What is expand (1+e^{2y})(e^y-(e^{2y})/2) ?

The solution to expand (1+e^{2y})(e^y-(e^{2y})/2) is e^y-(e^{2y})/2-(e^{4y})/2+e^{3y}

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expand (1+e^{2y})(e^y-(e^{2y})/2)

expand

(1 + e^{2 y}) (e^{y} - \frac{e ^{2 y}}{2})

e^{y} - \frac{e ^{2 y}}{2} - \frac{e ^{4 y}}{2} + e^{3 y}

Solution steps

(1 + e^{2 y}) (e^{y} - \frac{e ^{2 y}}{2})

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = e^{2 y}, c = e^{y}, d = - \frac{e ^{2 y}}{2}

= 1 \cdot e^{y} + 1 \cdot (- \frac{e ^{2 y}}{2}) + e^{2 y} e^{y} + e^{2 y} (- \frac{e ^{2 y}}{2})

Apply minus-plus rules

+ (- a) = - a

= 1 \cdot e^{y} - 1 \cdot \frac{e ^{2 y}}{2} + e^{y} e^{2 y} - e^{2 y} \frac{e ^{2 y}}{2}

Simplify

1 \cdot e^{y} - 1 \cdot \frac{e ^{2 y}}{2} + e^{y} e^{2 y} - e^{2 y} \frac{e ^{2 y}}{2} : e^{y} - \frac{e ^{2 y}}{2} - \frac{e ^{4 y}}{2} + e^{3 y}

= e^{y} - \frac{e ^{2 y}}{2} - \frac{e ^{4 y}}{2} + e^{3 y}

8 \cdot \frac{d}{d t} ((- 9 t - 6) e^{- 3 t} + (4 t + 12) e^{- 5 t})

\frac{5 ( x - 1 )}{6}

\frac{d}{d x} ((z) (z^{3} - 3 x^{2} y + 6 x y z))

(M - N)^{2} \cdot e^{- 2 k x} + 4 MN \cdot e^{- k x}

\frac{( u - 3 )}{u \cdot ( u + 1 )}

What is expand (1+e^{2y})(e^y-(e^{2y})/2) ?
The solution to expand (1+e^{2y})(e^y-(e^{2y})/2) is e^y-(e^{2y})/2-(e^{4y})/2+e^{3y}