Question

Solve the initial value problem \(y^{\prime \prime}-y^{\prime}-2 y=0, \quad y(0)=\alpha, \quad y^{\prime}(0)=2 .\) Then find \(\alpha\) so that the solution approaches zero as \(t \rightarrow \infty\).

Short answer

Answer: There is no value of \(\alpha\) that makes the solution approach \(0\) as \(t \rightarrow \infty\).

Step-by-step solution

Find the general solution to the differential equation.

To solve the differential equation \(y^{\prime \prime}-y^{\prime}-2y=0\), we will try the function \(y(t)=e^{rt}\) for some constant \(r\). Substituting this function into the differential equation, we obtain: \(r^2 e^{rt}-r e^{rt}-2 e^{rt}=0\) Factoring out the common term \(e^{rt}\), we get: \(e^{rt}(r^2-r-2)=0\) Since \(e^{rt}\) is never zero, the equation becomes: \(r^2-r-2=0\) Solving this quadratic equation, we find: \(r_1=2, r_2=-1\) Now we can write the general solution as: \(y(t)=c_1 e^{2t} + c_2 e^{-t}\)

Use the initial conditions to find the constants.

We are given the following initial conditions: \(y(0)=\alpha, \quad y^{\prime}(0)=2\) Using the initial condition for \(y(0)\) and plugging in \(t=0\) in the general solution: \(\alpha=c_1e^{2(0)}+c_2e^{-1(0)}=c_1 + c_2 \Rightarrow c_1 = \alpha - c_2\) Now, we will differentiate the general solution with respect to \(t\) to find \(y^{\prime}(t)\): \(y^{\prime}(t) = 2c_1 e^{2t} - c_2 e^{-t}\) Using the initial condition for \(y^{\prime}(0)\) and plugging in \(t=0\): \(2=2(\alpha-c_2)e^{2(0)}-c_2e^{-1(0)} = 2(\alpha-c_2) - c_2 \Rightarrow c_2 = \alpha - 1\) Now, we can substitute the value of \(c_2\) into the expression for \(c_1\): \(c_1 = \alpha - (\alpha - 1) = 1\) So, our solution is: \(y(t)=e^{2t}+(\alpha-1)e^{-t}\)

Find \(\alpha\) so that the solution approaches \(0\) as \(t \rightarrow \infty\).

In order to find \(\alpha\), we will consider the behavior of the solution as \(t\) goes to infinity. If we want the solution to approach \(0\), we have: \(\lim_{t\to\infty} (e^{2t}+(\alpha-1)e^{-t}) = 0\) As \(t\) goes to infinity, the term \(e^{2t}\) will grow without bound. In order to counteract this and make the limit go to zero, the term \((\alpha-1)e^{-t}\) must grow to negative infinity. However, this term has a decreasing exponential; therefore, the only possibility is for \(e^{2t}\) to be zero in the solution. But this cannot happen since exponentials are never zero. Thus, we conclude that there is no value of \(\alpha\) that would make the solution approach \(0\) as \(t \rightarrow \infty\).

Key concepts

Differential Equation

A differential equation is an equation that involves the derivatives of a function as well as the function itself. These equations are fundamental in expressing the relationships between variables and their rates of change. For instance, the expression \( y^{\prime \prime} - y^{\prime} - 2y = 0 \) is a second-order linear homogeneous differential equation. It describes a situation where the function \(y(t)\), its first derivative \( y^{\prime}(t) \), and second derivative \( y^{\prime \prime}(t) \) are related through specified coefficients and combined linearly.

When solving a differential equation, one aims to find the function \(y(t)\) that satisfies the given relation for all \(t\) within a certain interval. This process often unfolds through initial value problems where specific values are known at the outset, providing the necessary conditions to find a unique solution.

General Solution

The general solution to a differential equation represents the complete set of possible solutions before any initial conditions are applied. It encompasses all functions that satisfy the differential equation. For example, the general solution to the given equation \(y^{\prime \prime} - y^{\prime} - 2y = 0\) is \(y(t) = c_1 e^{2t} + c_2 e^{-t}\), where \(c_1\) and \(c_2\) are arbitrary constants. These constants are placeholders that will later be determined using specific initial values provided in the problem, a process which shapes the general solution into a particular solution that adheres to the given conditions.

The articulation of such a solution usually relies on well-established methods for different types of differential equations, in this case, the method of characteristic equation to find the necessary constants.

Exponential Function

The exponential function is a mathematical function of the form \( e^{rt}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828, \(r\) is a constant, and \(t\) is the variable. Exponential functions are widely used to describe growth or decay processes, such as population growth, radioactive decay, or in this case, the behavior of solutions to certain differential equations.

In our exercise, exponential functions \(e^{2t}\) and \(e^{-t}\) represent two solutions that correspond to the characteristic roots of the equation, and they play a crucial role in determining the behavior of the final solution over time. Particularly, \(e^{2t}\) describes exponential growth and \(e^{-t}\) describes exponential decay as \(t\) increases.

Characteristic Equation

The characteristic equation is an algebraic equation derived from a differential equation whose solutions are used to construct the general solution of the original differential equation. By assuming a trial solution of the form \( y(t) = e^{rt}\), and substituting it into the linear homogeneous differential equation, we obtain an algebraic equation in terms of \(r\).

In our problem, factoring out the common \(e^{rt}\) term from our trial solution in the differential equation leads to the characteristic equation \(r^2 - r - 2 = 0\). Solving this provides the roots \(r_1=2\) and \(r_2=-1\), which are critical in forming the linear combination that is the general solution to our differential equation. This process bridges the gap between purely algebraic methods and the field of differential equations.

Limit of a Function

The limit of a function as a variable approaches a particular value is a fundamental concept in calculus. It describes the behavior of a function as the variable gets arbitrarily close to a given point. For the context of our problem, we are interested in the limit of the solution \(y(t)\) as \(t\) approaches infinity. This can provide insight into the behavior of the system described by the differential equation over a long period.

Solving our initial value problem requires analyzing the limit of \( e^{2t} + (\alpha-1)e^{-t} \) as \(t\) goes to infinity. It turns out that no value of \(\alpha\) can make the first exponentially growing term disappear, thus there's no \(\alpha\) that will result in the solution tending towards zero. This conclusion is key to understanding that the given system, regardless of initial conditions, will not stabilize at zero over time.