What is laplacetransform 6cos(t-2) ?

The answer to laplacetransform 6cos(t-2) is (6(cos(2)s+sin(2)))/(s^2+1)

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laplacetransform 6cos(t-2)

laplace transform

6 cos (t - 2)

\frac{6 ( cos ( 2 ) s + sin ( 2 ) )}{s ^{2} + 1}

Solution steps

L {6 cos (t - 2)}

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)}

= 6 L {cos (t - 2)}

L {cos (t - 2)} : \frac{cos ( 2 ) s + sin ( 2 )}{s ^{2} + 1}

= 6 \cdot \frac{cos ( 2 ) s + sin ( 2 )}{s ^{2} + 1}

Refine

6 \frac{cos ( 2 ) s + sin ( 2 )}{s ^{2} + 1} : \frac{6 ( cos ( 2 ) s + sin ( 2 ) )}{s ^{2} + 1}

= \frac{6 ( cos ( 2 ) s + sin ( 2 ) )}{s ^{2} + 1}

\int sin (5 π x) d x

4 x^{\frac{1}{2}}

9 y^{^{''}} + 42 y^{^{'}} + 49 y = 0

\frac{d}{d x} (a + b x + c x^{2} + d x^{3})

x \to 7 lim (- 11)

What is laplacetransform 6cos(t-2) ?
The answer to laplacetransform 6cos(t-2) is (6(cos(2)s+sin(2)))/(s^2+1)