What is laplacetransform f(t)=4t^2-5sin(3t) ?

The answer to laplacetransform f(t)=4t^2-5sin(3t) is 8/(s^3)-(15)/(s^2+9)

Popular >

laplacetransform f(t)=4t^2-5sin(3t)

laplace transform

f (t) = 4 t^{2} - 5 sin (3 t)

\frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

Solution steps

L {4 t^{2} - 5 sin (3 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}} - 5 L {sin (3 t)}

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

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

= 4 \cdot \frac{2}{s ^{3}} - 5 \cdot \frac{3}{s ^{2} + 9}

Refine

4 \frac{2}{s ^{3}} - 5 \frac{3}{s ^{2} + 9} : \frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

= \frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

x \to - \infty lim (x^{1})

\int_{0}^{π} 11 sin^{2} (2 x) d x

\int (3 x - 4) d x

x^{4} - 13 x^{2} + 36

y = ln (8 x^{2} - 7 x + 9)

What is laplacetransform f(t)=4t^2-5sin(3t) ?
The answer to laplacetransform f(t)=4t^2-5sin(3t) is 8/(s^3)-(15)/(s^2+9)