Rhs ofics p0 p0 c1 f 4 p100 p0 t4 k inhomogeneous

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Unformatted text preview: p0 (t)4. K . inhomogeneous DE: + t = 0 ) c1 pp = Apply Substituting 0 (0) = 0 ) 4c1 + 3p2= 100 ) c2 = 4 (p0 4). 4. p into DE yields: 25 c = 0 ) p = 3 4 General solution of tinhomogeneous tDE: pt) ) = (pe 4t4)e 4ttsin(32) +t4. ( t + 4 c1 cos(3 ) + cte 4 sin(3t) + 4. Solution of IVP: p( ) = (p0 4)e cos(3 30 0 4t p (t = ( 4c +3c2 e 4t not t) (3c1 +4c monotoniClearly, for any p0 , limt! p(t) = 4. Note )that the1price)doescos(3approach 4 2 )e sin(3t). Applybut oscillatesparound 1 + 4 = p0 ) c1 = p0 4. cally, ICs: p(0) = 0 ) c it. 4 p0 (0) = 0 ) 4c1 + 3c2 = 0 ) c2 = 3 (p0 4). 4 Solution of IVP: p(t) = (p0 4)e 4t cos(3t) + 3 (p0 4)e 4t sin(3t) + 4. Clearly, for any p0 , limt! p(t) = 4. Note that the price does not approach 4 monotonically, but oscillates around it. 4. What does the Existence & Uniqueness Theorem (Theorem 3.8 in the Course Notes) predict about the solutions of the following initial value problems? Justify your answers. AMATH 350 2 Assignment 4 Solutions Fall 2010 dy 0 c y (0) Campbell 2010 = 1 (0) a) sin(x) 2 + cos(x)y = 0, S.A. = 0, AMATH 350 Assignment 4ySolutions dx...
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This document was uploaded on 02/09/2014.

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