laplacetransform te^{2t}sin(t)

解答

拉普拉斯变换

t e^{2 t} sin (t)

\frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}}

求解步骤

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

使用拉普拉斯变换表:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

对于

t e^{2 t} sin (t) : f (t) = e^{2 t} sin (t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{2 t} sin (t)})

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

\frac{d}{d s} (\frac{1}{( s - 2 ) ^{2} + 1}) = - \frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}}

= (- 1)^{1} (- \frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}})

= \frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}}

\frac{d y}{d x} + 2 y = e^{- x}

t a y l or X^{3} - X^{2} + 4 X + 5

t an g e n t f (x) = \frac{125}{x ^{2} + 24}, at x = - 1

\int \frac{1}{2 x ^{2} - 4 x + 20} d x

\frac{\partial}{\partial x} (x^{2} e^{y z^{2}})