What is laplacetransform 4t^2cos(2t)+5te^{-7t} ?

The answer to laplacetransform 4t^2cos(2t)+5te^{-7t} is -(8s(-s^2+12))/((s^2+4)^3)+5/((s+7)^2)

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laplacetransform 4t^2cos(2t)+5te^{-7t}

laplace transform

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

- \frac{8 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}} + \frac{5}{( s + 7 ) ^{2}}

Solution steps

L {4 t^{2} cos (2 t) + 5 e^{- 7 t} 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)}

= 4 L {t^{2} cos (2 t)} + 5 L {t e^{- 7 t}}

L {t^{2} cos (2 t)} : - \frac{2 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}

L {t e^{- 7 t}} : \frac{1}{( s + 7 ) ^{2}}

= 4 (- \frac{2 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}) + 5 \cdot \frac{1}{( s + 7 ) ^{2}}

Refine

4 (- \frac{2 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}) + 5 \frac{1}{( s + 7 ) ^{2}} : - \frac{8 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}} + \frac{5}{( s + 7 ) ^{2}}

= - \frac{8 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}} + \frac{5}{( s + 7 ) ^{2}}

\frac{d}{d x} (4 x^{9} + 2 x^{5})

d er i v a t i v e f (x) = \frac{c ^{5} - x ^{5}}{c ^{5} + c ^{5}}

f^{^{'}} (t) - f (t) = 6 t

y^{2} + x^{2} y^{^{'}} = x y y^{^{'}}

a re a y = x, y = (x - 1)^{2}

What is laplacetransform 4t^2cos(2t)+5te^{-7t} ?
The answer to laplacetransform 4t^2cos(2t)+5te^{-7t} is -(8s(-s^2+12))/((s^2+4)^3)+5/((s+7)^2)