/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 For each of the Problems 13-24 y... [FREE SOLUTION] | 91Ó°ÊÓ

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For each of the Problems 13-24 you should determine whether the problem needs to be solved using separation of variables or integrating factors (some of the problems may be solved using \mathrm{\\{} e i t h e r ~ m e t h o d ) . ~ T h e n ~ s o l v e ~ t h e ~ d i f f e r e n t i a l ~ e q u a t i o n . ~. $$ \frac{d y}{d t}=\frac{y}{t}-t^{2} $$

Short Answer

Expert verified
Using an integrating factor, the solution is \( y(t) = -\frac{t^3}{2} + Ct \).

Step by step solution

01

Identify the Form of the Differential Equation

The differential equation is \( \frac{dy}{dt} = \frac{y}{t} - t^2 \). Identify if it fits the form for separation of variables or the use of integrating factors.
02

Reformat the Equation for an Integrating Factor Solution

Reformat the equation to \( \frac{dy}{dt} - \frac{y}{t} = -t^2 \), which resembles a linear first-order differential equation in the form \( \frac{dy}{dt} + P(t) y = Q(t) \), where \( P(t) = -\frac{1}{t} \) and \( Q(t) = -t^2 \).
03

Calculate the Integrating Factor

The integrating factor \( \mu(t) \) is found using \( \mu(t) = e^{\int P(t) \, dt} = e^{-\int \frac{1}{t} \, dt} = e^{\ln|t|^{-1}} = \frac{1}{t} \).
04

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \frac{1}{t} \): \( \frac{1}{t} \frac{dy}{dt} - \frac{y}{t^2} = -t \).
05

Simplify to Recognize the Left Side as a Derivative

Recognize that the left side can be rewritten as the derivative of a product: \( \frac{d}{dt} \left( \frac{y}{t} \right) = -t \).
06

Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt} \left( \frac{y}{t} \right) dt = \int -t \, dt \]This gives: \( \frac{y}{t} = -\frac{t^2}{2} + C \), where \( C \) is the constant of integration.
07

Solve for \( y(t) \)

Multiply through by \( t \) to solve for \( y(t) \):\( y(t) = -\frac{t^3}{2} + Ct \).

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Key Concepts

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

Separation of Variables
When confronting a differential equation, one powerful method of solving it is by using the technique called separation of variables. This method is particularly useful when the variables involved can be expressed as a product of functions each depending solely on one variable. In simple terms, it helps us to separate the variables on opposite sides of the equation.

Here's a typical approach to applying this method:
  • Identify if the differential equation can be rearranged such that all terms involving one variable (say, \( y \)) are on one side, and all terms involving another variable (\( t \)) are on the other side.
  • Rewrite the equation in such a way that you can integrate both sides with respect to their respective variables.
  • Carry out the integration which often involves integrating both sides independently.
Using separation of variables is straightforward but requires that specific separation be possible. Unfortunately, it wouldn’t be helpful for the problem at hand where the equation involves mixed terms with \( y \) and \( t \). Therefore, another method should be considered, such as the use of an integrating factor.
Integrating Factors
Integrating factors are a clever trick introduced to transform a difficult differential equation into an expression that can be easily solved by direct integration. This approach is used primarily with linear first-order differential equations of the form \(\frac{dy}{dt} + P(t)y = Q(t)\). Here's how it works:

  • First, rewrite the given differential equation so that all the \( y \) terms and \( dy/dt \) terms are on one side, matching the standard form.
  • Find the integrating factor, \( \mu(t) \), which is essentially a function we multiply through the entire equation to facilitate integration. The formula is \( \mu(t) = e^{\int P(t) \, dt} \).
  • Multiply the whole differential equation by this integrating factor. This will convert the left-hand side into a derivative of a product, \( \frac{d}{dt}[\mu(t) y] \).
  • Now, integrate both sides to solve for \( y(t) \).
In the given example, the integrating factor is \( \frac{1}{t} \). By applying this factor, we transformed the differential equation into a solvable form, allowing us to integrate directly and find the desired solution.
Linear First-Order Differential Equations
Linear first-order differential equations appear widely in many fields, from engineering to finance. These equations take the form \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( y \) and its derivative are power to the first degree. The equation from the initial problem \( \frac{dy}{dt} = \frac{y}{t} - t^2 \) fits this category once reorganized.

To solve linear first-order differential equations, we often utilize integrating factors or separation of variables, depending on which applies best to the given situation. However, the method of integrating factors shines particularly here, as it simplifies and streamlines the process.

The problem given illustrates this approach as once rearranged, it reveals how the components fit the linear form, with quoting functions \( P(t) \) and \( Q(t) \) directly, allowing for a smooth path to solution using the integrating factor approach. This not only highlights the elegance of the method but also its adaptability to different types of equations more directly.

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Most popular questions from this chapter

Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=\frac{x^{3}}{x-1} $$

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d x}{d t}=x^{2}-h x $$

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=\frac{y-1}{y^{2}+1} $$

In Problems 13 and 14 we assumed that the per colony extinction rate was proportional to \(p .\) This means that the per colony extinction rate goes to 0 for small \(p .\) This may not be realisticsubpopulations may still go extinct even if they are not competing among themselves. One way to model this is to say that the per colony extinction rate is a function \(m(p)\) of \(p\). In Problems 15 and 16 we will assume that \(m(p)=a+b p\) for some constants \(a, b>0 .\) That is, the extinction rate increases with \(p\) because of competition between subpopulations, but \(m(p)\) does not vanish as \(p \rightarrow 0\). Then our model for proportion of occupied sites must be modified to: $$ \frac{d p}{d t}=c p(1-p)-(a+b p) p $$ where \(c, a, b\) are all positive constants. Assuming that the subpopulations obey the differential Equation (8.60) and the coefficients are \(a=1, b=2\), but \(c\) is allowed to take any value: (a) Find the equilibrium values of \(p\) (your answer will depend on the unknown coefficient \(c\) ). (b) What are the conditions on \(c\) for \(p\) to have a nontrivial equilibrium, that is, an equilibrium in which \(p \in(0,1] ?\) (c) Show that if your condition from (b) is met, then the nontrivial equilibrium is also stable.

One of the key ideas for sketching solutions from vector field plots is that a solution curve must be monotonic, that is, \(x(t)\) is either increasing or decreasing or constant but cannot switch from one behavior to another. We showed that a solution \(x(t)\) could not start by increasing and then switch to decreasing. Suppose that \(x(t)\) is a solution of the differential equation \(\frac{d x}{d t}=g(x)\) and that \(x(t)\) starts off decreasing with time. Show that \(x(t)\) cannot switch to increasing.

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