Internal problem ID [8602]
Internal file name [OUTPUT/7535_Sunday_June_05_2022_11_03_52_PM_39367346/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 266.
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
[`x=_G(y,y')`]
Unable to solve or complete the solution.
\[ \boxed {\left (y-x \right ) \sqrt {x^{2}+1}\, y^{\prime }-a \sqrt {\left (1+y^{2}\right )^{3}}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {a \sqrt {\left (y^{2}+1\right )^{3}}}{\left (y -x \right ) \sqrt {x^{2}+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}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{2} b_{4}+x y 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 {a \sqrt {\left (y^{2}+1\right )^{3}}\, \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{\left (y -x \right ) \sqrt {x^{2}+1}}-\frac {a^{2} \left (y^{2}+1\right )^{3} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{\left (y -x \right )^{2} \left (x^{2}+1\right )}-\left (\frac {a \sqrt {\left (y^{2}+1\right )^{3}}}{\left (y -x \right )^{2} \sqrt {x^{2}+1}}-\frac {a \sqrt {\left (y^{2}+1\right )^{3}}\, x}{\left (y -x \right ) \left (x^{2}+1\right )^{{3}/{2}}}\right ) \left (x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {3 a \left (y^{2}+1\right )^{2} y}{\sqrt {\left (y^{2}+1\right )^{3}}\, \left (y -x \right ) \sqrt {x^{2}+1}}-\frac {a \sqrt {\left (y^{2}+1\right )^{3}}}{\left (y -x \right )^{2} \sqrt {x^{2}+1}}\right ) \left (x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+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} Since the PDE has radicals, simplifying gives \[ \text {Expression too large to display} \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {\left (y^{2}+1\right )^{3}}, \sqrt {x^{2}+1}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {\left (y^{2}+1\right )^{3}} = v_{3}, \sqrt {x^{2}+1} = v_{4}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \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} \text {Expression too large to display} \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} b_{2}&=0\\ b_{5}&=0\\ a a_{3}&=0\\ a a_{6}&=0\\ -4 b_{2}&=0\\ -2 b_{2}&=0\\ 2 b_{2}&=0\\ 2 b_{4}&=0\\ 4 b_{4}&=0\\ -3 a a_{2}&=0\\ -2 a a_{6}&=0\\ 3 a b_{2}&=0\\ 6 a b_{2}&=0\\ 3 a b_{4}&=0\\ 6 a b_{4}&=0\\ -3 a^{2} a_{3}&=0\\ -a^{2} a_{3}&=0\\ -3 a^{2} a_{5}&=0\\ -a^{2} a_{5}&=0\\ -6 a^{2} a_{6}&=0\\ -2 a^{2} a_{6}&=0\\ -4 b_{4}+b_{5}&=0\\ 2 b_{4}-2 b_{5}&=0\\ 2 b_{5}-8 b_{4}&=0\\ -a a_{1}+a b_{1}&=0\\ -3 a a_{2}-3 a a_{3}&=0\\ -3 a a_{2}+a a_{3}&=0\\ -6 a a_{3}+6 a b_{2}&=0\\ -3 a^{2} a_{3}+b_{2}&=0\\ -3 a^{2} a_{3}+2 b_{2}&=0\\ -a^{2} a_{3}+b_{2}&=0\\ -2 a a_{3}-a b_{3}&=0\\ -a a_{5}-3 a a_{6}&=0\\ -a a_{5}-a a_{6}&=0\\ -a^{2} a_{5}+2 b_{4}&=0\\ -6 a^{2} a_{6}+b_{5}&=0\\ -6 a^{2} a_{6}+2 b_{5}&=0\\ a b_{2}-a b_{3}&=0\\ -3 a b_{4}+3 a b_{5}&=0\\ -2 a b_{4}+2 a b_{5}&=0\\ a b_{4}-a b_{5}&=0\\ -3 a a_{2}-9 a a_{3}+3 a b_{2}&=0\\ -3 a a_{2}-3 a a_{3}+3 a b_{3}&=0\\ -a a_{2}-3 a a_{3}-2 a b_{3}&=0\\ -a a_{2}-a a_{3}-a b_{3}&=0\\ -a a_{2}-a a_{3}+2 a b_{3}&=0\\ -3 a a_{2}-3 a b_{2}+3 a b_{3}&=0\\ -a a_{2}-2 a b_{2}+2 a b_{3}&=0\\ -a a_{2}+a b_{2}-a b_{3}&=0\\ -a a_{2}+2 a b_{2}-2 a b_{3}&=0\\ -6 a a_{3}+12 a b_{2}+3 a b_{3}&=0\\ -2 a a_{3}+6 a b_{2}+2 a b_{3}&=0\\ 3 a a_{3}-3 a b_{2}+3 a b_{3}&=0\\ 3 a a_{3}-2 a b_{2}+2 a b_{3}&=0\\ -3 a^{2} a_{5}+2 b_{4}-2 b_{5}&=0\\ -3 a^{2} a_{5}+4 b_{4}-4 b_{5}&=0\\ -2 a^{2} a_{6}-4 b_{4}+b_{5}&=0\\ 3 a a_{1}-6 a a_{4}+3 a a_{6}+3 a b_{1}&=0\\ -6 a a_{1}+3 a a_{4}-2 a a_{6}+3 a b_{6}&=0\\ -3 a a_{1}-a a_{5}-a a_{6}+3 a b_{6}&=0\\ -3 a a_{2}-9 a a_{3}+6 a b_{2}+6 a b_{3}&=0\\ -3 a a_{2}+3 a a_{3}-6 a b_{2}+6 a b_{3}&=0\\ -a a_{2}-3 a a_{3}+3 a b_{2}+4 a b_{3}&=0\\ -a a_{2}+3 a a_{3}-4 a b_{2}+4 a b_{3}&=0\\ -3 