This is pretty basic as far as Laplace transform problems go. There's really not much to get stuck on. Still, here's my working.
Recall these properties of the Laplace transform L:
* My notation: L{x(t)} = X(s) by definition
* Linearity, so L{ax(t) + by(t)} = a X(s) + b Y(s) for any constants a and b
* L{0} = 0 from linearity
* L{x'(t)} = s X(s) + x(0)
* L{x''(t)} = s^2 X(s) - s x(0) - x'(0)
* L{e^(at)} = 1/(s-a)
Take the Laplace transform of your equation to get
L{y''(t) - 2y'(t) + 10y(t)}
= L{y''(t)} - 2 L{y'(t)} + 10 L{y(t)}
= [s^2 Y(s) - s y(0) - y'(0)] - 2 [s Y(s) - y(0)] + 10 Y(s)
= [s^2 - 2s + 10] Y(s) + [-s y(0) - y'(0) + 2y(0)]
= [s^2 - 2s + 10] Y(s) - [s + 6]
= L{0}
= 0
=> Y(s) = (s+6) / (s^2 - 2s + 10)
There are a few ways to take the inverse Laplace transform from here. You can factor the quadratic in the denominator (which has roots at s = 1 +/- 3i) and use partial fractions, which will give you something of the form 1/(s - a), which has inverse Laplace transform e^(at). You can also use the general form of the exponentially decaying sinusoids nearly directly. I'll pick the first method.
I won't write out the partial fractions steps, but after doing the computation one finds
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