laplacetransform e^{-t}(t^2+4t+5)

解答

拉普拉斯变换

e^{- t} (t^{2} + 4 t + 5)

\frac{5 s ^{2} + 14 s + 11}{( s + 1 ) ^{3}}

求解步骤

L {e^{- t} (t^{2} + 4 t + 5)}

应用变换法则: if

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

则

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

对于

e^{- t} (t^{2} + 4 t + 5) : f (t) = (t^{2} + 4 t + 5), a = - 1

= L {(t^{2} + 4 t + 5)} (s + 1)

L {(t^{2} + 4 t + 5)} : \frac{2}{s ^{3}} + \frac{4}{s ^{2}} + \frac{5}{s}

= \frac{2}{( s + 1 ) ^{3}} + \frac{4}{( s + 1 ) ^{2}} + \frac{5}{( s + 1 )}

= \frac{2}{( s + 1 ) ^{3}} + \frac{4}{( s + 1 ) ^{2}} + \frac{5}{s + 1}

化简

\frac{2}{( s + 1 ) ^{3}} + \frac{4}{( s + 1 ) ^{2}} + \frac{5}{s + 1} : \frac{5 s ^{2} + 14 s + 11}{( s + 1 ) ^{3}}

= \frac{5 s ^{2} + 14 s + 11}{( s + 1 ) ^{3}}

y^{^{'}} + 4 e^{x + y} = 0

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