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Unformatted text preview: MATH 152 L1 & L2 Applied Linear Algebra and Differential Equations Spring 2004-05 ¥ Week 03 Worksheet Solutions : First-Order Differential Equations Q1. (Existence and uniqueness for nonlinear equations) Consider the differential equation y = ( y- 5) 1 3 . (a) Find the equilibrium (i.e., constant) solution(s). Solution The constant solution is y ( t ) = 5. (b) Use the fact that the equation is separable to find a nonconstant solution satisfying the initial condition y (1) = 5. Solution The equation can be written as dy dt = ( y- 5) 1 3 , then by separating variables, we have Z 1 ( y- 5) 1 3 dy = Z dt, 3 2 ( y- 5) 2 3 = t + C. But y (1) = 5 = ⇒ C =- 1, then y ( t ) = 2 3 ( t- 1) ¶ 3 2 + 5. (c) Explain why Theorem 2.4.2 (page 27) does not apply to say there is a unique solution. Solution Note that the initial value problem has more than one solution. To explain why, we let f ( t,y ) = ( y- 5) 1 3 , then ∂f ∂y ( t,y ) = 1 3( y- 5) 2 3 is discontinuous when y = 5. That is, the necessary condition of Theorem 2.4.2 is not satisfied and hence this existence and uniqueness theorem does not apply to this initial value problem....
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