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Chapter 22: Problem 1

Find the Laplace transform of each of the following expressions: (a) \(t-3\) (b) \(2 t^{3}+5 t\) (c) \(7-3 t^{4}\) (d) \(\sin 2 t+2 \sin t\) (e) \(\cos t+t\)

Short Answer

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Question: Find the Laplace transform of the following functions: (a) \(t-3\), (b) \(2t^3 + 5t\), (c) \(7 - 3t^4\), (d) \(\sin 2t + 2\sin t\), and (e) \(\cos t + t\). Solution: (a) The Laplace transform of \(t-3\) is \(\frac{1}{s^2}+\frac{3}{s}\) (b) The Laplace transform of \(2t^3 + 5t\) is \(\frac{12}{s^4}+\frac{5}{s^2}\) (c) The Laplace transform of \(7-3t^4\) is \(\frac{7}{s}-\frac{72}{s^5}\) (d) The Laplace transform of \(\sin 2t + 2\sin t\) is \(\frac{2}{s^2+4}+\frac{2}{s^2+1}\) (e) The Laplace transform of \(\cos t + t\) is \(\frac{s}{s^2+1}+\frac{1}{s^2}\)

Step by step solution

01

Identify the Laplace transform formula for individual functions

In this case, we have a linear combination of two functions \(t\) and \(-3\). The Laplace transform is a linear operator, which means that \(\mathcal{L}\{t-3\} = \mathcal{L}\{t\} - \mathcal{L}\{-3\}\). To find the Laplace transform of \(t-3\), we will need to compute the Laplace transform of each function individually. The formula for the Laplace transform of polynomials is given by: $$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$ For the constant function, the Laplace transform is given by: $$\mathcal{L}\{c\}=\frac{c}{s}$$ Where c is a constant.
02

Compute the Laplace transform of each function

Applying the formulas above, we can compute the Laplace transform for t and -3: $$\mathcal{L}\{t\}=\frac{1!}{s^{1+1}}=\frac{1}{s^2}$$ $$\mathcal{L}\{-3\} = \frac{-3}{s}$$
03

Combine the Laplace transforms

Now, we can combine the results to find the Laplace transform of \(t-3\): $$\mathcal{L}\{t-3\} = \mathcal{L}\{t\} - \mathcal{L}\{-3\} = \frac{1}{s^2} - \frac{-3}{s} = \frac{1}{s^2} + \frac{3}{s}$$ (b) Laplace transform of \(2t^3 + 5t\):
04

Identify the Laplace transform formula for individual functions

In this case, we have the functions \(2t^3\) and \(5t\). Since the Laplace transform is a linear operator, we can compute the Laplace transform of each individual function and then combine them: $$\mathcal{L}\{2t^3 + 5t\} = 2\mathcal{L}\{t^3\} + 5\mathcal{L}\{t\}$$
05

Compute the Laplace transform of each function

Using the Laplace transform formula for polynomials, we find the Laplace transforms for \(t^3\) and \(t\): $$\mathcal{L}\{t^3\}=\frac{3!}{s^{3+1}}=\frac{6}{s^4}$$ $$\mathcal{L}\{t\}=\frac{1!}{s^{1+1}}=\frac{1}{s^2}$$
06

Combine the Laplace transforms

Now, we can combine the results with the constants to find the Laplace transform of \(2t^3 + 5t\): $$\mathcal{L}\{2t^3 + 5t\} = 2\cdot\frac{6}{s^4} + 5\cdot\frac{1}{s^2} = \frac{12}{s^4} + \frac{5}{s^2}$$ (c) Laplace transform of \(7 - 3t^4\): This problem is similar to parts (a) and (b), so we can directly write the Laplace transform using the polynomial formula: $$\mathcal{L}\{7 -3t^4\} = 7\mathcal{L}\{1\} - 3\mathcal{L}\{t^4\} = 7\cdot\frac{1}{s} - 3\cdot\frac{4!}{s^5} = \frac{7}{s} - \frac{72}{s^5}$$ (d) Laplace transform of \(\sin 2t + 2\sin t\):
07

