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Chapter 10: Problem 18

Use the linearity of \(L\) and the formulas derived in this section to determine \(L[f]\). $$f(t)=3 t^{2}-5 \cos 2 t+\sin 3 t$$

Short Answer

Expert verified
The short answer is: \[L[f(t)] = \frac{6}{s^3} - \frac{5s}{s^2+4} + \frac{3}{s^2+9} \]

Step by step solution

01

Find Laplace transform of \(g_1(t) = 3t^2 \)

Using the standard Laplace transform formula: \[L[t^n] = \frac{n!}{s^{n+1}}\] We can find the Laplace transform of the function \(g_1(t) = 3t^2\): \[L[g_1(t)] = L[3t^2] = 3L[t^2] = \frac{3 \cdot 2!}{s^{2+1}} = \frac{6}{s^3}\]
02

Find Laplace transform of \(g_2(t) = -5\cos(2t) \)

Using the standard Laplace transform formula for cosine functions: \[L[\cos(at)] = \frac{s}{s^2+a^2}\] We can find the Laplace transform of the function \(g_2(t) = -5\cos(2t)\): \[L[g_2(t)] = L[-5\cos(2t)] = -5L[\cos(2t)] = -5 \cdot \frac{s}{s^2+(2)^2} = \frac{-5s}{s^2+4}\]
03

Find Laplace transform of \(g_3(t) = \sin(3t) \)

Using the standard Laplace transform formula for sine functions: \[L[\sin(at)] = \frac{a}{s^2+a^2}\] We can find the Laplace transform of the function \(g_3(t) = \sin(3t)\): \[L[g_3(t)] = L[\sin(3t)] = \frac{3}{s^2+(3)^2} = \frac{3}{s^2+9}\]
04

Combine Laplace transforms using linearity property

Now that we have the Laplace transforms for all three functions, let's combine them using the linearity property of the Laplace transform. Since \(f(t) = g_1(t) + g_2(t) + g_3(t)\), we have: \[L[f(t)] = L[g_1(t)] + L[g_2(t)] + L[g_3(t)] = \frac{6}{s^3} + \frac{-5s}{s^2+4} + \frac{3}{s^2+9}\] So, the Laplace transform of the given function is: \[L[f(t)] = \frac{6}{s^3} - \frac{5s}{s^2+4} + \frac{3}{s^2+9}\]

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Most popular questions from this chapter

Determine the Laplace transform of \(f\). $$f(t)=e^{-2 t} \sin (t-\pi / 4)$$.

Solve the given initial-value problem. $$y^{\prime}+2 y=2 u_{1}(t), \quad y(0)=1$$.

Solve the given initial-value problem. $$y^{\prime \prime}-y^{\prime}-2 y=1-3 u_{2}(t), \quad y(0)=1, \quad y^{\prime}(0)=-2$$.

Solve the given Volterra integral equation. $$x(t)=2\left\\{1+\int_{0}^{t} \cos [2(t-\tau)] x(\tau) d \tau\right\\}$$

Use the Laplace transform to solve the given system of differential equations subject to the given initial conditions. $$\begin{array}{c} \frac{d x_{1}}{d t}=x_{1}+2 x_{2}, \quad \frac{d x_{2}}{d t}=2 x_{1}+x_{2} \\ x_{1}(0)=1, \quad \frac{d x_{1}}{d t}(0)=0 \end{array}$$
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