Express Laplace Transform of y(t) in given form.

[chinye11]

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) = \frac{2s+7}{s^{2}+2s+7} + \frac{e^{-s}}{s(s^{2}+2s+7)} + \frac{1}{s^{2}(s^{2}+2s+7)}+\frac{e^{-s}}{s^{2}(s^{2}+2s+7)}

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

\frac{2s+7}{s^{2}+2s+7} + \frac{1}{s^{2}(s^{2}+2s+7)}+\frac{e^{-s}}{s^{2}(s^{2}+2s+7)}

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

\frac{1}{s^{2}} -e^{s}(\frac{1}{s^{2}})?

 

[LCKurtz]

Yes, $r(t) = t(H(t) - H(t-1))$ but you are missing a term in the transform. You can check it yourself by just calculating the integral$$
\int_0^1 te^{-st}\, dt$$

 

[chinye11]

OK, Thanks very much, I realized that I had applied the shifting theorem to an equation of the form:
y(t) (H(t-a)) when y(t-a) (H(t-a)) is required

This dropped a H(t-1) term which was the missing term.