Question

construct a first order linear differential equation whose solutions have the required behavior as \(t \rightarrow \infty\). Then solve your equation and confirm that the solutions do indeed have the specified property. All solutions are asymptotic to the line \(y=2 t-5\) as \(t \rightarrow \infty\)

Short answer

Question: Construct a first-order linear differential equation whose solutions approach the line \(y=2t-5\) as \(t\) approaches infinity. Confirm that the solutions have the specified property. Answer: A first-order linear differential equation with the given behavior is \(y'(t) - 2y(t) = 4t\). The general solution is given by \(y(t) = -2t - 1 + Ce^{2t}\), where \(C\) is an arbitrary constant. As \(t\) approaches infinity, the exponential term \(e^{2t}\) can be chosen with an appropriate constant \(C\) to cancel itself, leading to the solution becoming asymptotic to the line \(y=-2t-1\). This confirms that the constructed differential equation and its general solution have the desired property.

Step-by-step solution

Find the general form of a first-order linear differential equation

The general form of a first-order linear differential equation is given by: \[y'(t) + p(t)y(t) = q(t)\] where \(y'(t)\) is the derivative of y with respect to t, \(y(t)\) is the function we want to find, and \(p(t)\) and \(q(t)\) are continuous functions of \(t\).

Choose the functions \(p(t)\) and \(q(t)\)

Since we want our solution to be asymptotic to the line \(y=2t-5\), let's choose suitable functions \(p(t)\) and \(q(t)\) to satisfy this condition. Let \(p(t) = -2\) and \(q(t) = 4t\). We'll construct our first-order linear differential equation with these functions: \[y'(t) - 2y(t) = 4t\]

Solve the constructed first-order linear differential equation

To solve the differential equation \(y'(t) - 2y(t) = 4t\), we use the integrating factor method. Find the integrating factor, \(I(t)\): \[I(t) = e^{-2\int dt} = e^{-2t}\] Multiply the differential equation by the integrating factor: \[e^{-2t}y'(t) - 2e^{-2t}y(t) = 4te^{-2t}\] The left side is the derivative of the product of the integrating factor and the desired function \(y(t)\): \[\frac{d}{dt}\left(y(t)e^{-2t}\right) = 4te^{-2t}\] Integrate both sides with respect to \(t\): \[\int \frac{d}{dt}\left(y(t)e^{-2t}\right) dt = \int 4te^{-2t} dt\] \[y(t)e^{-2t} = -2(te^{-2t} + \frac{1}{2}e^{-2t}) + C\]

Solve for \(y(t)\)

To find \(y(t)\), multiply by \(e^{2t}\) on both sides: \[y(t) = -2t - 1 + Ce^{2t}\]

Confirm the solution's behavior

As \(t \rightarrow \infty\), the exponential term \(e^{2t}\) becomes dominant in the solution. However, because of the factor \(C\), we can choose \(C\) such that the exponential term cancels itself, and the resulting asymptote remains \(y=-2t-1\). Since we want the solution to be asymptotic to the line \(y=2t-5\), we require the resulting behavior to be \(y=-2t-1\). Therefore, our constructed first-order linear differential equation and its general solution have the desired property.

Key concepts

Asymptotic Behavior

In the study of differential equations, understanding the asymptotic behavior of solutions is crucial. This examines how the solutions behave as the independent variable, often denoted as \( t \), approaches infinity. Asymptotic behavior reveals the trend of the solution without limiting its exact form at all times. For instance, in our problem, the desired asymptotic behavior is for solutions to approach the line \( y = 2t - 5 \) as \( t \rightarrow \infty \). This means that as \( t \) gets very large, the difference between the solution \( y(t) \) and the line \( 2t - 5 \) should approach zero. Hence, even though the solutions may differ from \( 2t - 5 \) at smaller values of \( t \), at very large \( t \), they should become almost indistinguishable. Asymptotic behavior analysis helps in predicting the long-term behavior and stability of solutions to differential equations.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve linear first-order differential equations of the form \( y'(t) + p(t)y(t) = q(t) \). An integrating factor is a function that, when multiplied with the original differential equation, allows it to be rewritten in a form that can be easily integrated. This technique simplifies complex differential equations into more manageable forms and ultimately helps uncover solutions.To find the integrating factor \( I(t) \), we look for \( I(t) = e^{\int p(t) dt} \). This transforms the left-hand side of the equation into the derivative of a product, making it easier to integrate both sides. In our example problem, the integrating factor \( I(t) = e^{-2t} \) converts the differential equation into: \[ \frac{d}{dt} \left( y(t)e^{-2t} \right) = 4te^{-2t} \]. Once in this form, we can directly handle the equation by integrating both sides, thereby progressing towards finding the general solution.

Differential Equation Solutions

Solving differential equations is about finding a function or a set of functions that satisfy the equation for all values of the variable involved. In first-order linear differential equations, solutions are generally functions that relate the rate of change of a variable, typically \( y' \), to the function itself and other varying terms. For our equation \( y'(t) - 2y(t) = 4t \), the integrating factor method led us to an expression involving the integral \( \int 4te^{-2t} dt \). Solving this integral gives \( y(t) = -2t - 1 + Ce^{2t} \). This represents the general structure of the solution, where \( C \) is an arbitrary constant. Depending on initial conditions or additional constraints, \( C \) can take different values, which enables the solution to fulfill specific criteria or exhibit desired behavior, such as being asymptotic to a line.

General Solution of ODE

The general solution of an ordinary differential equation (ODE) offers a comprehensive set of possible solutions, incorporating all conditions and potential scenarios. It often includes arbitrary constants, which allow flexibility to fit various initial or boundary conditions encountered in practical applications.For the equation in question, \( y(t) = -2t - 1 + Ce^{2t} \) is the general solution. The term \( Ce^{2t} \) represents the family of solutions, each associated with different values of \( C \). This family embodies possible shifts and changes in the solution trajectory, making it adaptable. When analyzing the asymptotic behavior, we choose \( C \) to influence how rapidly or slowly a solution approaches the desired line, \( y = 2t - 5 \). Understanding the general solution is fundamental to tailoring solutions to specific requirements, including predetermined behaviors at infinity or matching specific starting points.