The Laplace transform of the function (2t−3)et+23 is equal to: a) e23(2(s−13)2−3s−13) b) e23(2(s−13)2⋅3s−13) c) e23(6(s−13)2) d) (2(s−13)2+23)

Name: The Laplace transform of the function (2t−3)et+23 is equal to: a) e23(2(s−13)2−3s−13) b) e23(2(s−13)2⋅3s−13) c) e23(6(s−13)2) d) (2(s−13)2+23)
Uploaded: 2022-02-23T17:22:57+00:00
Description: The Laplace transform of the function (2t−3)et+23 is equal to: a) e23(2(s−13)2−3s−13) b) e23(2(s−13)2⋅3s−13) c) e23(6(s−13)2) d) (2(s−13)2+23)

Step 1 Given that (2t−3)et+23 To find the Laplace transformation of given function Step 2 Given (2t−3)et+23 =2tet+23−3et+23 Apply Laplace transform on both side L{(2t−3)et+23}=L{2tet+23−3et+23} =L{2tet+23}−L{3et+23} =2L{tet3⋅e23}−L{3et3⋅e23} =2e23L{tet3}−3e23L{et3} =2e231(s−13)2−3e231s−13 [because L{teat}=1(s−a)2L{eat}=1s−a] =e23[2(s−13)2−3s−13] L{(2t−3)et+23}=e23[2(s−13)2−3s−13] Answer A

Solved: "The Laplace transform of the function (2t−3)et+23 is..."

Alyce Wilkinson

Answered question

2021-01-15

The Laplace transform of the function

(2 t - 3) e^{\frac{t + 2}{3}}

is equal to:
a)

e^{\frac{2}{3}} (\frac{2}{{(s - \frac{1}{3})}^{2}} - \frac{3}{s - \frac{1}{3}})

e^{\frac{2}{3}} (\frac{2}{{(s - \frac{1}{3})}^{2}} \cdot \frac{3}{s - \frac{1}{3}})

e^{\frac{2}{3}} (\frac{6}{{(s - \frac{1}{3})}^{2}})

(\frac{2}{{(s - \frac{1}{3})}^{2} + \frac{2}{3}})

Answer & Explanation

hosentak

Skilled2021-01-16Added 100 answers

Step 1
Given that

(2 t - 3) e^{\frac{t + 2}{3}}

To find the Laplace transformation of given function
Step 2
Given

(2 t - 3) e^{\frac{t + 2}{3}}

= 2 t e^{\frac{t + 2}{3}} - 3 e^{\frac{t + 2}{3}}

Apply Laplace transform on both side

L {(2 t - 3) e^{\frac{t + 2}{3}}} = L {2 t e^{\frac{t + 2}{3}} - 3 e^{\frac{t + 2}{3}}}

= L {2 t e^{\frac{t + 2}{3}}} - L {3 e^{\frac{t + 2}{3}}}

= 2 L {t e^{\frac{t}{3}} \cdot e^{\frac{2}{3}}} - L {3 e^{\frac{t}{3}} \cdot e^{\frac{2}{3}}}

= 2 e^{\frac{2}{3}} L {t e^{\frac{t}{3}}} - 3 e^{\frac{2}{3}} L {e^{\frac{t}{3}}}

= 2 e^{\frac{2}{3}} \frac{1}{{(s - \frac{1}{3})}^{2}} - 3 e^{\frac{2}{3}} \frac{1}{s - \frac{1}{3}}

[because L {t e^{a t}} = \frac{1}{{(s - a)}^{2}} L {e^{a t}} = \frac{1}{s - a}]

= e^{\frac{2}{3}} [\frac{2}{{(s - \frac{1}{3})}^{2}} - \frac{3}{s - \frac{1}{3}}]

L {(2 t - 3) e^{\frac{t + 2}{3}}} = e^{\frac{2}{3}} [\frac{2}{{(s - \frac{1}{3})}^{2}} - \frac{3}{s - \frac{1}{3}}]

Answer A

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