Identify the Laplace transform formula for individual functions

The Laplace transform for sinusoidal functions is given by: $$\mathcal{L}\{\sin{at}\}=\frac{a}{s^2+a^2}$$ Using linearity, we can compute the Laplace transform of the given expression as: $$\mathcal{L}\{\sin 2t + 2\sin t\} = \mathcal{L}\{\sin 2t\} + 2\mathcal{L}\{\sin t\}$$
08

Compute the Laplace transform of each function

Substituting the appropriate values, we have: $$\mathcal{L}\{\sin 2t\} = \frac{2}{s^2+2^2} = \frac{2}{s^2+4}$$ $$\mathcal{L}\{\sin t\} = \frac{1}{s^2+1^2} = \frac{1}{s^2+1}$$
09

Combine the Laplace transforms

Now, combine the results: $$\mathcal{L}\{\sin 2t + 2\sin t\} = \frac{2}{s^2+4} + 2\cdot\frac{1}{s^2+1} = \frac{2}{s^2+4} + \frac{2}{s^2+1}$$ (e) Laplace transform of \(\cos t + t\):
10

Identify the Laplace transform formula for individual functions

The Laplace transform for cosine functions is given by: $$\mathcal{L}\{\cos{at}\}=\frac{s}{s^2+a^2}$$ Using linearity, we can compute the Laplace transform of the given expression as: $$\mathcal{L}\{\cos t + t\} = \mathcal{L}\{\cos t\} + \mathcal{L}\{t\}$$
11

Compute the Laplace transform of each function

Substituting the appropriate values, we have: $$\mathcal{L}\{\cos t\} = \frac{s}{s^2+1^2} = \frac{s}{s^2+1}$$ $$\mathcal{L}\{t\} = \frac{1!}{s^{1+1}} = \frac{1}{s^2}$$
12

Combine the Laplace transforms

Now, combine the results: $$\mathcal{L}\{\cos t + t\} = \frac{s}{s^2+1} + \frac{1}{s^2}$$

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Most popular questions from this chapter

From the definition of the Laplace transform, and using integration by parts, show that, $$ \mathcal{L}\left\\{f^{\prime \prime}(t)\right\\}=s^{2} F(s)-s f(0)-f^{\prime}(0) $$

From the definition of the Laplace transform, and using integration by parts, show that $$ \mathcal{L}\left\\{f^{\prime}(t)\right\\}=s F(s)-f(0) $$

The first shift theorem states that if \(\mathcal{L}(f(t)\\}=F(s)\), then $$ \mathcal{L}\left\\{\mathrm{e}^{-a t} f(t)\right\\}=F(s+a) $$ where \(a\) is a constant. (a) From the definition of the Laplace transform show that $$ \mathcal{L}\left\\{\mathrm{e}^{-a t} f(t)\right\\}=\int_{0}^{\infty} \mathrm{e}^{-(s+a) t} f(t) \mathrm{d} t $$ and hence prove the first shift theorem. (b) Use Table \(1.1\) in Block 1 and the first shift theorem to find \(\mathcal{L}\left\\{u(t-3) \mathrm{e}^{-7 t}\right\\}\) where \(u(t)\) is the unit step function.

Use Table \(1.1\) to find the Laplace transform of the following: (a) \(t^{5}\) (c) e \(^{9 t}\) (d) \(\mathrm{e}^{-6 t}\) (e) \(\sin 5 t\)

Find the inverse Laplace transform of each of the following expressions: (a) \(\frac{4}{s^{2}+4 s+5}\) (b) \(\frac{s+5}{s^{2}+10 s+29}\) (c) \(\frac{2 s-3}{s^{2}+6 s+10}\) (d) \(\frac{s^{2}+6 s-4}{\left(s^{2}+4\right)^{2}}\) (e) \(\frac{1}{4 s^{2}+4 s+1}\)
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