Internal problem ID [8674]
Internal file name [OUTPUT/7607_Sunday_June_05_2022_11_10_12_PM_21700480/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 338.
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
[[_homogeneous, `class A`], _dAlembert]
Unable to solve or complete the solution.
\[ \boxed {\left (\sqrt {x^{2}+y^{2}}\, y+\left (y^{2}-x^{2}\right ) \sin \left (\alpha \right )-2 x y \cos \left (\alpha \right )\right ) y^{\prime }+\sqrt {x^{2}+y^{2}}\, x +2 x y \sin \left (\alpha \right )+\left (y^{2}-x^{2}\right ) \cos \left (\alpha \right )=0} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {2 x y \sin \left (\alpha \right )-\cos \left (\alpha \right ) x^{2}+\cos \left (\alpha \right ) y^{2}+\sqrt {x^{2}+y^{2}}\, x}{-\sin \left (\alpha \right ) x^{2}+\sin \left (\alpha \right ) y^{2}-2 x y \cos \left (\alpha \right )+\sqrt {x^{2}+y^{2}}\, y}\\ 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 (2 x y \sin \left (\alpha \right )-\cos \left (\alpha \right ) x^{2}+\cos \left (\alpha \right ) y^{2}+\sqrt {x^{2}+y^{2}}\, x \right ) \left (b_{3}-a_{2}\right )}{-\sin \left (\alpha \right ) x^{2}+\sin \left (\alpha \right ) y^{2}-2 x y \cos \left (\alpha \right )+\sqrt {x^{2}+y^{2}}\, y}-\frac {\left (2 x y \sin \left (\alpha \right )-\cos \left (\alpha \right ) x^{2}+\cos \left (\alpha \right ) y^{2}+\sqrt {x^{2}+y^{2}}\, x \right )^{2} a_{3}}{\left (-\sin \left (\alpha \right ) x^{2}+\sin \left (\alpha \right ) y^{2}-2 x y \cos \left (\alpha \right )+\sqrt {x^{2}+y^{2}}\, y \right )^{2}}-\left (-\frac {2 y \sin \left (\alpha \right )-2 \cos \left (\alpha \right ) x +\frac {x^{2}}{\sqrt {x^{2}+y^{2}}}+\sqrt {x^{2}+y^{2}}}{-\sin \left (\alpha \right ) x^{2}+\sin \left (\alpha \right ) y^{2}-2 x y \cos \left (\alpha \right )+\sqrt {x^{2}+y^{2}}\, y}+\frac {\left (2 x y \sin \left (\alpha \right )-\cos \left (\alpha \right ) x^{2}+\cos \left (\alpha \right ) y^{2}+\sqrt {x^{2}+y^{2}}\, x \right ) \left (-2 \sin \left (\alpha \right ) x -2 \cos \left (\alpha \right ) y +\frac {y x}{\sqrt {x^{2}+y^{2}}}\right )}{\left (-\sin \left (\alpha \right ) x^{2}+\sin \left (\alpha \right ) y^{2}-2 x y \cos \left (\alpha \right )+\sqrt {x^{2}+y^{2}}\, y \right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {\frac {y x}{\sqrt {x^{2}+y^{2}}}+2 \sin \left (\alpha \right ) x +2 \cos \left (\alpha \right ) y}{-\sin \left (\alpha \right ) x^{2}+\sin \left (\alpha \right ) y^{2}-2 x y \cos \left (\alpha \right )+\sqrt {x^{2}+y^{2}}\, y}+\frac {\left (2 x y \sin \left (\alpha \right )-\cos \left (\alpha \right ) x^{2}+\cos \left (\alpha \right ) y^{2}+\sqrt {x^{2}+y^{2}}\, x \right ) \left (2 y \sin \left (\alpha \right )-2 \cos \left (\alpha \right ) x +\frac {y^{2}}{\sqrt {x^{2}+y^{2}}}+\sqrt {x^{2}+y^{2}}\right )}{\left (-\sin \left (\alpha \right ) x^{2}+\sin \left (\alpha \right ) y^{2}-2 x y \cos \left (\alpha \right )+\sqrt {x^{2}+y^{2}}\, y \right )^{2}}\right ) \left (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. \[ \left \{x, y, \sqrt {x^{2}+y^{2}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {x^{2}+y^{2}} = v_{3}\right \} \] 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 \[ \{v_{1}, v_{2}, v_{3}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (-2 a_{2} \sin \left (\alpha \right )+2 \sin \left (\alpha \right ) b_{3}+4 a_{3} \cos \left (\alpha \right )+6 \cos \left (\alpha \right ) b_{2}\right ) v_{1}^{5}+\left (-8 \sin \left (\alpha \right ) a_{3}-10 \sin \left (\alpha \right ) b_{2}-12 a_{2} \cos \left (\alpha \right )+12 \cos \left (\alpha \right ) b_{3}\right ) v_{1}^{4} v_{2}+\left (\sin \left (2 \alpha \right ) a_{2}-\sin \left (2 \alpha \right ) b_{3}-\cos \left (2 \alpha \right ) a_{3}-\cos \left (2 \alpha \right ) b_{2}-3 a_{3}-5 b_{2}\right ) v_{1}^{4} v_{3}+6 \cos \left (\alpha \right ) v_{1}^{4} b_{1}+\left (10 a_{2} \sin \left (\alpha \right )-10 \sin \left (\alpha \right ) b_{3}-6 a_{3} \cos \left (\alpha \right )-2 \cos \left (\alpha \right ) b_{2}\right ) v_{1}^{3} v_{2}^{2}+\left (4 \sin \left (2 \alpha \right ) a_{3}+4 \sin \left (2 \alpha \right ) b_{2}+4 \cos \left (2 \alpha \right ) a_{2}-4 \cos \left (2 \alpha \right ) b_{3}+8 a_{2}-8 b_{3}\right ) v_{1}^{3} v_{2} v_{3}+\left (-6 \sin \left (\alpha \right ) b_{1}-6 \cos \left (\alpha \right ) a_{1}\right ) v_{1}^{3} v_{2}-6 v_{3} v_{1}^{3} b_{1}+\left (-2 \sin \left (\alpha \right ) a_{3}-6 \sin \left (\alpha \right ) b_{2}-10 a_{2} \cos \left (\alpha \right )+10 \cos \left (\alpha \right ) b_{3}\right ) v_{1}^{2} v_{2}^{3}+\left (-6 \sin \left (2 \alpha \right ) a_{2}+6 \sin \left (2 \alpha \right ) b_{3}+6 \cos \left (2 \alpha \right ) a_{3}+6 \cos \left (2 \alpha \right ) b_{2}+2 a_{3}-2 b_{2}\right ) v_{1}^{2} v_{2}^{2} v_{3}+\left (6 \sin \left (\alpha \right ) a_{1}+6 \cos \left (\alpha \right ) b_{1}\right ) v_{1}^{2} v_{2}^{2}+6 v_{3} v_{1}^{2} v_{2} a_{1}+\left (12 a_{2} \sin \left (\alpha \right )-12 \sin \left (\alpha \right ) b_{3}-10 a_{3} \cos \left (\alpha \right )-8 \cos \left (\alpha \right ) b_{2}\right ) v_{1} v_{2}^{4}+\left (-4 \sin \left (2 \alpha \right ) a_{3}-4 \sin \left (2 \alpha \right ) b_{2}-4 \cos \left (2 \alpha \right ) a_{2}+4 \cos \left (2 \alpha \right ) b_{3}+8 a_{2}-8 b_{3}\right ) v_{1} v_{2}^{3} v_{3}+\left (-6 \sin \left (\alpha \right ) b_{1}-6 \cos \left (\alpha \right ) a_{1}\right ) v_{1} v_{2}^{3}-6 v_{3} v_{1} v_{2}^{2} b_{1}+\left (6 \sin \left (\alpha \right ) a_{3}+4 \sin \left (\alpha \right ) b_{2}+2 a_{2} \cos \left (\alpha \right )-2 \cos \left (\alpha \right ) b_{3}\right ) v_{2}^{5}+\left (\sin \left (2 \alpha \right ) a_{2}-\sin \left (2 \alpha \right ) b_{3}-\cos \left (2 \alpha \right ) a_{3}-\cos \left (2 \alpha \right ) b_{2}+5 a_{3}+3 b_{2}\right ) v_{2}^{4} v_{3}+6 \sin \left (\alpha \right ) v_{2}^{4} a_{1}+6 v_{3} v_{2}^{3} a_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} 6 a_{1}&=0\\ -6 b_{1}&=0\\ 6 \cos \left (\alpha \right ) b_{1}&=0\\ 6 \sin \left (\alpha \right ) a_{1}&=0\\ 6 \sin \left (\alpha \right ) a_{1}+6 \cos \left (\alpha \right ) b_{1}&=0\\ -6 \sin \left (\alpha \right ) b_{1}-6 \cos \left (\alpha \right ) a_{1}&=0\\ -2 a_{2} \sin \left (\alpha \right )+2 \sin \left (\alpha \right ) b_{3}+4 a_{3} \cos \left (\alpha \right )+6 \cos \left (\alpha \right ) b_{2}&=0\\ 10 a_{2} \sin \left (\alpha \right )-10 \sin \left (\alpha \right ) b_{3}-6 a_{3} \cos \left (\alpha \right )-2 \cos \left (\alpha \right ) b_{2}&=0\\ 12 a_{2} \sin \left (\alpha \right )-12 \sin \left (\alpha \right ) b_{3}-10 a_{3} \cos \left (\alpha \right )-8 \cos \left (\alpha \right ) b_{2}&=0\\ -8 \sin \left (\alpha \right ) a_{3}-10 \sin \left (\alpha \right ) b_{2}-12 a_{2} \cos \left (\alpha \right )+12 \cos \left (\alpha \right ) b_{3}&=0\\ -2 \sin \left (\alpha \right ) a_{3}-6 \sin \left (\alpha \right ) b_{2}-10 a_{2} \cos \left (\alpha \right )+10 \cos \left (\alpha \right ) b_{3}&=0\\ 6 \sin \left (\alpha \right ) a_{3}+4 \sin \left (\alpha \right ) b_{2}+2 a_{2} \cos \left (\alpha \right )-2 \cos \left (\alpha \right ) b_{3}&=0\\ -6 \sin \left (2 \alpha \right ) a_{2}+6 \sin \left (2 \alpha \right ) b_{3}+6 \cos \left (2 \alpha \right ) a_{3}+6 \cos \left (2 \alpha \right ) b_{2}+2 a_{3}-2 b_{2}&=0\\ \sin \left (2 \alpha \right ) a_{2}-\sin \left (2 \alpha \right ) b_{3}-\cos \left (2 \alpha \right ) a_{3}-\cos \left (2 \alpha \right ) b_{2}-3 a_{3}-5 b_{2}&=0\\ \sin \left (2 \alpha \right ) a_{2}-\sin \left (2 \alpha \right ) b_{3}-\cos \left (2 \alpha \right ) a_{3}-\cos \left (2 \alpha \right ) b_{2}+5 a_{3}+3 b_{2}&=0\\ -4 \sin \left (2 \alpha \right ) a_{3}-4 \sin \left (2 \alpha \right ) b_{2}-4 \cos \left (2 \alpha \right ) a_{2}+4 \cos \left (2 \alpha \right ) b_{3}+8 a_{2}-8 b_{3}&=0\\ 4 \sin \left (2 \alpha \right ) a_{3}+4 \sin \left (2 \alpha \right ) b_{2}+4 \cos \left (2 \alpha \right ) a_{2}-4 \cos \left (2 \alpha \right ) b_{3}+8 a_{2}-8 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{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 \\ \eta &= 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\sqrt {x^{2}+y^{2}}\, y+\left (y^{2}-x^{2}\right ) \sin \left (\alpha \right )-2 x y \cos \left (\alpha \right )\right ) y^{\prime }+\sqrt {x^{2}+y^{2}}\, x +2 x y \sin \left (\alpha \right )+\left (y^{2}-x^{2}\right ) \cos \left (\alpha \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-\sqrt {x^{2}+y^{2}}\, x -2 x y \sin \left (\alpha \right )-\left (y^{2}-x^{2}\right ) \cos \left (\alpha \right )}{\sqrt {x^{2}+y^{2}}\, y+\left (y^{2}-x^{2}\right ) \sin \left (\alpha \right )-2 x y \cos \left (\alpha \right )} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous G 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful <- homogeneous successful`
✓ Solution by Maple
Time used: 1.578 (sec). Leaf size: 132
dsolve((y(x)*(y(x)^2+x^2)^(1/2)+(y(x)^2-x^2)*sin(alpha)-2*x*y(x)*cos(alpha))*diff(y(x),x)+x*(y(x)^2+x^2)^(1/2)+2*x*y(x)*sin(alpha)+(y(x)^2-x^2)*cos(alpha) = 0,y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {RootOf}\left (-\ln \left (x \right )-\left (\int _{}^{\textit {\_Z}}-\frac {-\textit {\_a}^{3} \cos \left (2 \alpha \right )-3 \textit {\_a}^{2} \sin \left (2 \alpha \right )-\textit {\_a}^{3}+3 \textit {\_a} \cos \left (2 \alpha \right )+\sin \left (2 \alpha \right )+\sqrt {2}\, \sqrt {\left (\textit {\_a}^{2}+1\right ) \left (\textit {\_a}^{2}+1+\textit {\_a}^{2} \cos \left (2 \alpha \right )+2 \textit {\_a} \sin \left (2 \alpha \right )-\cos \left (2 \alpha \right )\right )}-\textit {\_a}}{\left (\textit {\_a}^{2}+1\right ) \left (\textit {\_a}^{2}+1+\textit {\_a}^{2} \cos \left (2 \alpha \right )+2 \textit {\_a} \sin \left (2 \alpha \right )-\cos \left (2 \alpha \right )\right )}d \textit {\_a} \right )+c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 5.901 (sec). Leaf size: 116
DSolve[2*x*Sin[\[Alpha]]*y[x] + Cos[\[Alpha]]*(-x^2 + y[x]^2) + x*Sqrt[x^2 + y[x]^2] + (-2*x*Cos[\[Alpha]]*y[x] + Sin[\[Alpha]]*(-x^2 + y[x]^2) + y[x]*Sqrt[x^2 + y[x]^2])*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\sqrt {\cos ^2(\alpha )} \sec (\alpha ) \left (\log \left (\cos (\alpha ) \left (\sin (\alpha )+\frac {\cos (\alpha ) y(x)}{x}\right )\right )-\log \left (\frac {1}{2} \left (\cos (2 \alpha )-2 \sqrt {\cos ^2(\alpha )} \sqrt {\frac {y(x)^2}{x^2}+1}-\frac {\sin (2 \alpha ) y(x)}{x}+1\right )\right )\right )+\log \left (\frac {y(x)^2}{x^2}+1\right )-\frac {1}{2} \log \left (\left (\sin (\alpha )+\frac {\cos (\alpha ) y(x)}{x}\right )^2\right )=-\log (x)+c_1,y(x)\right ] \]