Economics Dynamics Problems 77

# Economics Dynamics Problems 77 - = c 1 e r (0) + c 2 e s...

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Continuous dynamic systems 61 which is referred to as the auxiliary equation of the homogeneous equation. The quadratic has two solutions r = b + b 2 4 ac 2 a , s = b b 2 4 ac 2 a If b 2 > 4 ac the roots r and s are real and distinct; if b 2 = 4 ac the roots are real and equal; while if b 2 < 4 ac the roots are complex conjugate. There are, therefore, three types of solutions. Here we shall summarise them. 2.8.1 Real and distinct ( b 2 > 4 ac ) If the auxiliary equation has distinct real roots r and s , then e rt and e st are linearly independent solutions to the second-order linear homogeneous equation. The gen- eral solution is y ( t ) = c 1 e rt + c 2 e st where c 1 and c 2 are arbitrary constants. If y (0) and y ± (0) are the initial conditions when t = 0, then we can solve for c 1 and c 2 y (0)
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Unformatted text preview: = c 1 e r (0) + c 2 e s (0) = c 1 + c 2 y ± ( t ) = rc 1 e rt + sc 2 e st y ± (0) = rc 1 e r (0) + sc 2 e s (0) = rc 1 + sc 2 Hence c 1 = y ± (0) − sy (0) r − s , c 2 = y ± (0) − ry (0) s − r and the particular solution is y ( t ) = ± y ± (0) − sy (0) r − s ² e rt + ± y ± (0) − ry (0) s − r ² e st which satisFes the initial conditions y (0) and y ± (0). Example 2.17 Suppose d 2 y dt 2 + 4 dy dt − 5 y = Then the auxiliary equation is x 2 + 4 x − 5 = ( x + 5)( x − 1) = Hence, r = − 5 and s = 1, with the general solution y ( t ) = c 1 e − 5 t + c 2 e t...
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## This note was uploaded on 12/14/2011 for the course ECO 3023 taught by Professor Dr.gwartney during the Fall '11 term at FSU.

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