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ODE question

  • Thread starter longball
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  • #1
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setting a sinusoidal voltage term, the ODE can be written as

(d^2 i)/(d t^2 ) + 25i = A0 sin (ϖt)

assuming that ϖ^2 ≠ 25, determine the current i in terms of the parameters (ϖ and A0) and the variable t when the initial conditions are

i(0) = di/dt (0) = 0

i really don't have much of an idea what to do here.

so far i have found

L.F ic= Acos(5t)+Bsin(5t)

P.S ip = acos(ϖt) + bsin (ϖt)

ip' = -ϖasin (ϖt) + ϖbcos (ϖt)

ip'' = -2ϖacos (ϖt) - 2ϖbsin (ϖt)

therefore

[-2ϖacos (ϖt) - 2ϖbsin (ϖt)] + [acos(ϖt) + bsin (ϖt)] = A0 sin (ϖt)

any help or further guidance to the work i have done would be much appreciated
 

Answers and Replies

  • #2
938
9
Since the second derivative of sin is -sin, it would seem a good guess to try an ansatz [tex] i = A \sin 5t + B \cos 5t + C \sin \omega t [/tex]
 
  • #3
33,627
5,284
setting a sinusoidal voltage term, the ODE can be written as

(d^2 i)/(d t^2 ) + 25i = A0 sin (ϖt)

assuming that ϖ^2 ≠ 25, determine the current i in terms of the parameters (ϖ and A0) and the variable t when the initial conditions are

i(0) = di/dt (0) = 0

i really don't have much of an idea what to do here.

so far i have found

L.F ic= Acos(5t)+Bsin(5t)

P.S ip = acos(ϖt) + bsin (ϖt)
I don't know what L.F and P.S stand for, but these two functions are your complementary solution and particular solution. The general solution is the sum of these two functions. Use your initial conditions to solve for A and B in your complementary solution, and use the fact that ip'' + 25ip = A0sin(ϖt) to solve for the constants a and b.
ip' = -ϖasin (ϖt) + ϖbcos (ϖt)

ip'' = -2ϖacos (ϖt) - 2ϖbsin (ϖt)

therefore

[-2ϖacos (ϖt) - 2ϖbsin (ϖt)] + [acos(ϖt) + bsin (ϖt)] = A0 sin (ϖt)

any help or further guidance to the work i have done would be much appreciated
 

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