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Chapter 9: Problem 3

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

Short Answer

Expert verified
Answer: No, the equation is not separable since it cannot be rewritten in the form \(\frac{dy}{dt} = f(t)g(y)\) for any functions \(f(t)\) and \(g(y)\).

Step by step solution

01

Rewrite the equation in terms of the derivative of y with respect to t

Write the given equation in terms of the derivative of y with respect to t: \(\frac{dy}{dt} = 2y-t\).
02

Separate the terms involving y and t

We need to rewrite the equation in the form \(\frac{dy}{dt} = f(t)g(y)\). To do this, we'll first try to factor out the y and the t terms on the right-hand side of the equation: \((2y-t)\) does not seem to factor any further with any clear separable terms.
03

Conclusion

Since we cannot separate and rewrite the equation \(\frac{dy}{dt} = 2y-t\) as \(\frac{dy}{dt} = f(t)g(y)\), for any functions \(f(t)\) and \(g(y)\), the equation is not separable.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. They hold a central place in calculus, as they describe how functions change over time or space and are used to model a variety of real-world systems in physics, engineering, biology, and economics. A differential equation, at its simplest, looks something like \(\frac{dy}{dx}=f(x,y)\), where \(f\) is a given function, and \(y\) is the function to be determined. Solutions to differential equations can often depict physical phenomena such as motion, growth, or decay. In the given exercise involving \(y' = 2y - t\), we're dealing with a first-order differential equation, which incorporates a first derivative of an unknown function \(y\) with respect to a variable \(t\).
Calculus
Calculus is a branch of mathematics that studies continuous change, through derivatives and integrals.
Derivatives, one of the fundamental concepts of calculus, measure the rate at which a quantity changes.
When we take the derivative of a function with respect to a variable, we're determining how the function responds as that variable changes. Integrals, on the other hand, offer a way to determine the accumulation of quantities, such as area under a curve.

In our example, the use of derivatives allows us to express the rate at which \(y\) changes with respect to \(t\) in the form of a differential equation. Solving such equations often requires techniques from calculus, like separation of variables, integration, and in some cases, finding integrating factors.
Integrating Factors
An integrating factor is a function used to solve certain types of differential equations, particularly non-separable first-order linear equations. The purpose of an integrating factor is to multiply both sides of the differential equation to rearrange it into a form in which the left-hand side is the derivative of a product of two functions. This technique simplifies the equation, making it easier to solve.

More technically, if an equation has the form \(y' + p(t)y = q(t)\), we can often find an integrating factor, \(\mu(t)\), such that when we multiply the entire equation by \(\mu(t)\), it becomes \(\frac{d}{dt}[\mu(t)y(t)] = \mu(t)q(t)\), a form that is easy to integrate on both sides. However, in the case of our exercise where the differential equation is \(y' = 2y - t\), we can't apply this method directly since it's not in the correct form.
First-Order Differential Equations
First-order differential equations involve the first derivative of a function but no higher derivatives. They take the general form \(\frac{dy}{dx}=f(x,y)\), implying that the rate of change of \(y\) with respect to \(x\) depends on both \(x\) and \(y\) themselves.

Separable first-order equations are a subset where the function \(f(x,y)\) can be expressed as the product of two single-variable functions, specifically \(g(x)\) and \(h(y)\), allowing us to separate the variables onto opposite sides of the equation. This separated form enables us to integrate both sides with respect to their respective variables to find a solution.

However, as observed in our exercise with the equation \(\frac{dy}{dt} = 2y - t\), not all first-order equations are separable. When the equation cannot be written in a form allowing for the separation of variables, alternative techniques such as integrating factors or other methods must be used to solve them.

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Most popular questions from this chapter

Consider the differential equation \(y^{\prime \prime}(t)+k^{2} y(t)=0,\) where \(k\) is a positive real number. a. Verify by substitution that when \(k=1\), a solution of the equation is \(y(t)=C_{1} \sin t+C_{2} \cos t .\) You may assume this function is the general solution. b. Verify by substitution that when \(k=2\), the general solution of the equation is \(y(t)=C_{1} \sin 2 t+C_{2} \cos 2 t\) c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture.

In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=2 x-x y, y^{\prime}(t)=-y+x y$$

Direction field analysis Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a\), then the solution increases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a\), then the solution decreases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. Logistic population growth Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right)\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-r t}},\) where \(C\) is an arbitrary constant. b. Find the value of \(C\) that corresponds to the initial condition \(P(0)=50\) c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\) d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).

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. $$y^{\prime}(t)=12 y-18$$
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