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Question: What is the Laplace transform of \[t\cos at + \sin at\]?...

What is the Laplace transform of tcos⁡at+sin⁡att\cos at + \sin at?

Explanation

Solution

To find the Laplace transform of tcos⁡at+sin⁡att\cos at + \sin at, we will use the one of the properties of Laplace transformation i.e., . Using this we will find the Laplace transformation of sin⁡at\sin at and then using this we will find the Laplace transform of tcos⁡att\cos at and then we will add these two results to find the Laplace transformation of tcos⁡at+sin⁡att\cos at + \sin at.

Complete step by step answer:
We have to find the Laplace transform of tcos⁡at+sin⁡att\cos at + \sin at. For this we will use the rule for the Laplace transform of the derivative.
Here, L[f(t)]=F(s)L\left[ {f(t)} \right] = F(s)
Let f(t)=sin⁡at−−−(1)f(t) = \sin at - - - (1)
Putting t=0t = 0, we get
⇒f(0)=sin⁡(0)\Rightarrow f(0) = \sin \left( 0 \right)
On simplification we get,
⇒f(0)=0\Rightarrow f(0) = 0
On differentiating f(t)=sin⁡atf(t) = \sin at, we get
⇒f1(t)=acos⁡at\Rightarrow {f^1}(t) = a\cos at
Putting t=0t = 0, we get
⇒f1(0)=acos⁡(0)\Rightarrow {f^1}(0) = a\cos \left( 0 \right)
On simplification we get,
⇒f1(0)=a\Rightarrow {f^1}(0) = a
Now, on differentiating f1(t)=acos⁡at{f^1}(t) = a\cos at, we get
⇒f′′(t)=−a2sin⁡at−−−(2)\Rightarrow {f^{''}}(t) = - {a^2}\sin at - - - (2)
Using (1)(1) in (2)(2), we get
⇒f′′(t)=−a2f(t)\Rightarrow {f^{''}}(t) = - {a^2}f(t)
Taking the Laplace transformation of both the sides we get
⇒L[f′′(t)]=L[−a2f(t)]−−−(3)\Rightarrow L\left[ {{f^{''}}(t)} \right] = L\left[ { - {a^2}f(t)} \right] - - - (3)
We know L[f′(t)]=s2F(s)−sf(0)−f′(0)L\left[ {f'\left( t \right)} \right] = {s^2}F\left( s \right) - sf\left( 0 \right) - f'\left( 0 \right) . Using this we can write (3)(3) as
⇒s2F(s)−sf(0)−f1(0)=L[−a2f(t)]−−−(4)\Rightarrow {s^2}F(s) - sf(0) - {f^1}(0) = L\left[ { - {a^2}f(t)} \right] - - - (4)
As L[k×f(t)]=kL[f(t)]L\left[ {k \times f(t)} \right] = kL\left[ {f(t)} \right], where kk is a constant.
Therefore, we can write (4)(4) as
⇒s2F(s)−sf(0)−f1(0)=−a2L[f(t)]\Rightarrow {s^2}F(s) - sf(0) - {f^1}(0) = - {a^2}L\left[ {f(t)} \right]
Putting the value of f(0)f(0), f1(0){f^1}(0) and L[f(t)]=F(s)L\left[ {f(t)} \right] = F(s), we get
⇒s2F(s)−0−a=−a2F(s)\Rightarrow {s^2}F(s) - 0 - a = - {a^2}F(s)
On rearranging we get
⇒s2F(s)+a2F(s)=a\Rightarrow {s^2}F(s) + {a^2}F(s) = a
Taking F(s)F(s) common from the left hand side of the above equation, we get
⇒(s2+a2)F(s)=a\Rightarrow \left( {{s^2} + {a^2}} \right)F(s) = a
Dividing both the sides by (s2+a2)\left( {{s^2} + {a^2}} \right), we get
⇒F(s)=a(s2+a2)\Rightarrow F(s) = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}}
∴L[sin⁡at]=a(s2+a2)−−−(5)\therefore L\left[ {\sin at} \right] = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}} - - - (5)
Similarly, now suppose that we let f(t)=tcos⁡at−−−(6)f(t) = t\cos at - - - (6)
Putting t=0t = 0, we get
f(0)=(0)×cos⁡(0)f(0) = \left( 0 \right) \times \cos \left( 0 \right)
On simplification we get
⇒f(0)=0\Rightarrow f(0) = 0
On differentiating f(t)=tcos⁡atf(t) = t\cos at using the product rule of differentiation, we get
⇒f1(t)=(t)(−asin⁡at)+(1)(cos⁡at)\Rightarrow {f^1}(t) = \left( t \right)\left( { - a\sin at} \right) + \left( 1 \right)\left( {\cos at} \right)
On simplification we get
⇒f1(t)=−atsin⁡at+cos⁡at\Rightarrow {f^1}(t) = - at\sin at + \cos at
Again, on differentiating we get
⇒f′′(t)=(−at)(acos⁡at)+(−a)(sin⁡at)−asin⁡at\Rightarrow {f^{''}}(t) = \left( { - at} \right)\left( {a\cos at} \right) + \left( { - a} \right)\left( {\sin at} \right) - a\sin at
On simplification we get
⇒f′′(t)=−a2tcos⁡at−2asin⁡at\Rightarrow {f^{''}}(t) = - {a^2}t\cos at - 2a\sin at
Using (6)(6), we can write
