What is expand e^{t-pi}sin(2t)u(t-pi) ?

The solution to expand e^{t-pi}sin(2t)u(t-pi) is e^{t-pi}tusin(2t)-pie^{t-pi}usin(2t)

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expand e^{t-pi}sin(2t)u(t-pi)

expand

e^{t - π} sin (2 t) u (t - π)

e^{t - π} t u sin (2 t) - π e^{t - π} u sin (2 t)

Solution steps

e^{t - π} sin (2 t) u (t - π)

= e^{t - π} u sin (2 t) (t - π)

Apply the distributive law:

a (b - c) = ab - a c

a = e^{t - π} sin (2 t) u, b = t, c = π

= e^{t - π} sin (2 t) u t - e^{t - π} sin (2 t) u π

= e^{t - π} t u sin (2 t) - π e^{t - π} u sin (2 t)

2 (e^{2} - e)

- 2 (5 x + 3 y = 21)

ln (q) (- \frac{1}{a} - 1) - ln (- a)

\frac{20 ( 41 - 2 x )}{3}

(x - 3) (x + 1) \circ (6 x - 2)

What is expand e^{t-pi}sin(2t)u(t-pi) ?
The solution to expand e^{t-pi}sin(2t)u(t-pi) is e^{t-pi}tusin(2t)-pie^{t-pi}usin(2t)