Abstract
λ = 0, 1. The general solutions of Eqs. (2)–(5) can be found in [1]. One can readily see that Eqs. (2)–(5) have at most two isolated singular points. Obviously, nonisolated singular points of these equations fill a hyperbola, parabola, or two intersecting straight lines. By making the change of variables y = y1 − z1, z = y1 + z1 in Eq. (2), we obtain ( a1x+ λx + a2y 1 − a2z 1 ) dx+ 2 (b1y1 + b2xy1) dy1 − 2 (b1z1 + b2xz1) dz1 = 0. (2′)
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Schedule a callDisclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.
