laplacetransform tsin^2(t)

解答

拉普拉斯变换

t sin^{2} (t)

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

求解步骤

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

使用三角恒等式改写

L {\frac{1}{2} t - \frac{1}{2} t cos (2 t)}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= \frac{1}{2} L {t} - \frac{1}{2} L {t cos (2 t)}

L {t} : \frac{1}{s ^{2}}

L {t cos (2 t)} : - \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}

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

整理

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

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

\int sec^{2} (θ) tan^{4} (θ) d θ

2 y^{^{'}} = y^{2}, y (0) = 7

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 2 x}{x + 1})

\int sin^{2} (3 t) cos^{2} (3 t) d t

x \to 1 lim (x - 1 + ln (x))