laplacetransform e^{2(t-3)}(3t^4+2t^3+t)

解答

拉普拉斯变换

e^{2 (t - 3)} (3 t^{4} + 2 t^{3} + t)

\frac{s ^{3} - 6 s ^{2} + 24 s + 40}{e ^{6} ( s - 2 ) ^{5}}

求解步骤

L {e^{2 (t - 3)} (3 t^{4} + 2 t^{3} + t)}

展开

e^{2 (t - 3)} (3 t^{4} + 2 t^{3} + t) : 3 e^{2 t - 6} t^{4} + 2 e^{2 t - 6} t^{3} + e^{2 t - 6} t

= L {3 e^{2 t - 6} t^{4} + 2 e^{2 t - 6} t^{3} + e^{2 t - 6} 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)}

= 3 L {e^{2 t - 6} t^{4}} + 2 L {e^{2 t - 6} t^{3}} + L {e^{2 t - 6} t}

L {e^{2 t - 6} t^{4}} : \frac{24}{e ^{6} ( s - 2 ) ^{5}}

L {e^{2 t - 6} t^{3}} : \frac{6}{e ^{6} ( s - 2 ) ^{4}}

L {e^{2 t - 6} t} : \frac{1}{e ^{6} ( s - 2 ) ^{2}}

= 3 \cdot \frac{24}{e ^{6} ( s - 2 ) ^{5}} + 2 \cdot \frac{6}{e ^{6} ( s - 2 ) ^{4}} + \frac{1}{e ^{6} ( s - 2 ) ^{2}}

化简

3 \frac{24}{e ^{6} ( s - 2 ) ^{5}} + 2 \frac{6}{e ^{6} ( s - 2 ) ^{4}} + \frac{1}{e ^{6} ( s - 2 ) ^{2}} : \frac{s ^{3} - 6 s ^{2} + 24 s + 40}{e ^{6} ( s - 2 ) ^{5}}

= \frac{s ^{3} - 6 s ^{2} + 24 s + 40}{e ^{6} ( s - 2 ) ^{5}}

\frac{\partial}{\partial y} (cos (x - y) - x e^{y})

t an g e n t x^{2} + x y + 2 y^{2} = 16, (- 2, - 2)

\frac{\partial}{\partial x} ((5 x + 8 y)^{11})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{5 + x})

\int_{3}^{6} (2 x - 10) d x