hw1solution

# hw1solution - Homework 1 − Solution 1. For the given...

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Unformatted text preview: Homework 1 − Solution 1. For the given second order ordinary differential equation with initial conditions x ̈ 4 x ̇ 20 x 3sin5 t , x 1, x ̇ − 1 we first formulate the general solution as the superposition of homogeneous and particular solutions: x x h x p The homogeneous solution is established by solving the following homogeneous equation x ̈ h 4 x ̇ h 20 x h The homogeneous solution is in the exponential form x h ae t Substituting into the homogeneous equation results in the following characteristic equation 2 4 20 The roots of the characteristic equation are 1,2 − 2 4 i Since the two roots are complex conjugate pair, the homogeneous solution can be written in the following form: x h e − 2 t A 1 sin4 t A 2 cos4 t Now we need to find the particular solution: x ̈ p 4 x ̇ p 20 x p 3sin5 t Using the method of undetermined coefficients we assume the particular solution to be x p B 1 sin5 t B 2 cos5 t Plug it back into the differential equation and we have − 5 B 1 sin5 t B 2 cos5 t 20 B 1 cos5 t − B 2 sin5 t 3sin5 t Equate the coefficients for both the sine and cosine functions to solve the constants B 1 and B 2 : − 5 B 1 − 20 B 2 3 − 5 B 2 20 B 1 B 1 − 3 85 , B 2 − 12 85 The full solution can now be written as x e − 2 t A 1 sin4 t A 2 cos4 t − 3 85 sin5 t − 12 85 cos5 t We can now apply the initial conditions to solve the two remaining constants A 1 and A 2 : x 1 A 2 − 12 85 1 A 2 97 85 x ̇ − 1 − 2 A 2 4 A 1 − 3 17 − 1 A 1 31 85 Hence the total solution is x t 1 85 e − 2 t 31sin4 t 97cos4 t − 3 sin5 t 4cos5 t...
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## This note was uploaded on 04/30/2008 for the course ME 3504 taught by Professor Tschang during the Spring '08 term at Virginia Tech.

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hw1solution - Homework 1 − Solution 1. For the given...

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