L−1(2+2e−2s−4e−3ss(1−e−3s)) solve

asked by guest
on Mar 28, 2025 at 8:46 am



You asked:

Evaluate the expression: L−1(2+2e−2s−4e−3s⋅s(1−e−3s))L - 1 \left(2 + 2 e - 2 s - 4 e - 3 s \cdot s \left(1 - e - 3 s\right)\right)

MathBot Answer:

Evaluated



L−1(2+2e−2s−4e−3s⋅s(1−e−3s))=L+3s2(−3s−e+1)+2s−2+2e\displaystyle L - 1 \left(2 + 2 e - 2 s - 4 e - 3 s \cdot s \left(1 - e - 3 s\right)\right) = L + 3 s^{2} \left(- 3 s - e + 1\right) + 2 s - 2 + 2 e

Expanded

L−1(2+2e−2s−4e−3s⋅s(1−e−3s))=L−9s3−3s2e+3s2+2s−2+2eL - 1 \left(2 + 2 e - 2 s - 4 e - 3 s \cdot s \left(1 - e - 3 s\right)\right) = L - 9 s^{3} - 3 s^{2} e + 3 s^{2} + 2 s - 2 + 2 e

Factored

L−1(2+2e−2s−4e−3s⋅s(1−e−3s))=L−9s3−3s2e+3s2+2s−2+2eL - 1 \left(2 + 2 e - 2 s - 4 e - 3 s \cdot s \left(1 - e - 3 s\right)\right) = L - 9 s^{3} - 3 s^{2} e + 3 s^{2} + 2 s - 2 + 2 e

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

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