⇒f′′(t)=−a2f(t)−2asin⁡at−−−(7)\Rightarrow {f^{''}}(t) = - {a^2}f(t) - 2a\sin at - - - (7)
Putting t=0t = 0, we get
⇒f′′(0)=−a2f(0)−2asin⁡(0)\Rightarrow {f^{''}}(0) = - {a^2}f(0) - 2a\sin \left( 0 \right)
On simplification we get
⇒f′′(0)=0\Rightarrow {f^{''}}(0) = 0
On taking Laplace transformation of (7)(7), we get
⇒L[f′′(t)]=L[−a2f(t)−2asin⁡at]\Rightarrow L\left[ {{f^{''}}(t)} \right] = L\left[ { - {a^2}f(t) - 2a\sin at} \right]
As Laplace transform satisfy the linear property, so we can write the above equation as
⇒L[f′′(t)]=−a2L[f(t)]−2aL[sin⁡at]\Rightarrow L\left[ {{f^{''}}(t)} \right] = - {a^2}L\left[ {f(t)} \right] - 2aL\left[ {\sin at} \right]
Above we have proved that L[sin⁡at]=a(s2+a2)L\left[ {\sin at} \right] = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}}. Using this and putting the value of L[f′′(t)]L\left[ {{f^{''}}(t)} \right] and L[f(t)]L\left[ {f(t)} \right], we get
⇒s2F(s)−sf(0)−f1(0)=−a2F(s)−2a×as2+a2\Rightarrow {s^2}F(s) - sf(0) - {f^1}(0) = - {a^2}F(s) - 2a \times \dfrac{a}{{{s^2} + {a^2}}}
Putting the value of f(0)f(0) and f1(0){f^1}(0), we get
⇒s2F(s)−0−1=−a2F(s)−2a2s2+a2\Rightarrow {s^2}F(s) - 0 - 1 = - {a^2}F(s) - \dfrac{{2{a^2}}}{{{s^2} + {a^2}}}
On simplification, we get
⇒s2F(s)−1=−a2F(s)−2a2s2+a2\Rightarrow {s^2}F(s) - 1 = - {a^2}F(s) - \dfrac{{2{a^2}}}{{{s^2} + {a^2}}}
On rearranging we get
⇒s2F(s)+a2F(s)=1−2a2s2+a2\Rightarrow {s^2}F(s) + {a^2}F(s) = 1 - \dfrac{{2{a^2}}}{{{s^2} + {a^2}}}
On simplification and taking common, we get
⇒F(s)(s2+a2)=s2+a2−2a2s2+a2\Rightarrow F(s)\left( {{s^2} + {a^2}} \right) = \dfrac{{{s^2} + {a^2} - 2{a^2}}}{{{s^2} + {a^2}}}
On further simplification we get
⇒F(s)(s2+a2)=s2−a2s2+a2\Rightarrow F(s)\left( {{s^2} + {a^2}} \right) = \dfrac{{{s^2} - {a^2}}}{{{s^2} + {a^2}}}
Dividing both the sides by (s2+a2)\left( {{s^2} + {a^2}} \right) we get
⇒F(s)=s2−a2(s2+a2)2\Rightarrow F(s) = \dfrac{{{s^2} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}
∴L[tcos⁡at]=s2−a2(s2+a2)2−−−(8)\therefore L\left[ {t\cos at} \right] = \dfrac{{{s^2} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}} - - - (8)
Adding (5)(5) and (8)(8), we get
⇒L[sin⁡at]+L[tcos⁡at]=a(s2+a2)+s2−a2(s2+a2)2\Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}} + \dfrac{{{s^2} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}
On simplifying we get
⇒L[sin⁡at]+L[tcos⁡at]=a(s2+a2)+(s2−a2)(s2+a2)2\Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{{a\left( {{s^2} + {a^2}} \right) + \left( {{s^2} - {a^2}} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}
On further simplification, we get
⇒L[sin⁡at]+L[tcos⁡at]=as2+s2+a3−a2(s2+a2)2\Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{{a{s^2} + {s^2} + {a^3} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}
On taking common,
⇒L[sin⁡at]+L[tcos⁡at]=s2(a+1)+a2(a−1)(s2+a2)2\Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{{{s^2}\left( {a + 1} \right) + {a^2}\left( {a - 1} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}
On rewriting we get
⇒L[tcos⁡at+sin⁡at]=s2(a+1)+a2(a−1)(s2+a2)2\Rightarrow L\left[ {t\cos at + \sin at} \right] = \dfrac{{{s^2}\left( {a + 1} \right) + {a^2}\left( {a - 1} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}.

Therefore, the Laplace transform of tcos⁡at+sin⁡att\cos at + \sin at is s2(a+1)+a2(a−1)(s2+a2)2\dfrac{{{s^2}\left( {a + 1} \right) + {a^2}\left( {a - 1} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}.

Note: Laplace transform is an integral transform that converts a function of a real variable to a function of a complex variable. Laplace transform satisfy the linear property i.e., L[f(t)+g(t)]=L[f(t)]+L[g(t)]L\left[ {f\left( t \right) + g\left( t \right)} \right] = L\left[ {f\left( t \right)} \right] + L\left[ {g\left( t \right)} \right].