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Chapter 7: Problem 5

Use Laplace transforms to solve the initial value problems $$ x^{\prime \prime}+x=\sin 2 t ; x(0)=0=x^{\prime}(0) $$

Short Answer

Expert verified
The solution is \(x(t) = 0.5 \cos t + 0.25 \sin 2t\).

Step by step solution

01

Apply Laplace Transform

First, apply the Laplace Transform to the differential equation \[x'' + x = \sin(2t)\]To transform the entire equation, use properties of Laplace Transforms: - The Laplace Transform of \(x''\) is \(s^2 X(s) - s x(0) - x'(0)\)- The Laplace Transform of \(x\) is \(X(s)\)- The Laplace Transform of \(\sin(2t)\) is \(\frac{2}{s^2 + 4}\)Given initial conditions are \(x(0) = 0\) and \(x'(0) = 0\). These simplify the transform of the derivative:\[s^2 X(s) + X(s) = \frac{2}{s^2 + 4}\]
02

Solve for Laplace Transform of x(t)

Factor out \(X(s)\) from the left side of the transformed equation:\[X(s)(s^2 + 1) = \frac{2}{s^2 + 4}\]To solve for \(X(s)\), divide both sides by \(s^2 + 1\):\[X(s) = \frac{2}{(s^2 + 1)(s^2 + 4)}\]
03

Partial Fraction Decomposition

Decompose \(\frac{2}{(s^2+1)(s^2+4)}\) into partial fractions:Assume \(\frac{2}{(s^2 + 1)(s^2 + 4)} = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 4}\)Solve for constants A, B, C, D using methods such as equating coefficients or substituting convenient values for \(s\), which leads to:\(A = 0, B = 0.5, C = 0\), and \(D = 0.5\).Now we have:\[X(s) = \frac{0.5}{s^2 + 1} + \frac{0.5}{s^2 + 4}\]
04

Find the Inverse Laplace Transform

Find the inverse transforms:- \(\frac{0.5}{s^2 + 1}\) corresponds to \(0.5 \cos t\)- \(\frac{0.5}{s^2 + 4}\) corresponds to \(0.5\frac{1}{2}\sin(2t) = 0.25\sin(2t)\)The solution \(x(t)\) is:\[x(t) = 0.5\cos t + 0.25\sin 2t\]
05

Verify the Solution

Verify the solution by substituting back into the original differential equation:- Compute \(x'(t) = -0.5\sin t + 0.5\cos(2t)\)- Compute \(x''(t) = -0.5\cos t - \sin(2t)\)Verify that \(x''(t) + x(t) = \sin (2t)\):\(-0.5 \cos t - \sin(2t) + 0.5\cos t + 0.25\sin(2t)\), which simplifies to \(\sin(2t)\), confirming the solution.

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Key Concepts

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

Initial Value Problem
An initial value problem (often abbreviated as IVP) is a type of differential equation along with specified values at a certain point, usually the starting point, which is time zero. For example, in our problem, the differential equation is a second-order type: \[x'' + x = \sin 2t\]. The initial conditions are \(x(0) = 0\) and \(x'(0) = 0\). These conditions mean that at time \(t = 0\), the function \(x(t)\) is zero and its derivative is also zero. This setup allows us to solve the differential equation uniquely by moving forward in time from a known state.
  • An initial value problem gives us a specific solution: we know exactly how the system behaves from an initial point.
  • Using the Laplace transform is a common method to solve these types of problems because it translates differential equations into algebraic equations, which are usually easier to handle.
It is important to apply the initial conditions after the Laplace transformation to simplify the mathematical process.
Second-Order Differential Equations
Second-order differential equations involve second derivatives, like \(x''\) in our instance. They can model a variety of physical systems, such as oscillating springs, circuits, or waves. The general form of a linear second-order differential equation is:\[a x'' + b x' + c x = f(t)\].In our problem, it's \[x'' + x = \sin 2t\], which is a simple form since it lacks a first derivative term \(x'\). The given equation describes a system where the acceleration plus the position equals a sinusoidal forcing function, \(\sin 2t\).
  • The goal is to find a function \(x(t)\) that satisfies this equation for all time \(t\).
  • When initial values are provided, it helps us find the constants involved, giving a complete solution.
Effectively solving these equations often involves transforming them to a simpler form, which is why techniques like the Laplace Transform come into play.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to simplify the process of finding inverse Laplace transforms. In our problem, we reach a point where we need to perform decomposition:\[X(s) = \frac{2}{(s^2 + 1)(s^2 + 4)}\].Partial fractions break down this expression into simpler fractions, making it easier to find the inverse transforms:\[X(s) = \frac{0.5}{s^2 + 1} + \frac{0.5}{s^2 + 4}\].
  • This process involves assuming a form \(\frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 4}\) and solving for the constants \(A, B, C, \) and \(D\).
  • It essentially transforms a complex expression into a sum of simpler terms, each representing a known form for inverse Laplace Transforms.
By using partial fractions, calculating the inverse transform becomes straightforward, as each term corresponds to an easily recognizable inverse transformation.
Inverse Laplace Transform
The inverse Laplace transform is used to convert a function in the Laplace domain back to the time domain. After performing partial fraction decomposition, the task now is to apply the inverse transform:
  • For \(\frac{0.5}{s^2 + 1}\), the inverse Laplace transform is \(0.5\cos t\).
  • For \(\frac{0.5}{s^2 + 4}\), the inverse transform is \(0.25\sin(2t)\).
This process allows us to rediscover \(x(t)\) given as \[x(t) = 0.5\cos t + 0.25\sin(2t)\].Applying the inverse Laplace transform is critical because it gives the exact solution to our original differential equation. This step translates algebraic expressions back into a function of time, providing a complete understanding of the behavior modeled by the differential equation.

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Most popular questions from this chapter

In Problems 31 through 35, the values of mass \(m\), spring constant \(k\), dashpot resistance \(c\), and force \(f(t)\) are given for \(a\) mass-spring-dashpot system with external forcing function. Solve the initial value problem $$ m x^{\prime \prime}+c x^{\prime}+k x=f(t), \quad x(0)=x^{\prime}(0)=0 $$ and construct the graph of the position function \(x(t) .\) \(m=1, k=9, c=0 ; f(t)=\sin t\) if \(0 \leqq t \leqq 2 \pi, f(t)=0\) if \(t>2 \pi\)

Find the Laplace transforms of the functions given in Problems 11 through \(22 .\) $$ f(t)=2 \text { if } 0 \leqq t<3 ; f(t)=0 \text { if } t \geqq 3 $$

Use Laplace transforms to solve the initial value problems. \(x^{\prime \prime}-6 x^{\prime}+8 x=2 ; x(0)=x^{\prime}(0)=0\)

Find the convolution \(f(t) * g(t)\) in Problems. $$ f(t)=g(t)=\sin t $$

Find the inverse Laplace transform \(f(t)\) of each function given in Problems 1 through \(10 .\) Then sketch the graph of \(f\). $$ F(s)=\frac{2 s\left(e^{-\pi s}-e^{-2 \pi s}\right)}{s^{2}+4} $$
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