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

The solution to expand e^{(t-a)^2}u(t-a) is e^{t^2-2at+a^2}tu-ae^{t^2-2at+a^2}u

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expand e^{(t-a)^2}u(t-a)

expand

e^{(t - a)^{2}} u (t - a)

e^{t^{2} - 2 a t + a^{2}} t u - a e^{t^{2} - 2 a t + a^{2}} u

Solution steps

e^{(t - a)^{2}} u (t - a)

Apply the distributive law:

a (b - c) = ab - a c

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

= e^{(t - a)^{2}} u t - e^{(t - a)^{2}} u a

= e^{(t - a)^{2}} t u - a e^{(t - a)^{2}} u

Simplify

e^{(t - a)^{2}} t u - a e^{(t - a)^{2}} u : e^{t^{2} - 2 a t + a^{2}} t u - a e^{t^{2} - 2 a t + a^{2}} u

= e^{t^{2} - 2 a t + a^{2}} t u - a e^{t^{2} - 2 a t + a^{2}} u

\frac{π ( 2 ln ( 2 ) - 1 )}{4}

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\frac{6 x ^{3} + 18 x ^{2} + 11 x + 2}{x ( 6 x ^{2} + 5 x + 1 )}

- \frac{2}{4} (1 - s) (1 - t) + (- \frac{1}{4}) (1 + s) (1 - t)

What is expand e^{(t-a)^2}u(t-a) ?
The solution to expand e^{(t-a)^2}u(t-a) is e^{t^2-2at+a^2}tu-ae^{t^2-2at+a^2}u