What is laplacetransform 7sin(4t+8) ?

The answer to laplacetransform 7sin(4t+8) is (7(sin(8)s+4cos(8)))/(s^2+16)

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laplacetransform 7sin(4t+8)

laplace transform

7 sin (4 t + 8)

\frac{7 ( sin ( 8 ) s + 4 cos ( 8 ) )}{s ^{2} + 16}

Solution steps

L {7 sin (4 t + 8)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 7 L {sin (4 t + 8)}

L {sin (4 t + 8)} : \frac{sin ( 8 ) s + 4 cos ( 8 )}{s ^{2} + 16}

= 7 \cdot \frac{sin ( 8 ) s + 4 cos ( 8 )}{s ^{2} + 16}

Refine

7 \frac{sin ( 8 ) s + 4 cos ( 8 )}{s ^{2} + 16} : \frac{7 ( sin ( 8 ) s + 4 cos ( 8 ) )}{s ^{2} + 16}

= \frac{7 ( sin ( 8 ) s + 4 cos ( 8 ) )}{s ^{2} + 16}

\int \frac{1}{81 e ^{- 5 x} + e ^{5 x}} d x

\int_{0}^{5} y + 4 d y

y^{^{'}} - (\frac{7}{8}) y = - \frac{7}{8} e^{- 7 x} y^{9}

2 c

\frac{d y}{d x} = \frac{5 x \cdot sec ( \frac{y}{x} ) + y}{x}

What is laplacetransform 7sin(4t+8) ?
The answer to laplacetransform 7sin(4t+8) is (7(sin(8)s+4cos(8)))/(s^2+16)