Monday, July 29, 2013

Solutions to Rewriting Higher Order Differential Equations in Normal Form

1) y"  - 3y' + 4y = sin 3t

Solution:

Re-arrange to get:  y" = 3y - 4y + sin 3t

Let: y = x1,  y' = x2, then x2' = y"

Then we can rewrite the higher order DE pair as:

x2' = 3x2  - 4x1 + sin 3t  and x1' = x2 (since y = x1 and y' = x2))


2)   2 d2y/ dt2 + 4 dy/dt – 5y = 0


Divide through by 2 and transpose terms to get:

 d2y/ dt2 =   - 2 dy/dt + 5y/ 2

Let:     y = x1,  dy/dt  = y' = x2,  so that:    d2y/ dt2 =  x2'


Then:

x2'  = -2 x2  + 5 x1/2, and x1' = x2


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