Need help!!!. Laplace Convolution?

Question

applying convolution

y"+2y=3 sin (2t)

y(0) = 0 y'(0)=0

Answer 1

Apply L to both sides:

(s^2 Y(s) - 0s - 0) + 2Y(s) = 3 * 2/(s^2 + 2^2)

==> (s^2 + 2) Y(s) = 6/(s^2 + 4)

==> Y(s) = 6/[(s^2 + 2)(s^2 + 4)]



By partial fractions:

Y(s) = 3/(s^2 + 2) - 3/(s^2 + 4).



Inverting yields y(t) = (3/√2) sin(t√2) + (3/2) sin(2t).

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P.S.: Via convolution, we have

y(t) = L⁻¹ {6/[(s^2 + 2)(s^2 + 4)]}

......= L⁻¹ {1/(s^2 + 2)} * L⁻¹ {6/(s^2 + 4)}

......= (1/√2) sin(t√2) * (6/2) sin(2t)

......= ∫(s = 0 to t) (1/√2) sin(s√2) · (6/2) sin(2(t - s)) ds, by definition of convolution (*)

......= ∫(s = 0 to t) (-3/√2) sin(s√2) sin(2s - 2t) ds

......= ∫(s = 0 to t) (-3/√2) (1/2) [cos(s√2 - (2s - 2t)) - cos(s√2 + (2s - 2t))] ds, trig. identity

......= (-3/(2√2)) [sin(s√2 - (2s - 2t))/(√2 - 2) - sin(s√2 + (2s - 2t))/(√2 + 2)] {for s = 0 to t}

......= (-3/(2√2)) [sin(t√2) (1/(√2 - 2) - 1/(√2 + 2)) - sin(2t) (1/(√2 - 2) + 1/(√2 + 2))]

......= (-3/(2√2)) [-2 sin(t√2) - √2 sin(2t)], via common denominators

......= (3/√2) sin(t√2) + (3/2) sin(2t).



I hope this helps!

Answer 2

The Laplace remodel of a convolution is the made from the convolutions of the two purposes. it is regularly used in opposite. in case you comprehend the inverse transforms of F(s) and G(s) and that they are purposes f(x) and g(x), then the inverse remodel of F(s)G(s) is the convolution f*g(x)=int_0^x f(t)g(x-t)dt.

Answer 3

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