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solve y''+5y'+6y=2 where y(0)=y'(0)=0 using laplace transformation.
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Start by taking the Laplace transforms of both sides, using linearity. Let \[Y(s) = \mathcal{L}[y(t)]\] \(\mathcal{L}[y'']+5\mathcal{L}[y']+6\mathcal{L}[y]=\mathcal{L}[2]\) Now use your laplace table|dw:1457883208136:dw| see if you can set it up, look down at 35 - 37
So what I mean is \[\mathcal L [f''(t)] = s^2 Y(s)-s f(0)-f'(0)\] etc
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