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how that an implicit solution of 2x sin2(y) dx – (x² + 11) cos(y) dy = 0 X - given by In(x + 11) + csc(y) = C. cos(y) 2x ifferentiating In(x² + 11) + csc(v) = C we get dy = 0 or sin (y) x² + 11 dx x sin2(y) dx + |-cos(v)(2? +11) = 0. ind the constant solutions, if any, that were lost in the solution of the differential equation. (Let k represent an rbitrary integer.) In(x? + 11) + csc(y) *

how that an implicit solution of 2x sin2(y) dx – (x² + 11) cos(y) dy = 0 X - given by In(x + 11) + csc(y) = C. cos(y) 2x ifferentiating In(x² + 11) + csc(v) = C we get dy = 0 or sin (y) x² + 11 dx x sin2(y) dx + |-cos(v)(2? +11) = 0. ind the constant solutions, if any, that were lost in the solution of the differential equation. (Let k represent an rbitrary integer.) In(x? + 11) + csc(y) *

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Show that an implicit solution of 2x sin2(y) dx – (x² + 11) cos(Y) dy = 0 - is given by In(x² + 11) + csc(y) = C. cos(y) 2x dy Differentiating In(x2 + 11) + csc(y) = C we get = 0 or dx + x2 sin (y) + 11 2x sin2(y) dx + (-cos(y) (x² + 11) dy = 0. Find the constant solutions, if any, that were lost in the solution of the differential equation. (Let k represent an arbitrary integer.) y = In(? + 11) + csc(y) x
Transcribed Image Text:Show that an implicit solution of 2x sin2(y) dx – (x² + 11) cos(Y) dy = 0 - is given by In(x² + 11) + csc(y) = C. cos(y) 2x dy Differentiating In(x2 + 11) + csc(y) = C we get = 0 or dx + x2 sin (y) + 11 2x sin2(y) dx + (-cos(y) (x² + 11) dy = 0. Find the constant solutions, if any, that were lost in the solution of the differential equation. (Let k represent an arbitrary integer.) y = In(? + 11) + csc(y) x
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