Laplace transform to solve the initial-value problem y"+4y=sin(t)U(t-2*pi), y(0)=1, y'(0)=0? (2024)

You have:

y" + 4y = sin(t)*u(t - 2*π), with y(0)=1, y'(0)=0

If F(s) is the Laplace transform of f(t), then the "time-shift" property of the Laplace transform says that the Laplace transform of f(t - a)*u(t-a) = exp(-a*s)*F(s).

Your differential equation has the nono-hom*ogeneous term, sin(t)*u(t-2*π), which is not quite in the format we need to apply the time-shift property. Note, however, that sin(t) = sin(t + n*π), where n = 0, ±1, ±2, ±3...., so we can rewrite your differential equation as :

y" + 4y = sin(t - 2*π)*u(t - 2*π)

Now take the Laplace transform (using a table of transforms), letting Y(s) be the transform of y(t):

(s^2)*Y(s) - s*y(0) - y'(0) + 4*Y(s) = exp(-2π*s)*(1/(s^2 + 1))

Plug in the initial values: and simplify:

Y(s)*(s^2 + 4) - s = exp(-2π*s)*(1/(s^2 + 1))

Y(s) = exp(-2π*s)*(1/(s^2 + 4))*(1/(s^2 + 1)) + s/(s^2 + 4)

Y(s) = (1/3)*exp(-2π*s)*[1/(s^2 + 1) - 1/(s^2 + 4)] + s/(s^2 + 4)

Now take the inverse transform, again using the time-shift property:

y(t) = (1/3)*u(t - 2π)*[sin(t) - (1/2)*sin(2t)] + (1/2)*cos(2t)

Only partial help, but it might get you started.

Solving the left side is equivalent to finding the complementary function

L[y''(t) + 4y(t)] = s^2 Y - sy(0) - y'(0) + 4Y

substituting y(0) = 1 anf y'(0) = 0 and IF RHS was zero, (it is not), you would have

(s^2 + 4)Y - s = 0

Y = s/(s^2 = 4) ==> y(t) = cos(2t)

But of course the situation you actually have is

(s^2 + 4)Y - s = L[u(t - 2pi)*sin(t)]

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That means we need to find the LT of a product of a function and Heaviside or step function

A table of transforms quotes that

L[u(t)*f(t - 2pi)] = e^(-cs)*L[f(t)] = e^(-cs)*F(s) so we can say that

L[u(t)*sin(t - 2pi)] = e^(-2pi*s)*L[sin(t)] = e^(-2pi*s)*/(s^2 + 1)

but we want L[u(t - 2pi)*sin(t)] which just seems to need shifting.

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In this page from "Paul's online notes" he starts by explaining that in order to take the transform of a function that involves a Heaviside we’ve got to make sure the function has been properly shifted.

I suggest you take over here and follow his examples through to see the shifting idea.

http://tutorial.math.lamar.edu/Classes/DE/IVPWithStepFunction.aspx

There is another helpful page here about Unit Step Functions and Laplace Transforms

http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/DiffEq/TheUnitStepFunction.pdf

Regards - Ian

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