What is expand λe^{λt}g(e^t-λe^t) ?

The solution to expand λe^{λt}g(e^t-λe^t) is λe^{t+λt}g-λ^2e^{t+λt}g

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expand λe^{λt}g(e^t-λe^t)

expand

λ e^{λ t} g (e^{t} - λ e^{t})

λ e^{t + λ t} g - λ^{2} e^{t + λ t} g

Solution steps

λ e^{λ t} g (e^{t} - λ e^{t})

Apply the distributive law:

a (b - c) = ab - a c

a = λ e^{λ t} g, b = e^{t}, c = λ e^{t}

= λ e^{λ t} g e^{t} - λ e^{λ t} g λ e^{t}

= λ e^{t} e^{λ t} g - λλ e^{t} e^{λ t} g

Simplify

λ e^{t} e^{λ t} g - λλ e^{t} e^{λ t} g : λ e^{t + λ t} g - λ^{2} e^{t + λ t} g

= λ e^{t + λ t} g - λ^{2} e^{t + λ t} g

\frac{15 a}{s ( 3 s + 8 )}

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

x (1 + x ln (x y - 5))

\frac{5}{6} - 2

\frac{( 1 - v ^{2} ) d v}{v ( 1 + v ^{2} )}

What is expand λe^{λt}g(e^t-λe^t) ?
The solution to expand λe^{λt}g(e^t-λe^t) is λe^{t+λt}g-λ^2e^{t+λt}g