Worksheet 4 Solutions - MA 2930, Feb 16, 2011 Worksheet 4...

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MA 2930, Feb 16, 2011 Worksheet 4 1. We’ll use the method of successive approximation to solve the following initial value problem: y 0 = - y + t, y (0) = 0 (1) Starting with φ 0 ( t ) = 0 determine φ n ( t ) for n = 1 , 2 ,... . (2) Find lim n →∞ φ n ( t ) = φ ( t ) . This is the exact solution. (3) Verify your solution by solving the equation by another method. (1) The iterative formula for the successive approximations of the solu- tion, φ n ( x ), of the differential equation y 0 = f ( t,y ) with initial condition y ( t 0 ) = y 0 is φ n +1 ( t ) = y 0 + Z t t 0 f ( u,φ n ( u )) du Applying it to our case with φ 0 ( t ) = 0 we get φ 1 ( t ) = Z t 0 u du = t 2 / 2 φ 2 ( t ) = Z t 0 - u 2 / 2 + u du = - t 3 3 · 2 + t 2 2 φ 3 ( t ) = Z t 0 u 3 / 6 - u 2 / 2 + u du = t 4 4 · 3 · 2 - t 3 3 · 2 + t 2 2 etc. (2) It seems that as n increases we are getting more and more terms of the Taylor series of e - t except for the first two terms 1 and - t . So, seemingly lim t →∞ φ n ( t ) = e - t - 1 + t (3) Ours is a linear first order equation so we can solve it using the integrating factor e t : e t y = Z te t dt = ( t - 1) e t + c 1
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So, y = t - 1 + ce - t Upon applying the initial condition y (0) = 0 we get c = 1. So y = e - t - t +1, which verifies the guess for the limit of the approximations. 2.
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Worksheet 4 Solutions - MA 2930, Feb 16, 2011 Worksheet 4...

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