What is expand 2(e^x(Acos(x)+Bsin(x))) ?

The solution to expand 2(e^x(Acos(x)+Bsin(x))) is 2Ae^xcos(x)+2Be^xsin(x)

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expand 2(e^x(Acos(x)+Bsin(x)))

expand

2 (e^{x} (A cos (x) + B sin (x)))

2 A e^{x} cos (x) + 2 B e^{x} sin (x)

Solution steps

2 (e^{x} (A cos (x) + B sin (x)))

Remove parentheses:

(a) = a

= 2 e^{x} (A cos (x) + B sin (x))

Apply the distributive law:

a (b + c) = ab + a c

a = 2 e^{x}, b = A cos (x), c = B sin (x)

= 2 e^{x} A cos (x) + 2 e^{x} B sin (x)

= 2 A e^{x} cos (x) + 2 B e^{x} sin (x)

\frac{( - 6 + 3 y + 4 x ) ( 18 + 3 y - 4 x )}{9}

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

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

j (z - e^{T (j - 1)}) (z - e^{- T (j + 1)})

e^{2 x} (2 + e^{x})

What is expand 2(e^x(Acos(x)+Bsin(x))) ?
The solution to expand 2(e^x(Acos(x)+Bsin(x))) is 2Ae^xcos(x)+2Be^xsin(x)