac {-x b_{1}+y a_{1}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a^{2} y^{4} b_{3}+f \left (a \,y^{2}+x^{2}\right )^{2} a^{2} y^{2} b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a^{2} y^{3} b_{1}-f \left (a \,y^{2}+x^{2}\right )^{2} a \,x^{2} b_{2}+f \left (a \,y^{2}+x^{2}\right )^{2} a \,y^{2} a_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x^{3} y a_{3}-f \left (a \,y^{2}+x^{2}\right )^{2} a x b_{1}+f \left (a \,y^{2}+x^{2}\right )^{2} a y a_{1}-f \left (a \,y^{2}+x^{2}\right ) a \,y^{2} a_{2}+f \left (a \,y^{2}+x^{2}\right ) a \,y^{2} b_{3}+2 f \left (a \,y^{2}+x^{2}\right ) x y a_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a^{2} x \,y^{3} b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{3} y b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{2} y^{2} a_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{2} y^{2} b_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a x \,y^{3} a_{3}+2 f \left (a \,y^{2}+x^{2}\right )^{2} a x y a_{2}-2 f \left (a \,y^{2}+x^{2}\right )^{2} a x y b_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{2} y b_{1}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a x \,y^{2} a_{1}+2 f \left (a \,y^{2}+x^{2}\right ) a x y b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x^{4} a_{2}-f \left (a \,y^{2}+x^{2}\right )^{2} x^{2} a_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x^{3} a_{1}+f \left (a \,y^{2}+x^{2}\right ) x^{2} a_{2}-f \left (a \,y^{2}+x^{2}\right ) x^{2} b_{3}}{{\left (x +a y f \left (a \,y^{2}+x^{2}\right )\right )}^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -x b_{1}+y a_{1}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a^{2} y^{4} b_{3}+f \left (a \,y^{2}+x^{2}\right )^{2} a^{2} y^{2} b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a^{2} y^{3} b_{1}-f \left (a \,y^{2}+x^{2}\right )^{2} a \,x^{2} b_{2}+f \left (a \,y^{2}+x^{2}\right )^{2} a \,y^{2} a_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x^{3} y a_{3}-f \left (a \,y^{2}+x^{2}\right )^{2} a x b_{1}+f \left (a \,y^{2}+x^{2}\right )^{2} a y a_{1}-f \left (a \,y^{2}+x^{2}\right ) a \,y^{2} a_{2}+f \left (a \,y^{2}+x^{2}\right ) a \,y^{2} b_{3}+2 f \left (a \,y^{2}+x^{2}\right ) x y a_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a^{2} x \,y^{3} b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{3} y b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{2} y^{2} a_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{2} y^{2} b_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a x \,y^{3} a_{3}+2 f \left (a \,y^{2}+x^{2}\right )^{2} a x y a_{2}-2 f \left (a \,y^{2}+x^{2}\right )^{2} a x y b_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a \,x^{2} y b_{1}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) a x \,y^{2} a_{1}+2 f \left (a \,y^{2}+x^{2}\right ) a x y b_{2}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x^{4} a_{2}-f \left (a \,y^{2}+x^{2}\right )^{2} x^{2} a_{3}+2 D\left (f \right )\left (a \,y^{2}+x^{2}\right ) x^{3} a_{1}+f \left (a \,y^{2}+x^{2}\right ) x^{2} a_{2}-f \left (a \,y^{2}+x^{2}\right ) x^{2} b_{3} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y, f \left (a \,y^{2}+x^{2}\right ), D\left (f \right )\left (a \,y^{2}+x^{2}\right )\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}, f \left (a \,y^{2}+x^{2}\right ) = v_{3}, D\left (f \right )\left (a \,y^{2}+x^{2}\right ) = v_{4}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 2 v_{4} a^{2} v_{1} v_{2}^{3} b_{2}+2 v_{4} a^{2} v_{2}^{4} b_{3}+2 