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**1. Homework Statement**

A particle of mass m is confined to a horizontal plane. It is elastically bound to its equilibrium position by an isotropic elastic force,

**F**=mω^2

**r**

Where

**r**is the displacement of the particle from equilibrium and ω is a real constant parameter.

From Newton's law, we obtain the equation of motion:

**r**'' + ω^2

**r**= 0.

In rect. coordinates

**r**(t) =

**i**x(t) +

**j**y(t).

Obtain expressions for x(t) and y(t).

**2. Homework Equations**

**3. The Attempt at a Solution**

The x-component of force :

Fx = -x(t)mω^2 = mx''(t)

x''(t) + x(t)ω^2 = 0

The solution to this DE is x(t) = c1cos(ωt) + c2sin(ωt)

Then I'll apply the initial conditions given in the problem to obtain the constants.

Here's the part I'm unsure about

When I solve the differential equation for the y component I get

y(t) = a1cos(ωt) + a2sin(ωt). I'm thinking I have to multiply this by t so that the solutions are linearly independent:

y(t) = a1*tcos(ωt) + a2*tsin(ωt). Is this correct?