What is laplacetransform tsin(3t)+2tcos(t) ?

The answer to laplacetransform tsin(3t)+2tcos(t) is (6s)/((s^2+9)^2)-(2(-s^2+1))/((s^2+1)^2)

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laplacetransform tsin(3t)+2tcos(t)

laplace transform

t sin (3 t) + 2 t cos (t)

\frac{6 s}{( s ^{2} + 9 ) ^{2}} - \frac{2 ( - s ^{2} + 1 )}{( s ^{2} + 1 ) ^{2}}

Solution steps

L {t sin (3 t) + 2 t cos (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 {t sin (3 t)} + 2 L {t cos (t)}

L {t sin (3 t)} : \frac{6 s}{( s ^{2} + 9 ) ^{2}}

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

= \frac{6 s}{( s ^{2} + 9 ) ^{2}} + 2 (- \frac{- s ^{2} + 1}{( s ^{2} + 1 ) ^{2}})

Refine

\frac{6 s}{( s ^{2} + 9 ) ^{2}} + 2 (- \frac{- s ^{2} + 1}{( s ^{2} + 1 ) ^{2}}) : \frac{6 s}{( s ^{2} + 9 ) ^{2}} - \frac{2 ( - s ^{2} + 1 )}{( s ^{2} + 1 ) ^{2}}

= \frac{6 s}{( s ^{2} + 9 ) ^{2}} - \frac{2 ( - s ^{2} + 1 )}{( s ^{2} + 1 ) ^{2}}

f (x) = \frac{2 x + 1}{x + 2}, at x = 1

\frac{\partial}{\partial x} (- (5 x^{3} + 5 x y^{2}))

\frac{\partial}{\partial h} (x^{2} h)

\frac{1}{x + 2}

\int \frac{x ^{2} - 2}{x ^{2} + x} d x

What is laplacetransform tsin(3t)+2tcos(t) ?
The answer to laplacetransform tsin(3t)+2tcos(t) is (6s)/((s^2+9)^2)-(2(-s^2+1))/((s^2+1)^2)