v_{4} a^{2} v_{2}^{3} b_{1}+v_{3}^{2} a^{2} v_{2}^{2} b_{2}+2 v_{4} a v_{1}^{2} v_{2}^{2} a_{2}+2 v_{4} a v_{1} v_{2}^{3} a_{3}+2 v_{4} a v_{1}^{3} v_{2} b_{2}+2 v_{4} a v_{1}^{2} v_{2}^{2} b_{3}+2 v_{4} a v_{1} v_{2}^{2} a_{1}+2 v_{3}^{2} a v_{1} v_{2} a_{2}+v_{3}^{2} a v_{2}^{2} a_{3}+2 v_{4} a v_{1}^{2} v_{2} b_{1}-v_{3}^{2} a v_{1}^{2} b_{2}-2 v_{3}^{2} a v_{1} v_{2} b_{3}+2 v_{4} v_{1}^{4} a_{2}+2 v_{4} v_{1}^{3} v_{2} a_{3}+v_{3}^{2} a v_{2} a_{1}-v_{3} a v_{2}^{2} a_{2}-v_{3}^{2} a v_{1} b_{1}+2 v_{3} a v_{1} v_{2} b_{2}+v_{3} a v_{2}^{2} b_{3}+2 v_{4} v_{1}^{3} a_{1}-v_{3}^{2} v_{1}^{2} a_{3}+v_{3} v_{1}^{2} a_{2}+2 v_{3} v_{1} v_{2} a_{3}-v_{3} v_{1}^{2} b_{3}+v_{2} a_{1}-v_{1} b_{1} = 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} 2 v_{4} v_{1}^{4} a_{2}+\left (2 a b_{2}+2 a_{3}\right ) v_{1}^{3} v_{2} v_{4}+2 v_{4} v_{1}^{3} a_{1}+\left (2 a a_{2}+2 a b_{3}\right ) v_{1}^{2} v_{2}^{2} v_{4}+2 v_{4} a v_{1}^{2} v_{2} b_{1}+\left (-a b_{2}-a_{3}\right ) v_{1}^{2} v_{3}^{2}+\left (-b_{3}+a_{2}\right ) v_{1}^{2} v_{3}+\left (2 a^{2} b_{2}+2 a a_{3}\right ) v_{1} v_{2}^{3} v_{4}+2 v_{4} a v_{1} v_{2}^{2} a_{1}+\left (2 a a_{2}-2 a b_{3}\right ) v_{1} v_{2} v_{3}^{2}+\left (2 a b_{2}+2 a_{3}\right ) v_{1} v_{2} v_{3}-v_{3}^{2} a v_{1} b_{1}-v_{1} b_{1}+2 v_{4} a^{2} v_{2}^{4} b_{3}+2 v_{4} a^{2} v_{2}^{3} b_{1}+\left (a^{2} b_{2}+a a_{3}\right ) v_{2}^{2} v_{3}^{2}+\left (-a a_{2}+a b_{3}\right ) v_{2}^{2} v_{3}+v_{3}^{2} a v_{2} a_{1}+v_{2} a_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} a_{1}&=0\\ a a_{1}&=0\\ 2 a_{1}&=0\\ 2 a_{2}&=0\\ -b_{1}&=0\\ 2 a a_{1}&=0\\ -a b_{1}&=0\\ 2 a b_{1}&=0\\ 2 a^{2} b_{1}&=0\\ 2 a^{2} b_{3}&=0\\ -b_{3}+a_{2}&=0\\ -a a_{2}+a b_{3}&=0\\ 2 a a_{2}-2 a b_{3}&=0\\ 2 a a_{2}+2 a b_{3}&=0\\ -a b_{2}-a_{3}&=0\\ 2 a b_{2}+2 a_{3}&=0\\ a^{2} b_{2}+a a_{3}&=0\\ 2 a^{2} b_{2}+2 a a_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=-a 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 &= -a 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } y f \left (x^{2}+a y^{2}\right ) a +y^{\prime } x +x f \left (x^{2}+a y^{2}\right )-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y-x f \left (x^{2}+a y^{2}\right )}{x +a y f \left (x^{2}+a y^{2}\right )} \end {array} \]
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 -> Calling odsolve with the ODE`, diff(y(x), x) = -x/(a*y(x)), y(x)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful <- 1st order, canonical coordinates successful`
✓ Solution by Maple
Time used: 0.188 (sec). Leaf size: 52
dsolve(diff(y(x),x) - (y(x)-x*f(x^2+a*y(x)^2))/(x+a*y(x)*f(x^2+a*y(x)^2))=0,y(x), singsol=all)
\[ \frac {\arctan \left (\frac {\sqrt {a}\, x}{\sqrt {a^{2} y \left (x \right )^{2}}}\right )}{\sqrt {a}}-\frac {\left (\int _{}^{y \left (x \right )^{2}+\frac {x^{2}}{a}}\frac {f \left (\textit {\_a} a \right )}{\textit {\_a}}d \textit {\_a} \right )}{2}-c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 0.476 (sec). Leaf size: 184
DSolve[y'[x]- (y[x]-x*f[x^2+a*y[x]^2])/(x+a*y[x]*f[x^2+a*y[x]^2])==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {-f\left (x^2+a K[2]^2\right ) K[2] a^2-x a}{x^2+a K[2]^2}-\int _1^x\left (\frac {a-2 a^2 K[1] K[2] f'\left (K[1]^2+a K[2]^2\right )}{K[1]^2+a K[2]^2}-\frac {2 a K[2] \left (a K[2]-a f\left (K[1]^2+a K[2]^2\right ) K[1]\right )}{\left (K[1]^2+a K[2]^2\right )^2}\right )dK[1]\right )dK[2]+\int _1^x\frac {a y(x)-a f\left (K[1]^2+a y(x)^2\right ) K[1]}{K[1]^2+a y(x)^2}dK[1]=c_1,y(x)\right ] \]