By definition of the Laplace transform,
[tex]L(y''(t))=\displaystyle\int_0^\infty e^{-st}y''(t)\,\mathrm dt[/tex]
Integrate by parts:
[tex]u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt[/tex]
[tex]\mathrm dv=y''(t)\,\mathrm dt\implies v=y'(t)[/tex]
[tex]L(y''(t))=\displaystyle e^{-st}y'(t)\bigg|_0^\infty+s\int_0^\infty e^{-st}y'(t)\,\mathrm dt=sL(y'(t))-y'(0)[/tex]
Integrate by parts again to compute the transform of [tex]y'(t)[/tex]:
[tex]u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt[/tex]
[tex]\mathrm dv=y'(t)\,\mathrm dt\implies v=y(t)[/tex]
[tex]L(y'(t))=e^{-st}y(t)\bigg|_0^\infty+\displaystyle s\int_0^\infty e^{-st}y(t)\,\mathrm dt=s L(y(t))-y(0)[/tex]
Putting everything together, we have
[tex]L(y''(t))=s^2L(y(t))-sy(0)-y'(0)[/tex]
or in terms of the given initial conditions,
[tex]\boxed{L(y''(t))=s^2L(y(t))-sy_0-y_1}[/tex]