A04S - Math 138 Assignment 4 Solutions 1(a Using z = r cos...

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Assignment 4 Solutions 1. (a) Using z = r cos θ + ir sin θ , we get dz = - r sin θ + ir cos θ (b) Note that z = r cos θ + ir sin θ iz = ir cos θ - r sin θ Thus, comparing with part a) we have dz = iz i.e. f ( z ) = iz . (c) Using z ( θ ) = r cos θ + ir sin θ we see that z (0) = r so we are to solve the IVP dz = iz, z (0) = r. This is separable so we rewrite and solve as dz z = idθ ln | z | = + C | z | = e + C z = ± e C e z = Ae defining A = ± e C Next, using the IC z (0) = r gives A = r so our solution is z = re (d) Comparing each expression for z , that is z = r cos θ + ir sin θ and z = re we see that e = cos θ + i sin θ 2. Using the identities in the hint, y 0 = 1 x 0 , y 00 = - x 00 ( x 0 ) 3 we rewrite the DE as y 00 = yy 0 - x 00 ( x 0 ) 3 = y 1 x 0 x 00 = - y ( x 0 ) 2 Note that now the independent variable is y while x is the dependent variable. Also note that the function x ( y ) doesn’t appear explicitly. That is, only x 0 ( y ) and
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This note was uploaded on 02/14/2012 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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A04S - Math 138 Assignment 4 Solutions 1(a Using z = r cos...

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