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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Marked as Best Answer

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.

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