What is laplacetransform e^{2t}sin(j^2t) ?

The answer to laplacetransform e^{2t}sin(j^2t) is (j^2)/((s-2)^2+j^4)

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laplacetransform e^{2t}sin(j^2t)

laplace transform

e^{2 t} sin (j^{2} t)

\frac{j ^{2}}{( s - 2 ) ^{2} + j ^{4}}

Solution steps

L {e^{2 t} sin (j^{2} t)}

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

e^{2 t} sin (j^{2} t) : f (t) = sin (j^{2} t), a = 2

= L {sin (j^{2} t)} (s - 2)

L {sin (j^{2} t)} : \frac{j ^{2}}{s ^{2} + j ^{4}}

= \frac{j ^{2}}{( s - 2 ) ^{2} + j ^{4}}

\frac{\partial}{\partial x} ((9 + 20 x) y^{t a n (x)})

f (x) = - x^{2}, (- 3, - 9)

\int \frac{2}{( x - 2 ) ( x ^{4} + 4 )} d x

x \to - \infty lim (- 3 - x)

\int_{0}^{3} x^{2} + 16 d x

What is laplacetransform e^{2t}sin(j^2t) ?
The answer to laplacetransform e^{2t}sin(j^2t) is (j^2)/((s-2)^2+j^4)