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Chapter 7: Problem 27

\(z^{\prime \prime}+3 z^{\prime}+2 z=e^{-3 t} u(t-2)\) \(z(0)=2, \quad z^{\prime}(0)=-3\)

Short Answer

Expert verified
The solution to the given non-homogeneous differential equation with the given initial conditions will require solving the associated homogeneous equation, applying the Laplace transform to the non-homogeneous equation to find a particular solution, and lastly applying the initial conditions to find the specific solution. The exact solution would involve the constants \(C_1\) and \(C_2\) obtained from the initial conditions

Step by step solution

01

Solve the Homogeneous Equation

The homogeneous equation associated with the given differential equation is \(\(z'' + 3z' + 2z = 0\)\). The characteristic equation of this homogeneous equation is \(m^2 + 3m + 2 = 0\), which can be factored to \((m+1)(m+2)=0\), yielding roots \(m= -1, -2\). Therefore, the general solution of the homogeneous equation is \(z_h = C_1 e^{-t} + C_2 e^{-2t}\)
02

Apply Laplace Transform to the Non-Homogeneous Equation

Now, to find a particular solution for the non-homogeneous equation, apply the Laplace transform to both sides of the original equation, including the forcing function. This gives: \(\mathcal{L}\{z''\}+3\mathcal{L}\{z'\}+2\mathcal{L}\{z\}=\mathcal{L}\{e^{-3 t} u(t-2)\}\). Taking the Laplace transforms and using the properties of Laplace transforms, this initial value problem can then be converted into an algebraic equation in terms of \\(s\\), \(F(s)\), and the initial conditions.
03

Find the Particular Solution

Solve the algebraic equation from Step 2 for \(F(s)\). Then, use the inverse Laplace transform to find \(f(t)\) (the particular solution). After finding both the homogeneous and particular solutions, one could then write down the general solution to the original non-homogeneous differential equation.
04

Apply Initial Conditions

The last step is to apply the initial conditions given in the problem, \(z(0)=2, z'(0)=-3\), to the general solution. This is done by setting \(z(t) = 2\) and \(z'(t) = -3\) in the general solution, and solving simultaneously for the constants \(C_1\) and \(C_2\) in the general solution. This yields the specific solution to the initial value problem.

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Key Concepts

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

Homogeneous Differential Equation
A homogeneous differential equation has a specific form where all the terms involve the dependent variable and its derivatives. This means that there is no independent forcing term added to the equation. For the given exercise, the homogeneous differential equation can be written as \(z'' + 3z' + 2z = 0\). It is essential because it describes the system's natural behavior without external input.
  • To solve a homogeneous differential equation, we look for solutions of the form \(z = e^{mt}\), which leverage the properties of exponents to transform the problem into an algebraic equation.
  • The characteristic equation is derived by replacing the derivatives with terms involving \(m\), leading to \(m^2 + 3m + 2 = 0\).
  • Solving the characteristic equation provides the roots, here \(m = -1\) and \(m = -2\), that form the essential part of the general solution.
The general solution for the homogeneous equation, then, becomes \(z_h = C_1 e^{-t} + C_2 e^{-2t}\), where \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions.
Non-Homogeneous Differential Equation
Non-homogeneous differential equations are similar to their homogeneous counterparts but include an additional forcing function. This function represents external inputs or disturbances to the system. In our problem, the non-homogeneous equation is \(z'' + 3z' + 2z = e^{-3t} u(t-2)\). Here, \(e^{-3t} u(t-2)\) is the forcing function, which involves the unit step function \(u(t)\), accounting for a shift in time.
  • This equation requires a particular solution that responds to the external input.
  • The Laplace transform is a handy tool to convert the differential equation into an algebraic one by dealing with the forcing function systematically.
The combination of the particular solution from the non-homogeneous part and the general solution from the homogeneous equation will yield the complete solution to the problem.
Initial Value Problem
An initial value problem (IVP) is a differential equation coupled with specific conditions provided at the outset, such as the values of the function and its derivatives at a certain point. In our case, the initial conditions are \(z(0) = 2\) and \(z'(0) = -3\).
  • The initial conditions are vital because they allow the determination of the arbitrary constants in the general solution.
  • We start with the general solution, which includes arbitrary constants, and apply these conditions to solve for \(C_1\) and \(C_2\).
By incorporating these initial conditions into the general solution, we can arrive at a specific solution that satisfies the original problem's conditions.
Inverse Laplace Transform
The inverse Laplace transform is a crucial technique used to revert a function from the Laplace domain back to the time domain. This step is necessary to find the particular solution for the given differential equation by applying it to \(F(s)\), the transformed version of the non-homogeneous term.
  • Once the algebraic equation in Laplace form is solved for \(F(s)\), we apply the inverse Laplace transform to retrieve \(f(t)\).
  • This process often involves breaking down complex fractions, partial fraction decomposition, or using Laplace transform tables for known functions.
Using the inverse Laplace transform allows us to recover the time-dependent behavior of the system, providing a particular solution that accounts for the external input specified in the differential equation.

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Most popular questions from this chapter

\(y^{\prime \prime}+5 y^{\prime}+6 y=t u(t-2)\) \(y(0)=0, \quad y^{\prime}(0)=1\)

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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$$y^{\prime \prime}+3 y^{\prime}+2 y=g(t)$$ $$y(0)=2, \quad y^{\prime}(0)=-1$$ where $$g(t)=\left\\{\begin{array}{ll}{e^{-t},} & {0 \leq t< 3} \\ {1,} & {3< t}\end{array}\right.$$
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