laplacetransform e^{5(t-2)}(t^3+2t)^2

解答

拉普拉斯变换

e^{5 (t - 2)} (t^{3} + 2 t)^{2}

\frac{720}{e ^{10} ( s - 5 ) ^{7}} + \frac{96}{e ^{10} ( s - 5 ) ^{5}} + \frac{8}{e ^{10} ( s - 5 ) ^{3}}

求解步骤

L {e^{5 (t - 2)} (t^{3} + 2 t)^{2}}

展开

e^{5 (t - 2)} (t^{3} + 2 t)^{2} : e^{5 t - 10} t^{6} + 4 e^{5 t - 10} t^{4} + 4 e^{5 t - 10} t^{2}

= L {e^{5 t - 10} t^{6} + 4 e^{5 t - 10} t^{4} + 4 e^{5 t - 10} t^{2}}

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)}

= L {e^{5 t - 10} t^{6}} + 4 L {e^{5 t - 10} t^{4}} + 4 L {e^{5 t - 10} t^{2}}

L {e^{5 t - 10} t^{6}} : \frac{720}{e ^{10} ( s - 5 ) ^{7}}

L {e^{5 t - 10} t^{4}} : \frac{24}{e ^{10} ( s - 5 ) ^{5}}

L {e^{5 t - 10} t^{2}} : \frac{2}{e ^{10} ( s - 5 ) ^{3}}

= \frac{720}{e ^{10} ( s - 5 ) ^{7}} + 4 \cdot \frac{24}{e ^{10} ( s - 5 ) ^{5}} + 4 \cdot \frac{2}{e ^{10} ( s - 5 ) ^{3}}

整理

\frac{720}{e ^{10} ( s - 5 ) ^{7}} + 4 \frac{24}{e ^{10} ( s - 5 ) ^{5}} + 4 \frac{2}{e ^{10} ( s - 5 ) ^{3}} : \frac{720}{e ^{10} ( s - 5 ) ^{7}} + \frac{96}{e ^{10} ( s - 5 ) ^{5}} + \frac{8}{e ^{10} ( s - 5 ) ^{3}}

= \frac{720}{e ^{10} ( s - 5 ) ^{7}} + \frac{96}{e ^{10} ( s - 5 ) ^{5}} + \frac{8}{e ^{10} ( s - 5 ) ^{3}}

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