Show that L{sin⁡kt}=ks2+k2 L{cos⁡kt}=ss2+k2 Use L{sin⁡kt}=∫0∞e−stsin⁡ktdt 1)L{1}= 2)L{tn}= 3)L{ekt}= 4)L{sin⁡kt}= 5)L{cos⁡kt}= 6)L{sin⁡hkt}= 7)L{cos⁡hkt}=

Name: Show that L{sin⁡kt}=ks2+k2 L{cos⁡kt}=ss2+k2 Use L{sin⁡kt}=∫0∞e−stsin⁡ktdt 1)L{1}= 2)L{tn}= 3)L{ekt}= 4)L{sin⁡kt}= 5)L{cos⁡kt}= 6)L{sin⁡hkt}= 7)L{cos⁡hkt}=
Uploaded: 2022-02-23T17:22:57+00:00
Description: Show that L{sin⁡kt}=ks2+k2 L{cos⁡kt}=ss2+k2 Use L{sin⁡kt}=∫0∞e−stsin⁡ktdt 1)L{1}= 2)L{tn}= 3)L{ekt}= 4)L{sin⁡kt}= 5)L{cos⁡kt}= 6)L{sin⁡hkt}= 7)L{cos⁡hkt}=

Step 1 We have to lapace transform of sinkt and coskt Step 2 Show that L{sin⁡kt}=ks2+k2 L{cos⁡kt}=ss2+k2 Solution: We know that L{ f(t)}=∫0∞e−stf(t)dt Now, L{sin⁡kt}=∫0∞e−stsin⁡ktdt =L{sin⁡kt}=∫0∞e−stsin⁡ktdt =[e−st((−s)2+k2)(−ssin⁡kt−kcos⁡kt)]0∞ =0−e0(s2+k2)(−0−k)=ks2+k2 Hence L{sin⁡kt}=ks2+k2 Again L{cos⁡kt}=∫0∞e−stcos⁡ktdt =[e−st((−s)2+k2)(−scos⁡kt+ksin⁡kt)]0∞ =0−(e0s2+k2(−s+0))=ss2+k2 Hence , L{cos⁡kt}=s(s2+k2) Note: (1)∫eaxsin⁡bxdx=eaxa2+b2(asin⁡bx−bcos⁡bx) (2)∫eaxcos⁡bxdx=eaxa2+b2(acos⁡bx+bsin⁡bx) (3)L{1}=13 (4)L{tn}=n!sn+1 (5)

Expert answers to "Show that L{sin⁡kt}=ks2+k2 L{cos⁡kt}=ss2+k2 Use L{sin⁡kt}=∫0∞e−stsin⁡ktdt 1)L{1}= 2)L{tn}=..."

alesterp

Answered question

2021-09-13

Show that $L {\sin k t} = \frac{k}{s^{2} + k^{2}}$
$L {\cos k t} = \frac{s}{s^{2} + k^{2}}$
Use $L {\sin k t} = \int_{0}^{\infty} e^{- s t} \sin k t d t$
1) $L {1} =$
2) $L {t^{n}} =$
3) $L {e^{k t}} =$
4) $L {\sin k t} =$
5) $L {\cos k t} =$
6) $L {\sin h k t} =$
7) $L {\cos h k t} =$

Answer & Explanation

Aamina Herring

Skilled2021-09-14Added 85 answers

Step 1
We have to lapace transform of sinkt and coskt
Step 2
Show that $L {\sin k t} = \frac{k}{s^{2} + k^{2}}$
$L {\cos k t} = \frac{s}{s^{2} + k^{2}}$
Solution: We know that
$L {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t$
Now, $L {\sin k t} = \int_{0}^{\infty} e^{- s t} \sin k t d t$
$= L {\sin k t} = \int_{0}^{\infty} e^{- s t} \sin k t d t$
$= {[\frac{e^{- s t}}{({(- s)}^{2} + k^{2})} (- s \sin k t - k \cos k t)]}_{0}^{\infty}$
$= 0 - \frac{e^{0}}{(s^{2} + k^{2})} (- 0 - k) = \frac{k}{s^{2} + k^{2}}$
Hence $L {\sin k t} = \frac{k}{s^{2} + k^{2}}$
Again $L {\cos k t} = \int_{0}^{\infty} e^{- s t} \cos k t d t$
$= {[\frac{e^{- s t}}{({(- s)}^{2} + k^{2})} (- s \cos k t + k \sin k t)]}_{0}^{\infty}$
$= 0 - (\frac{e^{0}}{s^{2} + k^{2}} (- s + 0)) = \frac{s}{s^{2} + k^{2}}$
Hence , $L {\cos k t} = \frac{s}{(s^{2} + k^{2})}$
Note:
(1) $\int e^{a x} \sin b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \sin b x - b \cos b x)$
(2) $\int e^{a x} \cos b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \cos b x + b \sin b x)$
(3) $L {1} = \frac{1}{3}$
(4) $L {t^{n}} = \frac{n!}{s^{n + 1}}$
(5)

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