First we solve the quadratic equation 2 dy (d) dx to obtain the two possible values dy y± √√√x² + y² dx X This can be rewritten as X SO X = y X dy dx so it is homogeneous. Putting y = xv gives +v=v± √1+v², dv dx Y = + 1+ X dx X = dy dx - 2y- = ± ± which gives 2cy= =S₁ Thus In x + lnc = ±sinh v or v =±sinh (ln(cr)). Hence -1 eln(cx) how come ?? - e x = 0 dv 1+ v² +v². 2 2 (²) ², = ± (c²x² - 1). - ln(cx)
First we solve the quadratic equation 2 dy (d) dx to obtain the two possible values dy y± √√√x² + y² dx X This can be rewritten as X SO X = y X dy dx so it is homogeneous. Putting y = xv gives +v=v± √1+v², dv dx Y = + 1+ X dx X = dy dx - 2y- = ± ± which gives 2cy= =S₁ Thus In x + lnc = ±sinh v or v =±sinh (ln(cr)). Hence -1 eln(cx) how come ?? - e x = 0 dv 1+ v² +v². 2 2 (²) ², = ± (c²x² - 1). - ln(cx)
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter1: Equations And Graphs
Section: Chapter Questions
Problem 15CC
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