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Show that x=Be^(µt) is a solution of mx ̈+kx=0
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it's not. differentiate it.
I know i differentiated it and didn't get the right answer.
I've been given some more information: B=Ae^jϕ
and µ=α+jω
better. but have you differentiated that?? it's suddenly become more complicated!
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I'm afraid to try, but i'll give it a go.
lol! \[x=B e^{µ t} \ : \ B=A e^{jϕ} \ \ : \ µ=α+j ω \\ x=A e^{jϕ} e^{(α+j ω) t}\] \[x = A e^{α t+j ( ω t + \phi ) }\] looks interesting!!
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