What is laplacetransform xe^{3x+4}-x^2 ?

The answer to laplacetransform xe^{3x+4}-x^2 is (e^4)/((s-3)^2)-2/(s^3)

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laplacetransform xe^{3x+4}-x^2

laplace transform

x e^{3 x + 4} - x^{2}

\frac{e ^{4}}{( s - 3 ) ^{2}} - \frac{2}{s ^{3}}

Solution steps

L {e^{3 x + 4} x - x^{2}}

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 {x e^{3 x + 4}} - L {x^{2}}

L {x e^{3 x + 4}} : \frac{e ^{4}}{( s - 3 ) ^{2}}

L {x^{2}} : \frac{2}{s ^{3}}

= \frac{e ^{4}}{( s - 3 ) ^{2}} - \frac{2}{s ^{3}}

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

y^{^{''}} + 12 y^{^{'}} + 100 y = 0

\int tan^{3} (2 t) sec^{4} (2 t) d t

\int \frac{4}{x ^{2} - 8 x + 12} d x

y^{^{''}} + y^{^{'}} - 6 y = 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 0

What is laplacetransform xe^{3x+4}-x^2 ?
The answer to laplacetransform xe^{3x+4}-x^2 is (e^4)/((s-3)^2)-2/(s^3)