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EXERCISE 3. Use Partial Fractions and the table to find the inverse Laplace transform of (i) D(s) = 9 + 3s / ((s - 1)(s + 1)(s + 2)), (ii) G(s) = 1 / (s^4 - 16), (iii) F(s) = s / ((s - 1)(s^2 - 2s + 2)). EXERCISE 4. Find the Laplace transform of ?_0^t(u^3 + sin(2u))du without computing the integral. EXERCISE 5. Use the formula L(tf(t)) = -F'(s) to find the Laplace transform of g(t) = t cos(?t). Deduce the next to last row of the left column of the table. EXERCISE 6. Use the s-shift theorem to find the Laplace transform of the functions (i) g(t) = (t^3 - 3t + 2)e^-2t, (ii) d(t) = e^4t(t - cos t). Conceptual understanding EXERCISE 7. Solve the initial value problem y'' - 4y = 16sin(2t), y(0) = 0, y'(0) = 0. Hint: After applying the Laplace transform, recall Exercise 3(ii) .

          EXERCISE 3. Use Partial Fractions and the table to find the inverse Laplace transform of
(i) D(s) = 9 + 3s / ((s - 1)(s + 1)(s + 2)), (ii) G(s) = 1 / (s^4 - 16), (iii) F(s) = s / ((s - 1)(s^2 - 2s + 2)).
EXERCISE 4. Find the Laplace transform of ?_0^t(u^3 + sin(2u))du without computing the integral.
EXERCISE 5. Use the formula L(tf(t)) = -F'(s) to find the Laplace transform of g(t) =
t cos(?t). Deduce the next to last row of the left column of the table.
EXERCISE 6. Use the s-shift theorem to find the Laplace transform of the functions
(i) g(t) = (t^3 - 3t + 2)e^-2t, (ii) d(t) = e^4t(t - cos t).
Conceptual understanding
EXERCISE 7. Solve the initial value problem
y'' - 4y = 16sin(2t), y(0) = 0, y'(0) = 0.
Hint: After applying the Laplace transform, recall Exercise 3(ii) .

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EXERCISE 3. Use Partial Fractions and the table to find the inverse Laplace transform of
(i) D(s) = 9 + 3s / ((s - 1)(s + 1)(s + 2)), (ii) G(s) = 1 / (s^4 - 16), (iii) F(s) = s / ((s - 1)(s^2 - 2s + 2)).
EXERCISE 4. Find the Laplace transform of ?0^t(u^3 + sin(2u))du without computing the integral.
EXERCISE 5. Use the formula L(tf(t)) = -F'(s) to find the Laplace transform of g(t) =
t cos(?t). Deduce the next to last row of the left column of the table.
EXERCISE 6. Use the s-shift theorem to find the Laplace transform of the functions
(i) g(t) = (t^3 - 3t + 2)e^-2t, (ii) d(t) = e^4t(t - cos t).
Conceptual understanding
EXERCISE 7. Solve the initial value problem
y” - 4y = 16sin(2t), y(0) = 0, y'(0) = 0.
Hint: After applying the Laplace transform, recall Exercise 3(ii) .

Added by Jessica D.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Adi S

06/13/2022

Step 1

Step 1: Given \( G(s) = \frac{1}{(s^2 - 4)(s^2 + 4)} \), we can rewrite it as \( G(s) = \frac{1}{(s^2 - 2^2)(s^2 + 2^2)} \). Show more…

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EXERCISE 3. Use Partial Fractions and the table to find the inverse Laplace transform of (i) D(s) = (9 + 3s) / ((s - 1)(s + 1)(s + 2)), (ii) G(s) = 1 / (s^4 - 16), (iii) F(s) = s / ((s - 1)(s^2 - 2s + 2)). EXERCISE 4. Find the Laplace transform of ∫_0^t (u^3 + sin(2u)) du without computing the integral. EXERCISE 5. Use the formula L(tf(t)) = -F'(s) to find the Laplace transform of g(t) = t cos(ωt). Deduce the next to last row of the left column of the table. EXERCISE 6. Use the s-shift theorem to find the Laplace transform of the functions (i) g(t) = (t^3 - 3t + 2)e^-2t, (ii) d(t) = e^4t (t - cos t). Conceptual understanding EXERCISE 7. Solve the initial value problem y'' - 4y = 16 sin(2t), y(0) = 0, y'(0) = 0. Hint: After applying the Laplace transform, recall Exercise 3(ii).

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00:01 Here we have been given the function g of f is equal to 1 divided by s rest of 4 minus 16.

00:12 Now we can simplify and write this as 1 divided by s square against square minus the square of 4, which will be equal to clearly 1 divided by s square plus 4 as 1 1 bracket and s square minus 4 as second bracket which is equal to 1 divided by 8 into bracket 1 divided by s squared minus 4 minus 1 divided by s 2 plus 4.

00:44 Here we have simply converted the product into a minus side.

00:50 Now we have g of s is equal to 1 divided by 8 multiplied by 1 divided by s 2 minus 4 minus 1 divided by 8 into 1 divided by s square plus 4.

01:07 Now we have small g of t is equal to 1 divided by 8 inverse la plus function of 1 divided by s 2 minus 1 divided by 4 minus 1 divided by 8 la plus inverse of 1 divided by s square plus 4.

01:28 Now we have g of t which is the inverse lap loss transformation as 1 divided by 8 into hyperbolic sign of 2t divided by 2 minus 1 divided by 8 into sign of 2t divided by 2 which simplifying further can be written as 1 divided by 16 into bracket hyperbore sine of 2d minus sign of 2d.

02:05 Now the required inverse laplace transformation is given by this.

02:12 Now moving further to the next part we have the equation double derivative of y plus 4y is equal to 16 sine 2 t.

02:27 Now name this equation as 1 also we have been given the initial value conditions as y of 0 is equal to 0 and y dash of 0 is equal to 0.

02:37 Now taking laplace transform on both the sides of equation 1 we will have l of y double dash minus 4 laplus of y is equal to 16 laplace transformation of sine of 2 t...

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