What is laplacetransform ((2te^{-3t}))/5+3t*sin(t) ?

The answer to laplacetransform ((2te^{-3t}))/5+3t*sin(t) is 2/(5(s+3)^2)+(6s)/((s^2+1)^2)

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laplacetransform ((2te^{-3t}))/5+3t*sin(t)

laplace transform

\frac{( 2 t e ^{- 3 t} )}{5} + 3 t \cdot sin (t)

\frac{2}{5 ( s + 3 ) ^{2}} + \frac{6 s}{( s ^{2} + 1 ) ^{2}}

Solution steps

L {\frac{( 2 e ^{- 3 t} t )}{5} + 3 t sin (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 {\frac{2 t e ^{- 3 t}}{5}} + 3 L {t sin (t)}

L {\frac{2 t e ^{- 3 t}}{5}} : \frac{2}{5 ( s + 3 ) ^{2}}

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

= \frac{2}{5 ( s + 3 ) ^{2}} + 3 \cdot \frac{2 s}{( s ^{2} + 1 ) ^{2}}

Refine

\frac{2}{5 ( s + 3 ) ^{2}} + 3 \frac{2 s}{( s ^{2} + 1 ) ^{2}} : \frac{2}{5 ( s + 3 ) ^{2}} + \frac{6 s}{( s ^{2} + 1 ) ^{2}}

= \frac{2}{5 ( s + 3 ) ^{2}} + \frac{6 s}{( s ^{2} + 1 ) ^{2}}

(13 + x^{2}) y^{^{'}} - 2 x y = 0

\frac{\partial}{\partial x} (y x + 2 x + g (y))

y^{^{'}} = - x y

f (x) = x^{3} - 2 x + 2, at x = 3

\frac{\partial}{\partial x} (ln (a x + b y) + c^{x + y})

What is laplacetransform ((2te^{-3t}))/5+3t*sin(t) ?
The answer to laplacetransform ((2te^{-3t}))/5+3t*sin(t) is 2/(5(s+3)^2)+(6s)/((s^2+1)^2)