What is the solution to the linear equation? –12 + 3b – 1 = –5 – b
2 answers:
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Answer:
a. L{t} = 1/s² b. L{1} = 1/s
Step-by-step explanation:
Here is the complete question
The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0
Solution
a. L{t}
L{t} = ∫₀⁰⁰
Integrating by parts ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = and v =
and du/dt = dt/dt = 1
So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w
So, ∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + [
]₀⁰⁰
= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]
= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]
= -1/s[(0 - 0] - 1/s²[0 - 1]
= -1/s[(0] - 1/s²[- 1]
= 0 + 1/s²
= 1/s²
L{t} = 1/s²
b. L{1}
L{1} = ∫₀⁰⁰
= []₀⁰⁰
= -1/s[exp(-∞s) - exp(-0s)]
= -1/s[exp(-∞) - exp(-0)]
= -1/s[0 - 1]
= -1/s(-1)
= 1/s
L{1} = 1/s