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Unformatted text preview: Engineering Analysis 4 Workshop questions - week of 10/19/08 1. Consider a linear second order equation with constant coefficients, ay 00 + by + cy = 0 . (1) Suppose someone tells you that they believe that y = x 2 is a solution to this equation. Explain why he must be wrong. Answer - You can see this two ways. Since the function y ( x ) = x 2 satisfies the conditions y (0) = y (0) = 0 , it cannot be a solution to equation (1) because the zero solution satisfies these initial conditions and the solution must be unique. You can also see this by the general solution using the characteristic equation. Any such solution must be of the form y = e rx where r is a root of the characteristic equation ar 2 + br + c = 0 . If the roots are real and distinct you get exponential solutions. If the roots are real and re- peated you get an exponential solution multiplied by x . If the roots are complex, you get an exponential solution multiplied by a trig function. For all cases you cannot get a solution that looks like x 2 ....
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This note was uploaded on 09/08/2010 for the course GEN_ENG 205 taught by Professor Taflove during the Spring '08 term at Northwestern.
- Spring '08