Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Second-order linear ODE

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?

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

(edited 9 years ago)

These are solutions to the equation for simple harmonic motion I do believe. So you can read a little more into it using that.

OP

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?

Original post by halpme

Did you forget to square root your solutions?

He did yes

Original post by halpme

Did you forget to square root your solutions?

sorry, you can say that
I meant w² = ...

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

(edited 9 years ago)

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