What is expand e^{at}(\H(t-1)-\H(t-2)) ?

The solution to expand e^{at}(\H(t-1)-\H(t-2)) is e^{at}\H(t-1)-e^{at}\H(t-2)

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expand e^{at}(\H(t-1)-\H(t-2))

expand

e^{a t} (H (t - 1) - H (t - 2))

e^{a t} H (t - 1) - e^{a t} H (t - 2)

Solution steps

e^{a t} (H (t - 1) - H (t - 2))

Apply the distributive law:

a (b - c) = ab - a c

a = e^{a t}, b = H (t - 1), c = H (t - 2)

= e^{a t} H (t - 1) - e^{a t} H (t - 2)

(y e^{x y} + 3 x^{2}) d x + (x e^{x y} - cos (y)) d y

f (x) = (e^{x} + 2 x)^{4}

[(2 x^{2} + x) (5 x)]^{^{'}}

cos (10 5^{\circ})

\frac{n}{( nb - 1 - ib ) i}

What is expand e^{at}(\H(t-1)-\H(t-2)) ?
The solution to expand e^{at}(\H(t-1)-\H(t-2)) is e^{at}\H(t-1)-e^{at}\H(t-2)