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Chapter 8: Problem 43

Find the inverse Laplace transform of: $$\frac{p^{2}}{\left(p^{2}+a^{2}\right)^{2}}$$

Short Answer

Expert verified
The inverse Laplace transform of \[ \frac{p^{2}}{(p^{2}+a^{2})^{2}} \] is \[ t \sin(at) \].

Step by step solution

01

Identify the Function

Consider the given function for which the inverse Laplace transform needs to be found: \[\frac{p^{2}}{(p^{2}+a^{2})^{2}}\]
02

Use Partial Fraction Decomposition

Express the function in a more suitable form if possible. However, in this case, it is already in a form that directly allows us to use known inverse Laplace transforms.
03

Recognize the Known Inverse Transform

Recognize that the given function corresponds to a known inverse Laplace transform. The standard form is \[\mathcal{L}^{-1}\left(\frac{p^{2}}{(p^{2}+a^{2})^{2}}\right) = t \sin(at)\]
04

Apply the Inverse Transform

Based on our recognition of the standard form:\[\mathcal{L}^{-1}\left(\frac{p^{2}}{(p^{2}+a^{2})^{2}}\right) = t \sin(at)\]

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Key Concepts

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

partial fraction decomposition
Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions. This technique simplifies the process of finding inverse Laplace transforms by making the expressions easier to handle. However, in the given exercise, the function $$\frac{p^{2}}{(p^{2}+a^{2})^{2}}$$ is already in a suitable form. When faced with functions that aren't directly recognizable, partial fraction decomposition can be a helpful strategy to transform them into simpler terms.
  • Step-by-step approach:
  • Write the complex function as a sum of simpler fractions.
  • Simplify each fraction separately.
Though we did not need this procedure for the current problem, understanding it is crucial for handling more complex functions.
Laplace transform
The Laplace transform is a powerful tool in mathematics and engineering, widely used to solve differential equations and transform complex functions into simpler algebraic forms. The forward Laplace transform of a function \( f(t) \), denoted by \( \mathcal{L}\{ f(t) \} \), converts it into a function \( F(p) \) of a complex variable \( p \). This process simplifies the operations of differentiation and integration, turning them into algebraic manipulations.
  • Definition: \[ \mathcal{L}\{ f(t) \} = \int_{0}^{\infty} e^{-pt} f(t) \, dt \]
  • Commonly used in solving linear time-invariant systems.
  • Simplifies the process of analysis and design of systems.
In the given exercise, the function we dealt with was already in the Laplace domain.
inverse Laplace transform properties
Understanding the properties of the inverse Laplace transform is crucial for converting functions back from the Laplace domain to the time domain. Recognizing patterns and standard forms can greatly speed up the process. In our exercise, the function $$\frac{p^{2}}{(p^{2}+a^{2})^{2}}$$ matches a known inverse Laplace transform
  • Property: \( \mathcal{L}^{-1}\left( \frac{p^{2}}{(p^{2}+a^{2})^{2}} \right) = t \sin(at) \)
By identifying this pattern, we can easily apply the inverse transform and find the solution. Knowing properties and common transforms can enhance one's ability to solve complex problems efficiently.

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Most popular questions from this chapter

If an incompressible fluid flows in a corner bounded by walls meeting at the origin at an angle of \(60^{\circ},\) the streamlines of the flow satisfy the equation \(2 x y d x+\left(x^{2}-y^{2}\right) d y\) \(=0 .\) Find the streamlines.

For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves. \(y^{\prime}-x y=x, \quad y=1\) when \(x=0\)

Consider the differential equation \((D-a)(D-b) y=P_{n}(x),\) where \(P_{n}(x)\) is a polynomial of degree \(n\). Show that a particular solution of this equation is given by (6.24) with \(c=0 ;\) that is, \(y_{p}\) is \(\left\\{\begin{array}{l}\text { a polynomial } Q_{n}(x) \text { of degree } n \text { if } a \text { and } b \text { are both different from zero; } \\ x Q_{n}(x) \text { if } a \neq 0, \text { but } b=0 \\ x^{2} Q_{n}(x) \quad \text { if } a=b=0\end{array}\right.\) Hint: To show that \(Q_{n}(x)=\sum a_{n} x^{n}\) is a solution of the differential equation for a given \(P_{n}=\sum b_{n} x^{n},\) you have only to show that the coefficients \(a_{n}\) can be found so that \((D-a)(D-b) Q_{n}(x) \equiv P_{n}(x) .\) Equate coefficients of \(x^{n}, x^{n-1}, \cdots,\) to see that this is always possible if \(a \neq b\). For \(b=0,\) the differential equation becomes \((D-a) D y=P_{n} ;\) what is \(D y\) if \(y=x Q_{n} ?\) Similarly, consider \(D^{2} y\) if \(y=x^{2} Q_{n}\).

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x(\ln y) y^{\prime}-y \ln x=0$$

(a) Show that $$\begin{aligned} D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\ D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y \end{aligned}$$ and so on; that is, for any positive integral \(n\) $$D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y$$ Thus show that if \(L(D)\) is any polynomial in the operator \(D,\) then $$L(D)\left(e^{a x} y\right)=e^{a x} L(D+a) y$$ This is called the exponential shift. (b) Use (a) to show that $$\begin{aligned} (D-1)^{3}\left(e^{x} y\right) &=e^{x} D^{3} y \\ \left(D^{2}+D-6\right)\left(e^{-3 x} y\right) &=e^{-3 x}\left(D^{2}-5 D\right) y \end{aligned}$$ (c) Replace \(D\) by \(D-a,\) to obtain $$e^{a x} P(D) y=P(D-a) e^{a x} y$$ This is called the inverse exponential shift. (d) Using (c), we can change a differential equation whose right-hand side is an exponential times a polynomial, to one whose right-hand side is just a polynomial. For example, consider \(\left(D^{2}-D-6\right) y=10 x e^{3 x} ;\) multiplying both sides by \(e^{-3 x}\) and using \((\mathrm{c}),\) we get $$\begin{aligned} e^{-3 x}\left(D^{2}-D-6\right) y &=\left[(D+3)^{2}-(D+3)-6\right] y e^{-3 x} \\\ &=\left(D^{2}+5 D\right) y e^{-3 x}=10 x \end{aligned}$$ Show that a solution of \(\left(D^{2}+5 D\right) u=10 x\) is \(u=x^{2}-\frac{2}{5} x ;\) then \(y e^{-3 x}=x^{2}-\frac{2}{5} x\) or \(y=e^{3 x}\left(x^{2}-\frac{2}{5} x\right) .\) Use this method to solve Problems 23 to 26
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