Find the Inverse Laplace transform of

Question

$$\mathscr{L}^{-1} \frac{(s+λ)}{((s+λ)^2+μ^2)(s^2+1)}$$

Here, λ, μ are constants. I tried to decompose into partial fractions and solve but failed badly. $$\frac{(s+λ)}{((s+λ)^2+μ^2)(s^2+1)}=\frac{(Cs+D)}{((s+λ)^2+μ^2)}+\frac{(Es+F)}{(s^2+1)}$$ $$(s+λ)=(Cs+D)(s^2+1)+(Es+F)((s+λ)^2+μ^2)$$ By equating the coefficients i got, $$C+E=0~~.~.~.~.~.~.~.~.~.~(i)$$ $$D+Eλ+F=0~~.~.~.~.~.~.~.~.~.~(ii)$$ $$C+E(λ^2+μ^2)+Fλ=1~~.~.~.~.~.~.~.~.~.~(iii)$$ $$D+F(λ^2+μ^2)=λ~~.~.~.~.~.~.~.~.~.~(iv)$$ Can anyone help me with this?

Answer 1

Use convolution theorem.

$x(t) = \mathscr{L}^{-1}\left\{\dfrac{s+\lambda}{((s+\lambda)^2+\mu^2)(s^2+1)}\right\} = \mathscr{L}^{-1}\left\{\dfrac{s+\lambda}{(s+\lambda)^2+\mu^2}\right\}*\mathscr{L}^{-1}\left\{\dfrac{1}{s^2+1}\right\}$

$x(t) =\left(e^{-\lambda t}\cos(\mu t)\right)*\sin(t) = \int_0^t e^{-\lambda y}\cos(\mu y) \sin(t-y)dy$.

You may then use the following formulas to evaluate the integral.

$\cos a \sin b = \dfrac{1}{2}\left[\sin(a+b)-\sin(a-b)\right]$
$\int e^{az}\sin(bz)dz = \dfrac{e^{az}}{a^2+b^2}\left[a\sin(bz)-b\cos(bz)\right]$

Answer 2

hint

use this one $$\frac{(s+λ)}{((s+λ)^2+μ^2)(s^2+1)}=\frac{A}{(s+λ)+iμ} + \frac{B}{(s+λ)-iμ} + \frac{C}{s+i} + \frac{D}{s-i}$$