What is laplacetransform 6e^{5t}cos(2t)-e^{7t} ?

The answer to laplacetransform 6e^{5t}cos(2t)-e^{7t} is (6(s-5))/((s-5)^2+4)-1/(s-7)

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laplacetransform 6e^{5t}cos(2t)-e^{7t}

laplace transform

6 e^{5 t} cos (2 t) - e^{7 t}

\frac{6 ( s - 5 )}{( s - 5 ) ^{2} + 4} - \frac{1}{s - 7}

Solution steps

L {6 e^{5 t} cos (2 t) - e^{7 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)}

= 6 L {e^{5 t} cos (2 t)} - L {e^{7 t}}

L {e^{5 t} cos (2 t)} : \frac{s - 5}{( s - 5 ) ^{2} + 4}

L {e^{7 t}} : \frac{1}{s - 7}

= 6 \cdot \frac{s - 5}{( s - 5 ) ^{2} + 4} - \frac{1}{s - 7}

Refine

6 \frac{s - 5}{( s - 5 ) ^{2} + 4} - \frac{1}{s - 7} : \frac{6 ( s - 5 )}{( s - 5 ) ^{2} + 4} - \frac{1}{s - 7}

= \frac{6 ( s - 5 )}{( s - 5 ) ^{2} + 4} - \frac{1}{s - 7}

t an g e n t f (x) = x^{2} + 3, (- 5, 28)

10 y^{^{''}} - 10 y^{^{'}} + 5 y = 0

\int cos (\frac{nπ x}{4}) d x

\int_{- 3}^{3} ∣ x - 1 ∣ d x

\int (x + 2)^{2} (1 - x)^{5} d x

What is laplacetransform 6e^{5t}cos(2t)-e^{7t} ?
The answer to laplacetransform 6e^{5t}cos(2t)-e^{7t} is (6(s-5))/((s-5)^2+4)-1/(s-7)