What is e^{(-at)}sin(bt)(t^2+5) ?

The solution to e^{(-at)}sin(bt)(t^2+5) is (sin(tb)(t^2+5))/(e^{at)}

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simplify e^{(-at)}sin(bt)(t^2+5)

simplify

e^{(- a t)} sin (b t) (t^{2} + 5)

\frac{sin ( t b ) ( t ^{2} + 5 )}{e ^{a t}}

Solution steps

e^{(- a t)} sin (b t) (t^{2} + 5)

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{a t}} sin (t b) (t^{2} + 5)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sin ( b t ) ( t ^{2} + 5 )}{e ^{a t}}

Multiply:

1 \cdot sin (b t) = sin (b t)

= \frac{sin ( t b ) ( t ^{2} + 5 )}{e ^{a t}}

\frac{5 ^{8}}{7 ^{8}}

\frac{1}{x ^{2} ( 1 - e ^{- 2 x} )} (x^{3} e^{- x})

e^{x} \cdot e^{4 x}

What is e^{(-at)}sin(bt)(t^2+5) ?
The solution to e^{(-at)}sin(bt)(t^2+5) is (sin(tb)(t^2+5))/(e^{at)}