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$$\mathscr{L}\left\\{t^{4} e^{5 t}-e^{t} \cos \sqrt{7} t\right\\}$$

Short Answer

Expert verified
The Laplace transform of the function is \(L\{t^{4} e^{5 t}-e^{t} \cos \sqrt{7} t\} = \frac{4!}{(s-5)^5} - \frac{s-1}{(s-1)^2+7}\)

Step by step solution

01

- Compute Laplace Transform for \(t^{4} e^{5 t}\)

In the first term, assume \(f(t) = t^4\) and \(a = 5\). We use \(t^n\) to get the Laplace Transform of \(f(t)\) which is \(\frac{n!}{s^{n+1}}\), where \(n=4\). Thus, \(F(s) = \frac{4!}{s^5}\). Now, apply the formula \(L\{e^{at}f(t)\} = F(s-a)\) we get \(L\{e^{5t} t^4\} = F(s-5) = \frac{4!}{(s-5)^5}\).
02

- Compute Laplace Transform for \(e^{t} \cos \sqrt{7} t\)

In the second term, assume \(f(t) = cos(\sqrt{7}t)\) and \(a = 1\). We know that \(L\{\cos(at)\} = \frac{s}{s^2+a^2}\), so \(F(s) = \frac{s}{s^2+7}\). Now, apply the formula \(L\{e^{at}f(t)\} = F(s-a)\) we get \(L\{e^{t} \cos \sqrt{7} t\} = F(s-1) = \frac{s-1}{(s-1)^2+7}\)
03

- Final Answer

The final answer is the combination of both term's Laplace Transforms. We obtain \(L\{t^{4} e^{5 t}-e^{t} \cos \sqrt{7} t\} = \frac{4!}{(s-5)^5} - \frac{s-1}{(s-1)^2+7}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are a mathematical way to describe rapidly increasing or decreasing behavior. They take the form of \( e^{at} \), where \( t \) represents time and \( a \) determines the rate of growth or decay. If \( a \) is positive, the function grows over time; if negative, it decays.

In the Laplace transform, exponential functions are vital because they can shift the spectrum of the original function \( f(t) \). The transformation \( L\{e^{at}f(t)\} = F(s-a) \) shows how exponentials shift the Laplace domain, adjusting the 's' variable appropriately.

This shifting property makes it easier to manage and transform complex functions, like those involving exponential growth or decay in real-world scenarios, such as populations, radioactive decay, or interest calculations over time.
Trigonometric Functions
Trigonometric functions like sine and cosine are essential in modeling periodic phenomena, such as sound waves, light waves, and pendulum motion. In this context, we encounter \( \cos(\sqrt{7}t) \), where \( \sqrt{7} \) is the angular frequency.

When transformed via Laplace, these functions help in handling oscillatory behaviors. The formula \( L\{\cos(at)\} = \frac{s}{s^2+a^2} \) is particularly useful. It converts the oscillatory function into a rational expression in terms of \( s \), which is easier to solve or manipulate.

Applying this to a different frequency, and with exponential components, allows for analyzing systems with damping (through the exponential part) and predicting their future behavior.
Differential Equations
Differential equations involve functions and their derivatives, describing how a quantity changes over time or space. They are heavily used in engineering, physics, and various scientific fields due to their ability to model dynamic systems such as motion, heat, or even financial markets.

The magic of the Laplace Transform here is its ability to convert these equations into algebraic equations. By transforming the differentiation problem into an "s" domain problem, solving becomes more manageable. Once solved in this new domain, the solution can be transformed back to give explicit answers in terms of time with less hassle. This technique greatly simplifies handling systems involving exponential and oscillatory inputs simultaneously.
Mathematics Education
Understanding the Laplace Transform is a cornerstone of advanced mathematics education, especially for those exploring differential equations and systems analysis. It's a powerful tool that simplifies complex calculus problems into manageable algebraic solutions.

For students, mastering the Laplace Transform means they can tackle real-world problems involving control systems, signal processing, and electrical circuits with confidence. This makes it not only an academic exercise but also a practical skill in modern technology and engineering disciplines.

Educators emphasize conceptual understanding alongside procedural fluency. By illustrating both the theoretical underpinnings and practical applications, students can better grasp how and why the Laplace Transform works, which in turn enriches their problem-solving toolkit.

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Most popular questions from this chapter

The current \(I(t)\) in an \(R L C\) series circuit is governed by the initial value problem $$I^{\prime \prime}(t)+2 I^{\prime}(t)+2 I(t)=g(t)$$ $$I(0)=10, \quad I^{\prime}(0)=0$$ where $$g(t) :=\left\\{\begin{array}{ll}{20,} & {0< t <3 \pi} \\ {0,} & {3 \pi< t <4 \pi} \\ {20,} & {4 \pi< t}\end{array}\right.$$

$$w^{\prime \prime}+w=\delta(t-\pi) ; \quad w(0)=0, \quad w^{\prime}(0)=0$$

Figure 7.29 shows a beam of length 2\(\lambda\) that is imbed- ded in a support on the left side and free on the right. The vertical deflection of the beam a distance \(x\) from the support is denoted by \(y(x) .\) If the beam has a concen- trated load L acting on it in the center of the beam, then the deflection must satisfy the symbolic boundary value problem $$E I y^{(4)}(x)=L \delta(x-\lambda)$$ $$y(0)=y^{\prime}(0)=y^{\prime \prime}(2 \lambda)=y^{\prime \prime \prime}(2 \lambda)=0$$ where \(E,\) the modulus of elasticity, and \(I,\) the moment of inertia, are constants. Find a formula for the dis- placement \(y(x)\) in terms of the constants \(\lambda, L, E,\) and I. [Hint: Let \(y^{\prime \prime}(0)=A\) and \(y^{\prime \prime \prime}(0)=B .\) First solve the fourth-order symbolic initial value problem and then use the conditions \(y^{\prime \prime}(2 \lambda)=y^{\prime \prime \prime}(2 \lambda)=0\) to determine \(A\) and \(B . ]\)

17\. $$y(t)+\int_{0}^{t}(t-v) y(v) d v=1$$

Find an expansion for \(\ln \left[1+\left(1 / s^{2}\right)\right]\) in powers of 1\(/ s .\) Assuming the inverse Laplace transform can be com- puted term by term, show that \(\mathscr{L}^{-1}\left\\{\ln \left(1+\frac{1}{s^{2}}\right)\right\\}(t)=\frac{2}{t}(1-\cos t)\)

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