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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 \( y_h = Ce^{-4t} \).
Solving an IVP requires combining both the complementary and particular solutions and using the initial conditions to find the specific solution for \( y \). This approach ensures the solution passes through the specified point at the initial time, adhering to the problem's constraints.
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.
Linear non-homogeneous equations contrast with homogeneous equations, where the right-hand side equals zero. Here, a notable strategy is to find both the complementary and particular solutions. Combining these solutions will yield the general solution to the equation.
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.
This method works well when the non-homogeneous part of the equation is a simple polynomial, exponential, sine, or cosine function. By assuming a solution that mirrors the non-homogeneous part, it becomes manageable to find the necessary coefficients.
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: \( y = y_h + y_p \).
This combined solution represents the complete response of the system described by the differential equation, capturing both the natural behavior (complementary) and the forced response (particular) induced by external inputs.

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Most popular questions from this chapter

The built-in plotting routines of MATLAB's solvers were illustrated in Exercise 3. They work equally well with systems. For example, create function yprime \(=\) heart \((t, y)\) yprime \(=\operatorname{zeros}(2,1)\); \(\operatorname{yprime}(1)=\mathrm{y}(2)\); yprime \((2)=-16 * y(1)+4 * \sin (2 * t)\); and save as heart.m. a) Enter ode45(@heart, \([0,2 * \mathrm{pi}],[0 ; 2])\) and note that MATLAB dynamically plots both \(y_{1}\) and \(y_{2}\) versus \(t\). b) To change the output routine, enter options = odeset ('OutputFcn', 'odephas ' \(^{\prime}\) ), followed by ode45 (@heart, [0,2*pi], [0;2], options). Note that MATLAB dynamically plots \(y_{2}\) versus \(y_{1}\) in the phase plane.

\(x^{\prime}+x+x^{3}=\cos ^{2} t\), with \(x(0)=-3,-2, \ldots, 3\).

In the 1920 's, the Italian mathematician Umberto Volterra proposed the following mathematical model of a predator-prey situation to explain why, during the first World War, a larger percentage of the catch of Italian fishermen consisted of sharks and other fish eating fish than was true both before and after the war. Let \(x(t)\) denote the population of the prey, and let \(y(t)\) denote the population of the predators. In the absence of the predators, the prey population would have a birth rate greater than its death rate, and consequently would grow according to the exponential model of population growth, i.e. the growth rate of the population would be proportional to the population itself. The presence of the predator population has the effect of reducing the growth rate, and this reduction depends on the number of encounters between individuals of the two species. Since it is reasonable to assume that the number of such encounters is proportional to the number of individuals of each population, the reduction in the growth rate is also proportional to the product of the two populations, i.e., there are constants \(a\) and \(b\) such that $$ x^{\prime}=a x-b x y $$ Since the predator population depends on the prey population for its food supply it is natural to assume that in the absence of the prey population, the predator population would actually decrease, i.e. the growth rate would be negative. Furthermore the (negative) growth rate is proportional to the population. The presence of the prey population would provide a source of food, so it would increase the growth rate of the predator species. By the same reasoning used for the prey species, this increase would be proportional to the product of the two populations. Thus, there are constants \(c\) and \(d\) such that $$ y^{\prime}=-c y+d x y . $$ a) A typical example would be with the constants given by \(a=0.4, b=0.01, c=0.3\), and \(d=0.005\). Start with initial conditions \(x_{1}(0)=50\) and \(x_{2}(0)=30\), and compute the solution to (8.22) and (8.23) over the interval \([0,100]\). Prepare both a time plot and a phase plane plot. After Volterra had obtained his model of the predator-prey populations, he improved it to include the effect of "fishing," or more generally of a removal of individuals of the two populations which does not discriminate between the two species. The effect would be a reduction in the growth rate for each of the populations by an amount which is proportional to the individual populations. Furthermore, if the removal is truly indiscriminate, the proportionality constant will be the same in each case. Thus, the model in equations \((8.22)\) and (8.23) must be changed to $$ \begin{aligned} &x^{\prime}=a x-b x y-e x \\ &y^{\prime}=-c y+d x y-e y \end{aligned} $$ where \(e\) is another constant. b) To see the effect of indiscriminate reduction, compute the solutions to the system in \((8.24)\) when \(e=0\), \(0.01,0.02,0.03\), and \(0.04\), and the other constants are the same as they were in part a). Plot the five solutions on the same phase plane, and label them properly. c) Can you use the plot you constructed in part b) to explain why the fishermen caught more sharks during World War I? You can assume that because of the war they did less fishing.

Separate variables and find an explicit solution of \(x^{\prime}=t /(1+x), x(0)=1\). a) Use ode45 to find the numerical solution of the initial value problem on the interval \([0,3]\). Store the solution in the variables \(t\) and \(x_{-}\)ode45. b) Use the explicit solution to find exact solution values at each point in the vector \(t\) obtained in part a). Store the results in the variable \(x_{-}\)exact. Compare the solutions graphically with plot \(\left(t\right.\), \(x_{-}\)exact, \(t\), x_ode45, 'r. '). Obtain a printout with a legend. c) For a finer graphical image of the error plot \(\mathrm{x}_{-}\)exact - \(\mathrm{x}_{-}\)ode45 versus \(\mathrm{t}\) in a new figure. Pay special attention to the dimensions on the \(y\)-axis.

\(x_{1}{ }^{\prime}=x_{2}\) and \(x_{2}{ }^{\prime}=-x_{1}+x_{2}\left(1-3 x_{1}^{2}-3 x_{2}^{2}\right)\), with \(x_{1}(0)=0.2\) and \(x_{2}(0)=0.2\) on \([0,20]\).

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