# Find inverse Laplace transform of (s)/(s^2+s+1)

Solve the following equation:
${y}^{″}+{y}^{\prime }+y=1+{e}^{-t}$
After applying the Laplace transform and using partial fractions I have terms of the form:
$\frac{1}{{s}^{2}+s+1}$
and
$\frac{s}{{s}^{2}+s+1}$
which I don't know how to reverse transform.
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r2t1orrso
You didn't give the initial conditions...a bit hard to check your solution
You have that
$\frac{s}{{s}^{2}+s+1}=\frac{s}{\left({s}^{2}+s+1/4+3/4\right)}=\frac{s}{\left(s+1/2{\right)}^{2}+3/4}$
$\frac{s}{\left(s+1/2{\right)}^{2}+3/4}=\frac{s+1/2}{\left(s+1/2{\right)}^{2}+3/4}-\frac{1/2}{\left(s+1/2{\right)}^{2}+3/4}$
And look at the sine and cosine function in the table...
$\mathcal{L}\left({e}^{at}\mathrm{cos}\left(\beta t\right)\right)=\frac{s-a}{\left(s-a{\right)}^{2}+{\beta }^{2}}$
$\mathcal{L}\left({e}^{at}\mathrm{sin}\left(\beta t\right)\right)=\frac{\beta }{\left(s-a{\right)}^{2}+{\beta }^{2}}$