Laplace Transform, Finding solution: y′′+4y′+4y=f(t)

[jakejakejake]

y′′+4y′+4y=f(t)
where f(t)=cos(ωt) if 0<t<π and f(t)=0 if t>π?
The initial conditions are y(0) = 0 , y'(0) = 1

I know that f(t)=cos(ωt)−uπ(t)cos(ωt), the heaviside equation.

AND ω is allowed to vary, supposed to find the general solution, i.e. f(t) in terms of ω

I think that after applying Laplace to both sides, I get: (s + 2)² * F(s) - 1 = s / [ s² + w²] e^(-πs) * [s / (s² + ω²)] But I'm still not sure where to go from here...

Thanks in advance!

 

[mathmari]

y′′+4y′+4y=f(t)
where f(t)=cos(ωt) if 0<t<π and f(t)=0 if t>π?
The initial conditions are y(0) = 0 , y'(0) = 1

I know that f(t)=cos(ωt)−uπ(t)cos(ωt), the heaviside equation.

AND ω is allowed to vary, supposed to find the general solution, i.e. f(t) in terms of ω

I think that after applying Laplace to both sides, I get: (s + 2)² * F(s) - 1 = s / [ s² + w²] e^(-πs) * [s / (s² + ω²)] But I'm still not sure where to go from here...

Thanks in advance!

$$f(t)=\cos{( \omega t)} - u_{\pi}(t) \cos{( \omega t)}$$$$\mathcal{L}(y)=Y(s)$$

$$\mathcal{L}(y')=sY(s)-y(0)=sY(s)$$

$$\mathcal{L}(y'')=s^2Y(s)-sy(0)-y'(0)=s^2Y(s)-1$$

$$\mathcal{L}(f(t))=\mathcal{L}(\cos{( \omega t)}-u_{\pi}(t) \cos{(\omega t)} ) = \\
\mathcal{L}(\cos{( \omega t)})-\mathcal{L}(u_{\pi}(t) \cos{(\omega t)}) = \\
\frac{s}{s^2+\omega^2}-e^{-\pi s} \mathcal{L}(\cos{(\omega (t+\pi))}) =
\\ \frac{s}{s^2+\omega^2}-e^{-\pi s} \mathcal{L}(\cos{(\omega t+\omega \pi)}) = \\
\frac{s}{s^2+\omega^2}-e^{-\pi s} \mathcal{L}(\cos{(\omega t)} \cos{(\omega \pi)}-\sin{(\omega t)} \sin{(\omega \pi)})= \\
\frac{s}{s^2+\omega^2}-e^{-\pi s} \cos{(\omega \pi)}\mathcal{L}(\cos{(\omega t)}) +e^{-\pi s}\sin{(\omega \pi)}\mathcal{L}(\sin{(\omega t)} ) = \\
\frac{s}{s^2+\omega^2}-e^{-\pi s} \cos{(\omega \pi)}\frac{s}{s^2+\omega^2} +e^{-\pi s}\sin{(\omega \pi)}\frac{\omega}{s^2+\omega^2}$$Therefore,

$$y''+4y'+4y=f(t) \overset{\mathcal{L}}{\Rightarrow } \\ s^2Y(s)-1+4sY(s)+4Y(s) =
\frac{s}{s^2+\omega^2}-e^{-\pi s} \cos{(\omega \pi)}\frac{s}{s^2+\omega^2} +e^{-\pi s}\sin{(\omega \pi)}\frac{\omega}{s^2+\omega^2}$$

$$\Rightarrow (s^2+4s+4)Y(s) = \frac{s}{s^2+\omega^2}-e^{-\pi s} \cos{(\omega \pi)}\frac{s}{s^2+\omega^2} +e^{-\pi s}\sin{(\omega \pi)}\frac{\omega}{s^2+\omega^2}+1$$

$$\Rightarrow (s+2)^2Y(s) = \frac{s}{s^2+\omega^2}-e^{-\pi s} \cos{(\omega \pi)}\frac{s}{s^2+\omega^2} +e^{-\pi s}\sin{(\omega \pi)}\frac{\omega}{s^2+\omega^2}+1$$

$$\Rightarrow Y(s)= \frac{1}{(s+2)^2} \left ( \frac{s}{s^2+\omega^2}-e^{-\pi s} \cos{(\omega \pi)}\frac{s}{s^2+\omega^2} +e^{-\pi s}\sin{(\omega \pi)}\frac{\omega}{s^2+\omega^2}+1\right )$$

$$\Rightarrow Y(s) =\frac{1}{(s+2)^2} \frac{s}{s^2+\omega^2}-e^{-\pi s} \frac{1}{(s+2)^2}\cos{(\omega \pi)}\frac{s}{s^2+\omega^2} +e^{-\pi s}\frac{1}{(s+2)^2} \sin{(\omega \pi)}\frac{\omega}{s^2+\omega^2}+\frac{1}{(s+2)^2}$$

Using partial fractions apply the inverse Laplace Transformation.