204Lecture152008Web - Economics 204 Lecture 15Friday,...

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Economics 204 Lecture 15–Friday, August 15, 2008 Second Order Linear Diferential Equations Consider the second order diferential equation y 0 = cy + by 0 with b, c R . Rewrite this as a frst order linear diferential equation in two variables: ¯ y ( t )= y ( t ) y 0 ( t ) ¯ y 0 ( t y 0 ( t ) y 0 ( t ) = 01 cb y ( t ) y 0 ( t ) = ¯ y The eigenvalues are b ± b 2 +4 c 2 , the roots oF the equation λ 2 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 o± the Diferential Equations Handout. Example 1 Consider the second order linear diferential equation y 0 =2 y + y 0 As above, let ¯ y = y y 0 1
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so the equation becomes ¯ y 0 = 01 21 ¯ 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) From this information alone, we know the qualitative properties 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 solutions flow o± to in²nity; along the eigenvector v 2 , the solutions converge to zero. Solutions flow in directions consistent with flows along asymptotes On the y -axis, we have y 0 = 0, which means that everywhere on the y -axis (except at the stationary point 0), the solution must have a vertical tangent. On the y 0 -axis, we have y =0 ,sowehave y 0 y + y 0 = y 0 Thus, above the y -axis, y 0 = y 0 > 0, so y 0 is increasing along the direction of the solution; below the y -axis, y 0 = y 0 < 0, so y 0 is decreasing along the direction of the solution. 2
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Along the line y 0 = 2 y , y 0 =2 y 2 y =0 ,so y 0 achieves a minimum or maximum where it crosses that line. General solution is given by y ( t ) y 0 ( t ) = Mtx U,V ( id ) e 2( t t 0 ) 0 0 e ( t t 0 ) V,U ( id ) y ( t 0 ) y 0 ( t 0 ) = 11 2 1 e 2( t t 0 ) 0 0 e ( t t 0 ) 1 / 31 / 3 2 / 3 1 / 3 y ( t 0 ) y 0 ( t 0 ) = 2 1 e 2( t t 0 ) 3 e 2( t t 0 ) 3 2 e ( t t 0 ) 3 e ( t t 0 ) 3 y ( t 0 ) y 0 ( t 0 ) = e 2( t t 0 ) +2 e ( t t 0 ) 3 e 2( t t 0 ) e ( t t 0 ) 3 2 e 2( t t 0 ) 2 e ( t t 0 ) 3 2 e 2( t t 0 ) + e ( t t 0 ) 3 y ( t 0 ) y 0 ( t 0 ) = y ( t 0 )+ y 0 ( t 0 ) 3 e 2( t t 0 ) + 2 y ( t 0 ) y 0 ( t 0 ) 3 e ( t t 0 ) 2 y ( t 0 )+2 y 0 ( t 0 ) 3 e 2( t t 0 ) + 2 y ( t 0 )+ y 0 ( t 0 ) 3 e ( t t 0 )
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204Lecture152008Web - Economics 204 Lecture 15Friday,...

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