a a_{4}-3 a a_{5}+3 a b_{1}+6 a b_{4}&=0\\ -a a_{4}-a a_{5}-a b_{5}+a b_{6}&=0\\ -9 a a_{1}+3 a a_{4}-a a_{5}-3 a a_{6}+6 a b_{6}&=0\\ a a_{1}-2 a a_{4}+3 a a_{6}-a b_{5}+a b_{6}&=0\\ -3 a a_{1}+a a_{4}+2 a b_{1}+a b_{4}-a b_{5}&=0\\ -2 a a_{1}+a a_{4}+a b_{1}+2 a b_{4}-2 a b_{5}&=0\\ a a_{1}-2 a a_{4}+3 a b_{1}+2 a b_{5}-2 a b_{6}&=0\\ -3 a a_{1}-3 a a_{5}-3 a a_{6}-3 a b_{1}+6 a b_{6}&=0\\ -a a_{1}-3 a a_{5}-3 a a_{6}-2 a b_{1}+3 a b_{6}&=0\\ 3 a a_{1}-9 a a_{4}-3 a a_{5}+3 a a_{6}+6 a b_{1}+3 a b_{4}&=0\\ a a_{1}-3 a a_{4}-a a_{5}+3 a a_{6}-2 a b_{5}+2 a b_{6}&=0\\ 3 a a_{1}-6 a a_{4}+a a_{6}+6 a b_{1}+3 a b_{5}-3 a b_{6}&=0\\ -3 a a_{4}-3 a a_{5}+6 a b_{1}+12 a b_{4}+3 a b_{5}-3 a b_{6}&=0\\ -a a_{4}-a a_{5}+3 a b_{1}+6 a b_{4}+2 a b_{5}-2 a b_{6}&=0\\ a a_{1}-3 a a_{4}-a a_{5}+6 a b_{1}+3 a b_{4}+4 a b_{5}-4 a b_{6}&=0\\ -6 a a_{1}+3 a a_{4}-6 a a_{6}-3 a b_{1}-6 a b_{4}+6 a b_{5}+6 a b_{6}&=0\\ -2 a a_{1}+a a_{4}-6 a a_{6}-2 a b_{1}-4 a b_{4}+4 a b_{5}+3 a b_{6}&=0\\ -9 a a_{1}+3 a a_{4}-3 a a_{5}-9 a a_{6}-6 a b_{1}-3 a b_{4}+3 a b_{5}+12 a b_{6}&=0\\ -3 a a_{1}+a a_{4}-3 a a_{5}-9 a a_{6}-4 a b_{1}-2 a b_{4}+2 a b_{5}+6 a b_{6}&=0\\ 3 a a_{1}-9 a a_{4}-3 a a_{5}+a a_{6}+12 a b_{1}+6 a b_{4}+6 a b_{5}-6 a b_{6}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=b_{6}\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=b_{6}\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=b_{6}\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=0\\ b_{6}&=b_{6} \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}+1 \\ \eta &= y^{2}+1 \\ \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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y-x \right ) \sqrt {x^{2}+1}\, y^{\prime }-a \sqrt {\left (1+y^{2}\right )^{3}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \sqrt {\left (1+y^{2}\right )^{3}}}{\left (y-x \right ) \sqrt {x^{2}+1}} \end {array} \]
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 trying exact Looking for potential symmetries trying inverse_Riccati trying an equivalence to an Abel ODE differential order: 1; trying a linearization to 2nd order --- trying a change of variables {x -> y(x), y(x) -> x} differential order: 1; trying a linearization to 2nd order trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3`[x^2+1, y^2+1]
✓ Solution by Maple
Time used: 0.234 (sec). Leaf size: 187
dsolve((y(x)-x)*sqrt(x^2+1)*diff(y(x),x)-a*sqrt((y(x)^2+1)^3)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {-x +\sqrt {-a^{2} \left (x^{2}+1\right )^{2} \left (a^{2}-1\right )}}{a^{2} x^{2}+a^{2}-1} \\ y \left (x \right ) &= \frac {-x -\sqrt {-a^{2} \left (x^{2}+1\right )^{2} \left (a^{2}-1\right )}}{a^{2} x^{2}+a^{2}-1} \\ \frac {\sqrt {2}\, \sqrt {\frac {a^{2}}{1+\cos \left (2 \arctan \left (x \right )-2 \arctan \left (y \left (x \right )\right )\right )}}\, \cos \left (\arctan \left (x \right )-\arctan \left (y \left (x \right )\right )\right ) \arctan \left (\frac {\cos \left (\arctan \left (x \right )-\arctan \left (y \left (x \right )\right )\right )}{\sqrt {a^{2}-1}}\right )+\arctan \left (\frac {\sqrt {a^{2}-1}\, \tan \left (\arctan \left (x \right )-\arctan \left (y \left (x \right )\right )\right )}{a}\right ) a -\sqrt {a^{2}-1}\, \left (c_{1} -\arctan \left (y \left (x \right )\right )\right )}{\sqrt {a^{2}-1}} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.935 (sec). Leaf size: 69
DSolve[(y[x]-x)*Sqrt[x^2+1]*y'[x]-a*Sqrt[(y[x]^2+1)^3]==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\left \{\frac {2 a \arctan \left (\frac {1-a \tan \left (\frac {K[1]}{2}\right )}{\sqrt {a^2-1}}\right )}{\sqrt {a^2-1}}+K[1]+\arctan (x)=c_1,y(x)=\frac {\tan (K[1])+x}{1-x \tan (K[1])}\right \},\{K[1],y(x)\}\right ] \]