What is laplacetransform 4te^{3t}sin(2t) ?

The answer to laplacetransform 4te^{3t}sin(2t) is (8(2s-6))/((s^2-6s+13)^2)

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laplacetransform 4te^{3t}sin(2t)

laplace transform

4 t e^{3 t} sin (2 t)

\frac{8 ( 2 s - 6 )}{( s ^{2} - 6 s + 13 ) ^{2}}

Solution steps

L {4 e^{3 t} t sin (2 t)}

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)}

= 4 L {t e^{3 t} sin (2 t)}

L {t e^{3 t} sin (2 t)} : \frac{2 ( 2 s - 6 )}{( s ^{2} - 6 s + 13 ) ^{2}}

= 4 \cdot \frac{2 ( 2 s - 6 )}{( s ^{2} - 6 s + 13 ) ^{2}}

Refine

4 \frac{2 ( 2 s - 6 )}{( s ^{2} - 6 s + 13 ) ^{2}} : \frac{8 ( 2 s - 6 )}{( s ^{2} - 6 s + 13 ) ^{2}}

= \frac{8 ( 2 s - 6 )}{( s ^{2} - 6 s + 13 ) ^{2}}

\int 18 - 18 sin (x) d x

x \to \infty lim (\frac{1}{x !})

- \frac{1}{x ^{\frac{2}{3}}}

\frac{d}{d x} (sinh^{2} (3 x))

y^{^{''}} + 4 y = 0, y (0) = 4, y^{^{'}} (0) = 1

What is laplacetransform 4te^{3t}sin(2t) ?
The answer to laplacetransform 4te^{3t}sin(2t) is (8(2s-6))/((s^2-6s+13)^2)