halpme
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Show that the following are solutions [(a),(b)] to the differential equation:
\frac{d^2y}{dx^2}=-\omega^2y
(a) y=2 e^{-2jx}
(b) y=-4 e^{5jx}

I am not actually sure where to start here, I am very confused. I am not given a definition of y in terms of x. First steps?
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TeeEm
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(Original post by halpme)
Show that the following are solutions [(a),(b)] to the differential equation:
\frac{d^2y}{dx^2}=-\omega^2y
(a) y=2 e^{-2jx}
(b) y=-4 e^{5jx}

I am not actually sure where to start here, I am very confused. I am not given a definition of y in terms of x. First steps?
differentiate each expression twice and substitute into the ODE
it should balance with w=4 in the first and w =25 in the second
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poorform
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These are solutions to the equation for simple harmonic motion I do believe. So you can read a little more into it using that.
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halpme
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(Original post by TeeEm)
differentiate each expression twice and substitute into the ODE
it should balance with w=4 in the first and w =25 in the second
Did you forget to square root your solutions?
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Phichi
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(Original post by halpme)
Did you forget to square root your solutions?
He did yes
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TeeEm
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(Original post by halpme)
Did you forget to square root your solutions?
sorry, you can say that
I meant w2 = ...
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Phichi
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(Original post by halpme)
Show that the following are solutions [(a),(b)] to the differential equation:
\frac{d^2y}{dx^2}=-\omega^2y
(a) y=2 e^{-2jx}
(b) y=-4 e^{5jx}

I am not actually sure where to start here, I am very confused. I am not given a definition of y in terms of x. First steps?
As mentioned in post 3 regarding S.H.M, if you are some what knowledgeable in physics you'd be able to see that the modulus of the exponent in both y expression is equal to \omega x in this case.

Hence from inspection, for part a)

|-2jx| = \omega x

\therefore \omega = 2

and for b)

|5jx| = \omega x

\therefore \omega = 5
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