08Ma2aPracSol6

# 08Ma2aPracSol6 - Math 2a Practical Fall 2008 Solutions to...

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Unformatted text preview: Math 2a Practical, Fall 2008: Solutions to Homework 6 You are bound by the honor code to not look at this file if you did not take Math 2a in the Fall of 2008 1. Write down a fundamental matrix for the solutions of the system y = 3 1 0 0 0 0 0 3 1 0 0 0 0 0 3 1 0 0 0 0 0 3 1 0 0 0 0 0 3 1 0 0 0 0 0 3 y and use it to solve the initial value problem y (0) = 1 2 3 4 5 6 . Solution. We solve the system in reverse order: y 6 = 3 y 6 ⇒ y 6 ( t ) = c 6 e 3 t . y 5 = 3 y 5 + y 6 = 3 y 5 + c 6 e 3 t . This is an easy first-order linear ODE. Its solution, for a given c 6 , is y 5 ( t ) = ( c 6 t + c 5 ) e 3 t . y 4 = 3 y 4 + y 5 = 3 y 4 + ( c 6 t + c 5 ) e 3 t . The homogeneous solution is given by y 4 ( t ) = c 4 e 3 t , and we use the method of undetermined coefficients to find a particular solution. Suppose a particular solution is given by y 4 ( t ) = ( At 2 + Bt ) e 3 t , then y 4 ( t ) = (2 At + B ) e 3 t + 3( At 2 + Bt ) e 3 t = (2 At + B ) e 3 t + 3 y 4 ( t ), and so ( c 6 t + c 5 ) e 3 t = y 4 ( t )- 3 y 4 ( t ) = (2 At + B ) e 3 t ⇒ A = c 6 / 2 ,B = c 5 . Thus for given c 6 and c 5 , the general solution for y 4 is y 4 ( t ) = ( c 6 2 t 2 + c 5 t + c 4 ) e 3 t . Similarly, we can find that y 3 ( t ) = ( c 6 3! t 3 + c 5 2! t 2 + c 4 t + c 3 ) e 3 t , y 2 ( t ) = ( c 6 4! t 4 + c 5 3! t 3 + c 4 2! t 2 + c 3 t + c 2 ) e 3 t , and y 1 ( t ) = ( c 6 5! t 5 + c 5 4! t 4 + c 4 3! t 3 + c 3 2! t 2 + c 2 t + c 1 ) e 3 t . In matrix form, we can write y 1 ( t ) y 2 ( t ) y 3 ( t ) y 4 ( t ) y 5 ( t ) y 6 ( t ) = e 3 t te 3 t 1 2! t 2 e 3 t 1 3! t 3 e 3 t 1 4! t 4 e 3 t 1 5! t 5 e 3 t e 3 t te 3 t 1 2! t 2 e 3 t 1 3! t 3 e 3 t 1 4! t 4 e 3 t e 3 t te 3 t 1 2! t 2 e 3 t 1 3!...
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## This note was uploaded on 01/08/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.

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08Ma2aPracSol6 - Math 2a Practical Fall 2008 Solutions to...

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