Express the solution of y"+ω^2*y=g(t); y(0)=0, y'(0)=1 in terms of a convolution integral.

Answer: y=(1/ω)sin(ωt)+(1/ω)∫ sin ω (t-torque) g(torque) d(torque) from 0 to t.

Express the solution of y"+ω^2*y=g(t); y(0)=0, y'(0)=1 in terms of a convolution integral.

Answer: y=(1/ω)sin(ωt)+(1/ω)∫ sin ω (t-torque) g(torque) d(torque) from 0 to t.

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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)} = s^{n}F(s) - s^{n-1}f(0) - s^{n-2}f'(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'' + ω^{2}y = g(t)} =>

£{y''} + ω^{2}£{y} = £{g(t)} =>

s^{2}Y(s) - sy(0) - y'(0) + ω^{2}Y(s) = G(s) =>

Y(s) = 1/(s^{2}+ω^{2}) + G(s)/(s^{2}+ω^{2}).

Now, let H(s) = 1/(s^{2}+ω^{2}), 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)∫_{0}^{t} sin(ω(t-τ))g(τ)dτ.

Regards,

Hassan H.