The Shifting Theorem
The Shifting Theorem :
If L{f(t)}= f̅(p), then
L{eᵃᵗf(t)} = f̅(p-a) , p>a
Proof :
By definition
∞
f̅(p) = ∫ e⁻ᵖᵗ f(t) dt
0
∞
Therefore, f̅(p-a) = ∫ e^-(p-a)t f(t) dt
0
∞
= ∫ e⁻ᵖᵗ[eᵃᵗ f(t)] dt
0
= L{eᵃᵗ f(t)}.
Corollary :
L{e⁻ᵃᵗ f(t)} = f̅(p+a); (p>-a).
This follows immediately from the above theorem by writing -a for a.
Example _1:
Find the transform of eᵃᵗ tⁿ .
Solution :
Since L{tⁿ} = n!/pⁿ⁺¹
by using the above theorem , we find
L{eᵃᵗ tⁿ} = n!/(p-a)ⁿ⁺¹
which is the required solution .
by using the above theorem , we find
L{eᵃᵗ tⁿ} = n!/(p-a)ⁿ⁺¹
which is the required solution .
Example_2 :
Find the Laplace Transformation of
f(t) = . e³ᵗ
Solution :
To find L{t^7/2 . e³ᵗ} first we have to find out the value of L{t^7/2}
Now , As we know that
L{tⁿ} = n!/pⁿ⁺¹
So L{t^7/2} = (7/2)!/p^[(7/2)+1]
⇒L{t^7/2} = (7/2)Γ(7/2) / p^(9/2)
= (7/2)Γ(5/2 + 1) / p^(9/2)
= (7/2)(5/2)Γ(5/2) / p^(9/2)
[∵Γ(n+1) = nΓ(n)]
= (7/2)(5/2)Γ(3/2 + 1) / p^(9/2)
= (7/2)(5/2)(3/2)Γ(3/2) / p^(9/2)
= (7/2)(5/2)(3/2)Γ(1/2 + 1)/ p^(9/2)
= (7/2)(5/2)(3/2)(1/2)Γ(1/2) / p^(9/2)
= (7/2)(5/2)(3/2)(1/2)√(π ) / p^(9/2)
[∵Γ(1/2) = √(π )]
= (7/2)Γ(5/2 + 1) / p^(9/2)
= (7/2)(5/2)Γ(5/2) / p^(9/2)
[∵Γ(n+1) = nΓ(n)]
= (7/2)(5/2)Γ(3/2 + 1) / p^(9/2)
= (7/2)(5/2)(3/2)Γ(3/2) / p^(9/2)
= (7/2)(5/2)(3/2)Γ(1/2 + 1)/ p^(9/2)
= (7/2)(5/2)(3/2)(1/2)Γ(1/2) / p^(9/2)
= (7/2)(5/2)(3/2)(1/2)√(π ) / p^(9/2)
[∵Γ(1/2) = √(π )]
⇒L{t^7/2} = 105√(π )/16 p^(9/2)= F(p)
∴ L{t^7/2 . e³ᵗ} = 105√(π )/16(p-a)^(9/2)
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