Chapter 8: Problem 4
In Exercises \(4-7\), plot the solutions to all of the initial value problems on one figure. Choose a solution interval starting at \(t=0\) that well illustrates any steady-state phenomena that you see. Exercise 1 might be helpful. \(y^{\prime}+4 y=2 \cos t+\sin 4 t\), with \(y(0)=-5,-4, \ldots, 5\).
Short Answer
Expert verified
Plot each solution for \( y(0) = -5, -4, \dots, 5 \) using the general solution with adjustments for initial conditions.
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( y^{\prime} + 4y = 2 \cos t + \sin 4t \). This is a first-order linear non-homogeneous ordinary differential equation.
02
Determine the General Solution
To solve the linear differential equation, find the complementary solution (solve the homogeneous equation) and the particular solution (for the non-homogeneous part). The homogeneous equation is \( y^{\prime} + 4y = 0 \), which has the solution \( y_h = Ce^{-4t} \) where \( C \) is a constant.
03
Find the Particular Solution
Use the method of undetermined coefficients to find a particular solution. Assume a solution of the form \( y_p = A \cos t + B \sin t + C\cos 4t + D\sin 4t\). Substitute \( y_p \) and its derivative into the non-homogeneous equation to solve for the coefficients \( A, B, C, D \).
04
Apply Initial Conditions
For each initial condition \( y(0) = -5, -4, \ldots, 5 \), substitute into the general solution \( y = y_h + y_p \) to solve for the constant \( C \).
05
Plot Each Solution
Using a graphing tool or software, plot each of the solutions \( y = y_h + y_p \) over a suitable interval starting at \( t = 0 \). This interval should be long enough to observe any steady-state behavior or damping oscillations with the various initial conditions provided.
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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 (IVP) in the context of differential equations is a problem where you are given a differential equation along with conditions at a particular value of the independent variable, typically denoted as \( t = 0 \). The goal is to find a specific solution to the differential equation that satisfies these given initial conditions.
- The initial value problem we're considering involves the differential equation \( y^{\prime} + 4y = 2 \cos t + \sin 4t \), with various initial conditions ranging from \( y(0) = -5 \) to \( y(0) = 5 \).
- Each initial condition provides a starting point for the solution, determining the constant \( C \) in the complementary solution \( y_h = Ce^{-4t} \).
Linear Non-Homogeneous Equation
A linear non-homogeneous differential equation is characterized by having terms that depend on the independent variable and do not all vanish as linear combinations of the dependent variable and its derivatives.
- The equation \( y^{\prime} + 4y = 2 \cos t + \sin 4t \) is non-homogeneous because of the presence of the functions \( 2 \cos t \) and \( \sin 4t \), which are not multiples of \( y \) or its derivatives.
- These additional functions on the right-hand side of the equation make it non-homogeneous, requiring techniques such as the method of undetermined coefficients to find a particular solution.
Undetermined Coefficients
The method of undetermined coefficients is a technique used to find a particular solution to a linear non-homogeneous differential equation.
- The approach involves assuming a form for the particular solution based on the non-homogeneous terms, which in our example are \( 2 \cos t \) and \( \sin 4t \).
- The assumed form for the particular solution in this exercise is \( y_p = A \cos t + B \sin t + C\cos 4t + D\sin 4t \).
- To find the coefficients \( A, B, C, \) and \( D \), substitute \( y_p \) and its derivatives into the original equation and solve the resulting system of equations.
Complementary and Particular Solutions
The solution to a non-homogeneous differential equation like \( y^{\prime} + 4y = 2 \cos t + \sin 4t \) comprises two parts: the complementary and particular solutions.
- The complementary solution, \( y_h \), solves the associated homogeneous equation \( y^{\prime} + 4y = 0 \). For this equation, the solution is \( y_h = Ce^{-4t} \).
- The particular solution, \( y_p \), solves the equation with the non-homogeneous terms included. In our case, this involves finding \( y_p = A \cos t + B \sin t + C\cos 4t + D\sin 4t \) through the method of undetermined coefficients.
- By adding these two solutions together, we obtain the general solution: \( y = y_h + y_p \).