What is laplacetransform (1-3sin(4t))^2 ?

The answer to laplacetransform (1-3sin(4t))^2 is 1/s-(24)/(s^2+16)+9/(2s)-(9s)/(2(s^2+64))

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laplacetransform (1-3sin(4t))^2

laplace transform

(1 - 3 sin (4 t))^{2}

\frac{1}{s} - \frac{24}{s ^{2} + 16} + \frac{9}{2 s} - \frac{9 s}{2 ( s ^{2} + 64 )}

Solution steps

L {(1 - 3 sin (4 t))^{2}}

Expand

(1 - 3 sin (4 t))^{2} : 1 - 6 sin (4 t) + 9 sin^{2} (4 t)

= L {1 - 6 sin (4 t) + 9 sin^{2} (4 t)}

Rewrite using trig identities

L {1 - 6 sin (4 t) + \frac{9}{2} - \frac{9}{2} cos (8 t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {1} - 6 L {sin (4 t)} + L {\frac{9}{2}} - \frac{9}{2} L {cos (8 t)}

L {1} : \frac{1}{s}

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

L {\frac{9}{2}} : \frac{9}{2 s}

L {cos (8 t)} : \frac{s}{s ^{2} + 64}

= \frac{1}{s} - 6 \cdot \frac{4}{s ^{2} + 16} + \frac{9}{2 s} - \frac{9}{2} \cdot \frac{s}{s ^{2} + 64}

Refine

\frac{1}{s} - 6 \frac{4}{s ^{2} + 16} + \frac{9}{2 s} - \frac{9}{2} \frac{s}{s ^{2} + 64} : \frac{1}{s} - \frac{24}{s ^{2} + 16} + \frac{9}{2 s} - \frac{9 s}{2 ( s ^{2} + 64 )}

= \frac{1}{s} - \frac{24}{s ^{2} + 16} + \frac{9}{2 s} - \frac{9 s}{2 ( s ^{2} + 64 )}

3 x - 1

\frac{\partial}{\partial x} (ln (16 x^{2} - 13 y^{2}))

f (y) = 2 x^{2} + 3 x y + 4 y^{2} + 5 x + 6 y - 7

(x^{2} + y^{2} - 3) d x = (y + x y) d y, y (0) = 1

\frac{d y}{d x} = 1 + e^{y - x + 5}

What is laplacetransform (1-3sin(4t))^2 ?
The answer to laplacetransform (1-3sin(4t))^2 is 1/s-(24)/(s^2+16)+9/(2s)-(9s)/(2(s^2+64))