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Unformatted text preview: z 2 = e r 2 t ) y 1 = x r 1 and y 2 = x r 2 . If r 1 = r 2 (real), then z 1 = e r 1 t and z 2 = te r 1 t ) y 1 = x r 1 and y 2 = x r 1 ln( x ). If r 1 ; r 2 = ® § i¯ (complex), then z 1 = e ®t cos( ¯t ) and z 2 = e ®t sin( ¯t ) ) y 1 = x ® cos[ ¯ ln( x )] and y 2 = x ® sin[ ¯ ln( x )]. For x < 0, let x = ¡ e t and y ( x ) = z ( t ). Then t = ln( ¡ x ), and the same equation for z ( t ) results. In either case, t = ln j x j . Since j x j = ½ x; x > ¡ x; x < ¾ , replacing x by j x j gives the solutions for any x 6 = 0. Thus, 2 If r 1 6 = r 2 are real, then y 1 = j x j r 1 and y 2 = j x j r 2 . If r 1 = r 2 (real), then y 1 = j x j r 1 and y 2 = x r 1 ln j x j . If r 1 ; r 2 = ® § i¯ (complex), then y 1 = j x j ® cos( ¯ ln j x j ) and y 2 = j x j ® sin( ¯ ln j x j )....
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 Winter '09
 Equations, Trigraph, DT DT DT, Axy, dt dz dt

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