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Differential EquationsHardLaplace II: Steps, Impulses & Convolution

Laplace II: Steps, Impulses & Convolution · Hard problem

The impulse response of y′′+4y′+13y=δ(t)y''+4y'+13y=\delta(t) (from rest) is h(t)=13e−2tsin⁡3th(t)=\dfrac13e^{-2t}\sin3t. Find the time t>0t>0 at which h(t)h(t) reaches its FIRST maximum.