Laplace transforms from the definition practice problems
The Laplace transform from its definition, as an improper integral. 22 problems with worked solutions; in 22 of them every
equation is proved by a computer algebra system.
13 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = \cos{\left(t \right)} \), for \( \displaystyle s > 0 \).
\[ \frac{\partial}{\partial t} \frac{\left(- s \cos{\left(t \right)} + \sin{\left(t \right)}\right) e^{- s t}}{s^{2} + 1} = e^{- s t} \cos{\left(t \right)} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > 0 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{s}{s^{2} + 1} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = e^{3 t} \), for \( \displaystyle s > 3 \).
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s - 3} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = \cos{\left(3 t \right)} \), for \( \displaystyle s > 0 \).
\[ \frac{\partial}{\partial t} \frac{\left(- s \cos{\left(3 t \right)} + 3 \sin{\left(3 t \right)}\right) e^{- s t}}{s^{2} + 9} = e^{- s t} \cos{\left(3 t \right)} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > 0 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
Reviewed
\[ - \left. \frac{\left(- s \cos{\left(3 t \right)} + 3 \sin{\left(3 t \right)}\right) e^{- s t}}{s^{2} + 9} \right|_{\substack{ t=0 }} = \frac{s}{s^{2} + 9} \]
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{s}{s^{2} + 9} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = \sin{\left(3 t \right)} \), for \( \displaystyle s > 0 \).
\[ \frac{\partial}{\partial t} \frac{\left(- s \sin{\left(3 t \right)} - 3 \cos{\left(3 t \right)}\right) e^{- s t}}{s^{2} + 9} = e^{- s t} \sin{\left(3 t \right)} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > 0 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
Reviewed
\[ - \left. \frac{\left(- s \sin{\left(3 t \right)} - 3 \cos{\left(3 t \right)}\right) e^{- s t}}{s^{2} + 9} \right|_{\substack{ t=0 }} = \frac{3}{s^{2} + 9} \]
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{3}{s^{2} + 9} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = t e^{2 t} \), for \( \displaystyle s > 2 \).
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = e^{- 3 t} \), for \( \displaystyle s > -3 \).
\[ \frac{\partial}{\partial t} \frac{e^{t \left(- s - 3\right)}}{- s - 3} = e^{- 3 t} e^{- s t} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > -3 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
Reviewed
\[ - \left. \frac{e^{t \left(- s - 3\right)}}{- s - 3} \right|_{\substack{ t=0 }} = \frac{1}{s + 3} \]
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s + 3} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = t e^{- t} \), for \( \displaystyle s > -1 \).
\[ \frac{\partial}{\partial t} \frac{\left(t \left(- s - 1\right) - 1\right) e^{t \left(- s - 1\right)}}{\left(- s - 1\right)^{2}} = t e^{- t} e^{- s t} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > -1 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
Reviewed
\[ - \left. \frac{\left(t \left(- s - 1\right) - 1\right) e^{t \left(- s - 1\right)}}{\left(- s - 1\right)^{2}} \right|_{\substack{ t=0 }} = \frac{1}{\left(s + 1\right)^{2}} \]
So the integral is 0 minus its value at t = 0.✓ Proved
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = e^{t} \), for \( \displaystyle s > 1 \).
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s - 1} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = t \), for \( \displaystyle s > 0 \).
\[ \frac{\partial}{\partial t} \frac{\left(- s t - 1\right) e^{- s t}}{s^{2}} = t e^{- s t} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > 0 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
Reviewed
\[ - \left. \frac{\left(- s t - 1\right) e^{- s t}}{s^{2}} \right|_{\substack{ t=0 }} = \frac{1}{s^{2}} \]
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s^{2}} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = e^{- t} \), for \( \displaystyle s > -1 \).
\[ \frac{\partial}{\partial t} \frac{e^{t \left(- s - 1\right)}}{- s - 1} = e^{- t} e^{- s t} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > -1 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
Reviewed
\[ - \left. \frac{e^{t \left(- s - 1\right)}}{- s - 1} \right|_{\substack{ t=0 }} = \frac{1}{s + 1} \]
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s + 1} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = \sin{\left(t \right)} \), for \( \displaystyle s > 0 \).
\[ \frac{\partial}{\partial t} \frac{\left(- s \sin{\left(t \right)} - \cos{\left(t \right)}\right) e^{- s t}}{s^{2} + 1} = e^{- s t} \sin{\left(t \right)} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > 0 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s^{2} + 1} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = t e^{t} \), for \( \displaystyle s > 1 \).
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = t e^{- 3 t} \), for \( \displaystyle s > -3 \).
\[ \frac{\partial}{\partial t} \frac{\left(t \left(- s - 3\right) - 1\right) e^{t \left(- s - 3\right)}}{\left(- s - 3\right)^{2}} = t e^{- 3 t} e^{- s t} \]
An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
For s > -3 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
Reviewed
\[ - \left. \frac{\left(t \left(- s - 3\right) - 1\right) e^{t \left(- s - 3\right)}}{\left(- s - 3\right)^{2}} \right|_{\substack{ t=0 }} = \frac{1}{\left(s + 3\right)^{2}} \]
So the integral is 0 minus its value at t = 0.✓ Proved
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
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