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Hello Sun,

The only difference between this problem and the one from yesterday is that you need to know how the Laplace transform behaves when applied to a derivative.  This should be something you have already covered, or at least it should be in your table of transforms, since you are taking a differential equations course.  Nevertheless, let us repeat the result (I'm stuck using the British Pound symbol for Laplace transform).  Here, I am letting capital letters represent the corresponding Laplace tranform, so for instance, F(s) = £{f(t)}.

£{f(n)(t)} = snF(s) - sn-1f(0) - sn-2f'(0) - ··· - sf(n-2)(0) - f(n-1)(0).

Using this result, the initial conditions, and the linearity of the Laplace transform, we obtain, upon transforming your DE, the following:

£{y'' + ω2y = g(t)}                             =>

£{y''} + ω2£{y} = £{g(t)}                 =>

s2Y(s) - sy(0) - y'(0) + ω2Y(s) = G(s)    =>

Y(s) = 1/(s22) + G(s)/(s22).

Now, let H(s) = 1/(s22), and note that £-1{H(s)} = (1/ω)sin(ωt).  Then, taking inverse Laplace transforms, and using linearity, we obtain

£-1{Y(s)} = y(t) = £-1{H(s)} + £-1{G(s)H(s)} = (1/ω)sin(ωt) + (1/ω)∫0t sin(ω(t-τ))g(τ)dτ.


Hassan H.