Question

In each exercise, discuss the behavior of the solution \(y(t)\) as \(t\) becomes large. Does \(\lim _{t \rightarrow \infty} y(t)\) exist? If so, what is the limit? \(\frac{y^{\prime}-e^{-t}+2}{y}=-2, \quad y(0)=-2\)

Short answer

If so, what is the limit? Answer: Yes, the limit exists. As t approaches infinity, the limit of the solution y(t) is 0.

Step-by-step solution

Rewrite the given equation

Rewrite the given differential equation in the form \(y'(t) = f(t,y(t))\) by isolating \(y'(t)\) and simplifying the expression. Doing so, we get: \(y'(t) = -2y(t) + e^{-t} + 2\)

Solve the homogeneous equation

Write down the homogeneous equation, i.e., the associated equation without the non-homogeneous term: \(y_h'(t) = -2y_h(t)\). Now, solve the homogeneous equation using separation of variables: \(\frac{1}{y_h(t)} dy_h(t) = -2 dt\). Integrating both sides, we get: \(\ln\left\lvert{y_h(t)}\right\rvert=-2t+C_1\). Taking exponential of both sides and introducing the integration constant \(C_1\), we get the general solution of the homogeneous equation: \(y_h(t) = C_1e^{-2t}\).

Find a particular solution

Use the method of variation of parameters to find a particular solution to the non-homogeneous equation. We aim to find a function \(u(t)\) that satisfies: \(u(t) y_h(t) = y_p(t)\). Then, differentiate the equation to get: \(u'(t) y_h(t) + u(t) y_h'(t) = y_p'(t)\). Now, compute \(y_p'(t)\) using the original non-homogeneous equation: \(y_p'(t) = -2y_p(t) + e^{-t} + 2\). By substitution, we have \(u'(t) y_h(t) + u(t) (-2y_h(t)) = -2u(t) y_h(t) + e^{-t} + 2\). Cancel out the term \(-2u(t) y_h(t)\) on both sides and divide by \(y_h(t)\): \(u'(t) = \frac{e^{-t} + 2}{C_1e^{-2t}}\). Integrate both sides: \(u(t) = \int \frac{e^{-t} + 2}{C_1e^{-2t}} dt = \frac{1}{C_1} \int (e^t + 2e^{t}) dt\). To find the final answer, we compute the integral: \(u(t) = \frac{1}{C_1}(e^t + 2e^t) + C_2\). Now, the particular solution is given by \(y_p(t) = u(t)y_h(t)\), where \(y_h(t)\) is the general solution of the homogeneous equation: \(y_p(t) = (e^t + 2e^t + C_2 C_1) e^{-2t}\).

Apply the initial condition

Given that \(y(0) = -2\), we have \(-2 = (e^0 + 2e^0 + C_2 C_1) e^0\). Solve for \(C_2 C_1\): \(C_2 C_1 = -5\).

Analyze the behavior of the solution as t approaches infinity

We have found the particular solution: \(y(t) = (e^t + 2e^t - 5)e^{-2t}\). Now, we analyze the behavior of the solution as \(t \rightarrow \infty\). Since the terms with \(e^t\) will cancel out, and the terms with \(e^{-t}\) will approach zero, the limit is given by: \(\lim_{t \rightarrow \infty} y(t) = \lim_{t \rightarrow \infty} (e^t + 2e^t - 5)e^{-2t} = (1 + 2 - 5)(0) = 0\). Therefore, the limit exists, and it is equal to 0.

Key concepts

Homogeneous Differential Equations

Understanding homogeneous differential equations is crucial for students tackling higher-level math. These refer to differential equations where every term is a function of the variable and its derivatives. Typically, they take the form \(ay'' + by' + cy = 0\) where \(a\), \(b\), and \(c\) are constants. To solve them, one common method is separation of variables, where we isolate the terms involving the derivative \(y'\) on one side and the function \(y\) on the other.

Once the variables are separated, we integrate both sides with respect to their respective variables. This process yields a general solution involving an arbitrary constant, representing the infinite family of solutions that a homogeneous differential equation possesses. In our problem, after separating variables, we obtained the solution \(y_h(t) = C_1e^{-2t}\), which represents the general solution for the homogeneous part of the equation.

Non-Homogeneous Differential Equations

While homogeneous differential equations are vital, students must also understand non-homogeneous differential equations to solve more complex problems. These equations have a similar form to homogeneous versions, but include an additional function that does not merely consist of the variable and its derivatives multiplied by constants. In other words, we have something like \(ay'' + by' + cy = g(x)\), where \(g(x)\) is a non-zero function that brings in non-homogeneity to the equation.

For our textbook problem, the non-homogeneous term comes from \(e^{-t} + 2\). Solving such equations often involves finding a particular solution that satisfies the non-homogeneous equation and then adding it to the general solution of the homogeneous equation. This comprehensive solution covers the behaviors of all potential specific scenarios described by the differential equation.

Method of Variation of Parameters

One advanced method for solving non-homogeneous differential equations is the method of variation of parameters. This technique involves introducing new functions, typically \(u(t)\), to replace the constants in the general solution to the homogeneous equation. The goal is to find a particular solution to the non-homogeneous equation, which complements the general solution of the homogeneous equation.

Students should note that this method involves differentiating the \(u(t)y_h(t)\) product and replacing the \(y'\) and \(y\) terms into the original non-homogeneous equation. After some algebraic manipulations, we find that \(u(t)\) satisfies a simpler differential equation, which can then be integrated to find the function \(u(t)\) itself. This strategy is exemplified in our exercise by integrating to find \(u(t)\) and consequently obtaining the particular solution \(y_p(t)\).

Limit of a Function

A fundamental concept in calculus is the limit of a function, which describes the behavior of a function as the input approaches a certain value. In the context of differential equations, understanding the limit helps students discuss the long-term behavior of solutions. For a function \(y(t)\), we denote the limit as \(t\) approaches infinity by \(\lim_{t \rightarrow \infty} y(t)\).

If the limit exists and is finite, it can signal the stabilization or eventual behavior of the system described by the differential equation. In practice, this often involves recognizing terms in the solution that will grow or decay without bound as \(t\) increases. In our exercise, we observed that the exponential terms cancel and the remaining term approaches zero, thus determining that the limit as \(t\) approaches infinity for \(y(t)\) is zero, indicating the solution will stabilize over time.