Express the solution?

Question

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.

Answer 1

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) - s^n-1f(0) - s^n-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/(s²+ω²) + G(s)/(s²+ω²).

Now, let H(s) = 1/(s²+ω²), 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.