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Is the equation \(t^{2} y^{\prime}(t)=\frac{t+4}{y^{2}}\) separable?

Short Answer

Expert verified
Answer: Yes, the equation is separable as it can be rewritten in the form \(M(t)dt + N(y)dy = 0\), where \(M(t) = \frac{1}{t^2}\) and \(N(y) = -\frac{t+4}{y^2}\).

Step by step solution

01

Rewrite the equation

First, we rewrite the given equation to isolate \(y'(t)\): $$y'(t) = \frac{t+4}{t^2 y^2}$$
02

Identify Functions

Now, we try to rewrite the equation in the form \(M(t)dt + N(y)dy = 0\). Let's consider \(M(t) = \frac{1}{t^2}\) and \(N(y) = -\frac{t+4}{y^2}\).
03

Rearrange the Equation

Now let's rearrange the equation to obtain the desired separable form: $$-\frac{t+4}{y^2}dy = \frac{1}{t^2}dt$$
04

Conclusion

Since we were able to rewrite the given equation in the form \(M(t)dt + N(y)dy = 0\), we can conclude that the equation is indeed separable.

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Key Concepts

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

Differential Equations
In the realm of mathematics, differential equations are powerful tools used to describe various phenomena in engineering, physics, economics, and other sciences. These equations involve functions and their derivatives, expressing the rate of change of certain quantities. When we talk about a differential equation, we refer to a mathematical equation that relates a function with its derivatives. Depending on the number of times derivatives are taken, differential equations can be classified into different orders, with the order being the highest derivative present in the equation.

For example, consider a population model where the rate of population growth at any time is proportional to the population at that time. Such a relationship can be expressed using a first-order differential equation. In this context, finding a solution to the differential equation means determining a function that describes the population at any given time. Solutions can provide predictive power, enabling professionals to forecast future events or understand the behavior of systems over time.
Variable Separation
The technique of variable separation is a fundamental method for solving first-order differential equations. The crux of this method involves rearranging the equation so that each variable and its corresponding differential are on opposite sides of the equation. This paves the way for integrating both sides separately.

Separation of variables can be visualized in steps. First, you identify and isolate the terms involving the dependent variable (usually 'y') and its differential ('dy'). Simultaneously, you gather the terms involving the independent variable (usually 't' or 'x') and its differential ('dt' or 'dx') on the other side. The separated equation can then be integrated with respect to its own variable, leading to a relationship between 'y' and 't', which can be solved explicitly in some cases, or implicitly left in terms of an equation relating 'y' and 't'.

Example:

If you have the equation \(\frac{dy}{dt} = g(t)h(y)\), after separating, you would integrate \(\int \frac{1}{h(y)}dy\) on one side and \(\int g(t)dt\) on the other.
Integrating Factors
Some differential equations can't be easily solved by simple separation of variables. In such scenarios, integrating factors are a helpful technique, particularly for linear first-order differential equations. An integrating factor is a function that is multiplied to both sides of a differential equation to facilitate its simplification and solution.

The goal is to transform the equation into an exact differential, where the left side becomes the derivative of a product of the integrating factor and the function we aim to find. The beautiful aspect of integrating factors is that for a differential equation of the form \(\frac{dy}{dt} + p(t)y = q(t)\), there's a systematic way to determine the integrating factor, usually denoted as \(\mu(t)\), given by \(\mu(t) = e^{\int p(t)dt}\). With the integrating factor at hand, multiplying it through the differential equation typically leads to a situation where the left side of the equation is a derivative of a product, making it ripe for direct integration.
First-Order Differential Equations
Among the hierarchy of differential equations, first-order differential equations are the starting point and are characterized by involving the first derivative of the unknown function but no higher derivatives. These equations take the form \(\frac{dy}{dx} = f(x, y)\), where \(f(x, y)\) can be any function involving 'x' and 'y'. The strategies we employ to solve them depend heavily on the form of \(f(x, y)\).

If the equation can be written such that \(f(x, y)\) equals a function of 'x' times a function of 'y', it's likely a candidate for the method of separation of variables, which is one of the simplest techniques to apply. Other first-order equations may require different approaches, such as using integrating factors or special substitutions that transform the equation into a separable one. The breadth of techniques available offers mathematicians and scientists a toolbox for addressing these equations, which model anything from radioactive decay to compound interest to fluid flow.

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Most popular questions from this chapter

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t .\) Use this idea to solve the following initial value problems. $$\left(t^{2}+1\right) y^{\prime}(t)+2 t y=3 t^{2}, y(2)=8$$

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t .\) Use this idea to solve the following initial value problems. $$e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}, y(0)=4$$

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one function, be sure to indicate which function corresponds to the solution of the initial value problem. $$z^{\prime}(x)=\frac{z^{2}+4}{x^{2}+16}, z(4)=2$$

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)=12 y-18$$

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

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