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Cristopher Knox 2022-06-30 Answered
y ( t ) + s i n ( t ) y ( t ) g ( t ) y ( t ) + g ( t ) y ( t ) = f ( t )
Write the following third order differential equation as an equivalent system of first order ordinary differential equations and write it in the form
d z d t = f ( t , z )
Here is what I have so far:
x 1 ( t ) = y ( t )
x 2 ( t ) = y ( t )
x 3 ( t ) = y ( t )
so I can then do,
x 1 ( t ) = y ( t )
x 2 ( t ) = y ( t )
x 3 ( t ) = y ( t ) = f ( t ) s i n ( t ) y ( t ) + g ( t ) y ( t ) g ( t ) y ( t )
x 3 ( t ) = f ( t ) s i n ( t ) x 3 ( t ) + g ( t ) x 2 ( t ) g ( t ) x 1 ( t )
and I think the dynamics function is:
f ( t , z ) = [ x 1 ( t ) , x 2 ( t ) , x 3 ( t ) ]
f ( t , z ) = [ x 2 ( t ) , x 3 ( t ) , f ( t ) s i n ( t ) x 3 ( t ) + g ( t ) x 2 ( t ) g ( t ) x 1 ( t ) ]
I'm not sure if I'm doing this correctly. What does difference does the f(t) and g(t) make in this case? what is z exactly?
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Answers (1)

Nathen Austin
Answered 2022-07-01 Author has 14 answers
Remember that your final goal is to obtain a system of FIRST order equations. So, any higher derivatives need to be rewritten appropriately.
Using your notation, we have
x 1 = x 2 x 2 = x 3 x 3 = f ( t ) sin ( t ) x 3 + g ( t ) x 2 g ( t ) x 1
Here z would represent the 3-vector ( x 1 x 2 x 3 )
Try and think of the entire set of expressions on the right hand side as the output of multivariable function from R 3 R 3 . (It's not the best idea to use f again in this context. Think of f(t,z) as f ( t , x 1 , x 2 , x 3 ))
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