Help with solving a IH equation

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SUMMARY

The discussion centers on solving the second-order linear differential equation d²i/dt² + 6di/dt + 25i = -292sin(4t). The proposed solution, i(t) = Ae^(-3t)cos(4t) + Be^(-3t)sin(4t) + 292/9(cos(4t)) + 146/9(sin(4t)), is confirmed to be reasonable. The solution is divided into the homogeneous part, ih(t), and the particular part, ip(t). To verify correctness, one must check that ip'' + 6ip' + 25ip equals -292sin(4t).

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Homework Statement



d2i/dt2 + 6di/dt + 25i = -292sin4t

Homework Equations


The Attempt at a Solution



i got a solution of
Ae^(-3t)cos4t + Be^(-3t)sin4t + 292/9(cos4t) + 146/9(sin4t)
want to know if I am right
 
Last edited:
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Your solution looks reasonable. Your solution can be divided into two parts: the solution to the homogeneous equation i'' + 6i' + 25i = 0, and the particular solution to the nonhomogeneous equation you have.

So your general solution is i(t) = ih(t) + ip(t), where ih(t) consists of the first two terms of your solution, and ip(t) consists of the last two terms.

Your homogeneous solution checks with what I got. To confirm that your general solution is correct, all you need to do is check that your particular solution actually works.

From ip(t) = (292/9)cos4t + (146/9)sin4t, calculate ip' and ip''. If ip'' + 6ip' + 25ip equals -292sin4t, your solution is correct.
 

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