- #1
chinye11
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Homework Statement
y(t) solves the following IVP
y''(t) + 2y'(t) + 10y(t) = r(t)
y(0) = 2
y'(0) = 3
r(t) =
0 if t < 0
t if 0 ≤ t ≤ 1
0 if t > 1
Demonstrate that the laplace transform of y(t) is
Y(s) = [itex]\frac{2s+7}{s^{2}+2s+7}[/itex] + [itex]\frac{e^{-s}}{s(s^{2}+2s+7)}[/itex] + [itex]\frac{1}{s^{2}(s^{2}+2s+7)}[/itex]+[itex]\frac{e^{-s}}{s^{2}(s^{2}+2s+7)}[/itex]
Homework Equations
H(t) is the heaviside step function which is 0 when t is less than 0 and 1 when t is greater than 1, (undefined at 0.)
Tables of common laplace transforms are available on the internet e.g. http://www.rapidtables.com/math/calculus/laplace_transform.htm
3. Attempt at a Solution
Okay so I use the laplace transform on both sides of the equation with
r(t) defined as t H(t) - t H(t-1)
When I calculate this I finish with
[itex]\frac{2s+7}{s^{2}+2s+7}[/itex] + [itex]\frac{1}{s^{2}(s^{2}+2s+7)}[/itex]+[itex]\frac{e^{-s}}{s^{2}(s^{2}+2s+7)}[/itex]
This is similar to the needed answer however I seem to have dropped a factor of se-s on the right side somewhere and I cannot see any mistakes in the algebra.
Is my definition of r(t) correct and am I correct to say that the laplace transform of t H(t) - t H(t-1) is
[itex]\frac{1}{s^{2}}[/itex] -e[itex]^{s}[/itex]([itex]\frac{1}{s^{2}}[/itex])?