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Unformatted text preview: Economics 204 Lecture 15–Friday, August 14, 2009 Revised 8/14/09, revisions indicated by ** and Sticky Notes Second Order Linear Differential Equations Consider the second order differential equation y = cy + by with b, c ∈ R . Rewrite this as a first order linear differential equation in two variables: ¯ y ( t ) = ⎛ ⎜ ⎜ ⎝ y ( t ) y ( t ) ⎞ ⎟ ⎟ ⎠ ¯ y ( t ) = ⎛ ⎜ ⎜ ⎝ y ( t ) y ( t ) ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 0 1 c b ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ y ( t ) y ( t ) ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 0 1 c b ⎞ ⎟ ⎟ ⎠ ¯ y The eigenvalues are b ± √ b 2 +4 c 2 , the roots of the equation λ 2 − bλ − c = 0. The qualitative behavior of the solutions can be explicitly described from the coeﬃcients b and c , by determining whether the eigenvalues are real or complex, and whether the real parts are negative, zero, or positive; see Section 6 of the Differential Equations Handout. Example 1 Consider the second order linear differential equation y = 2 y + y 1 • As above, let ¯ y = ⎛ ⎜ ⎜ ⎝ y y ⎞ ⎟ ⎟ ⎠ so the equation becomes ¯ y = ⎛ ⎜ ⎜ ⎝ 0 1 2 1 ⎞ ⎟ ⎟ ⎠ ¯ y • Eigenvalues are roots of the characteristic polynomial λ 2 − λ − 2 = 0 Eigenvalues and corresponding eigenvectors are given by λ 1 = 2 v 1 = (1 , 2) λ 2 = − 1 v 2 = (1 , − 1) • From this information alone, we know the qualitative proper ties of the solutions are as given in the phase plane diagram: – Solutions are roughly hyperbolic in shape with asymptotes along the eigenvectors. Along the eigenvector v 1 , the so lutions ﬂow off to infinity; along the eigenvector v 2 , the solutions converge to zero. – Solutions ﬂow in directions consistent with ﬂows along asymp totes – On the yaxis, we have y = 0, which means that everywhere on the yaxis (except at the stationary point 0), the solution must have a vertical tangent. – On the yaxis, we have y = 0, so we have y = 2 y + y = y 2 Thus, above the yaxis, y = y > 0, so y is increasing along the direction of the solution; below the yaxis, y = y < 0, so y is decreasing along the direction of the solution. – Along the line y = − 2 y , y = 2 y − 2 y = 0, so y achieves a minimum or maximum where it crosses that line. • General solution is given by ⎛ ⎜ ⎜ ⎝ y ( t ) y ( t ) ⎞ ⎟ ⎟ ⎠ = Mtx U,V ( id ) ⎛ ⎜ ⎜ ⎝ e 2( t − t ) e − ( t − t ) ⎞ ⎟ ⎟ ⎠ Mtx V,U ( id ) ⎛ ⎜ ⎜ ⎝ y ( t ) y ( t ) ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 1 1 2 − 1 ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ e 2( t − t ) e − ( t − t ) ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ 1 / 3 1 / 3 2 / 3 − 1 / 3 ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ y ( t ) y ( t ) ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 1 1 2 − 1 ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ e 2( t − t ) 3 e 2( t − t ) 3 2 e − ( t − t ) 3 − e − ( t − t ) 3 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ y ( t ) y ( t ) ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ e 2( t − t ) +2 e −...
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This note was uploaded on 10/04/2010 for the course ECON 204 taught by Professor Anderson during the Summer '08 term at Berkeley.
 Summer '08
 ANDERSON
 Economics

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