laplace transformation solve heaviside d.e. $y''+4y=U(t-4)$

Question

$y''+4y=U(t-4)$ so that $y(0)=3$ and $y'(0)=-2$

I have applied the transformation in both terms obtaining $Y=\frac{3s^2+10s+1-e^{4s}}{s(s+4)}$.

How can i solve it?

Answer 1

We are given:

$$y''+ 4y = U(t-4); y'(0)= -2, y(0)= 3$$

Taking the Laplace transform yields:

$$\mathscr{L}(y''+ 4y = U(t-4)) = (s^2 y(s) -s y(0) -y'(0)) +4 (y(s)) = \dfrac {e^{-4s}}{s}$$

Substituting the ICs yields:

$$(s^2 y(s) -3 s +2) + 4 y(s) = \dfrac {e^{-4s}}{s}$$

Solving for $y(s)$ yields:

$$y(s) = \dfrac{3s^2 -2s + e^{-4s}}{s(s^2+4)} = e^{-4s}\left(\frac{1}{4 s}-\frac{s}{4 (s^2+4)}\right)+ \dfrac{3s}{s^2+4} - \dfrac{2}{s^2+4} $$

Update

The inverse Laplace Transform (note that this can be simplified, but I am writing the four terms pre-simplifying so you can follow along):

$$y(t) = \frac{1}{4}U(t-4) -\frac{1}{4} U (t-4) \cos (2 (t-4)) + 3 \cos (2 t) -\sin (2 t)$$