Question

Find an equation of the form \(a y^{\prime \prime}+b y^{\prime}+c y=0\) for which all solutions approach a multiple of \(e^{-t}\) as \(t \rightarrow \infty\)

Short answer

Question: Find a second-order linear homogeneous ordinary differential equation (ODE) of the form \(ay''+by'+cy=0\) such that all its solutions approach a multiple of \(e^{-t}\) as \(t \rightarrow \infty\). Answer: The desired second-order linear homogeneous ODE can be written as: $$y''+(\alpha+\beta)y'+\alpha\beta y=0$$ or alternatively, $$y''+by'+cy=0$$ where \(b=\alpha+\beta\) and \(c=\alpha\beta\), and \(\alpha,\beta>0\).

Step-by-step solution

Set up the characteristic equation

Since the given ODE is a linear second-order homogeneous ODE, we can write down the associated characteristic equation in the form: $$ar^2+br+c=0$$ where \(r\) is the characteristic root.

Analyze given conditions of the solutions

The exercise says that all solutions should approach a multiple of \(e^{-t}\) as \(t \rightarrow \infty\). In order to make sure this happens, both roots of the characteristic equation must have negative real parts. So, let's assume the roots are in the form: $$r_1 = -\alpha, r_2 = -\beta$$ where \(\alpha, \beta > 0\).

Obtain the quadratic equation

To find the desired characteristic equation, we write the equation by considering it has roots \(r_1 = -\alpha\) and \(r_2 = -\beta\): $$(r+\alpha)(r+\beta)=0$$ Expanding the equation, we get: $$r^2+(\alpha+\beta)r+\alpha\beta = 0$$ By comparing the coefficients with the characteristic equation \(ar^2+br+c=0\), we can see that: $$a=1, \hspace{1cm} b=\alpha+\beta, \hspace{1cm} c=\alpha\beta$$

Write the differential equation

Now, we can write down the desired second-order linear homogeneous ODE using the values of \(a\), \(b\), and \(c\): $$y''+(\alpha+\beta)y'+\alpha\beta y=0$$ Alternatively, we can use the original form with a, b, and c constants: $$y''+by'+cy=0$$ where \(b=\alpha+\beta\) and \(c=\alpha\beta\), and \(\alpha,\beta>0\).

Key concepts

Characteristic Equation

Understanding the characteristic equation is the key to solving linear homogeneous differential equations, especially when dealing with second-order ODEs. The characteristic equation resembles a quadratic equation and is derived by assuming a solution of the form \( y = e^{rt} \), where \( r \) is the characteristic root. Essentially, it's the algebraic equation obtained from the differential equation by replacing each derivative of \( y \) with a power of \( r \). The solutions to the characteristic equation, which are the roots \( r_1 \) and \( r_2 \), determine the behavior of the original differential equation's solutions.

For the given problem, the characteristic equation can be set up as \( ar^2 + br + c = 0 \), following the standard quadratic form. This equation helps you determine the values of the constants from the desired behavior of the differential equation solutions. In our case, the solutions should approach a multiple of \( e^{-t} \), emphasizing the importance of the roots' sign and magnitude.

Second-Order ODE

A second-order ODE contains derivatives of the function up to the second order but no higher. It is crucial to understand the general form \( a y'' + b y' + c y = 0 \) where \( y' \) and \( y'' \) represent the first and second derivatives of \( y \) with respect to \( t \), and \( a \), \( b \), and \( c \) are coefficients which can impact the solution's behavior.

In applications, second-order ODEs typically model systems with acceleration, such as vibrations in mechanical systems, or in our case, functions whose long-term behavior we are interested in. These equations require special techniques for solutions, one of which involves solving for the roots of the characteristic equation related to the ODE.

Exponential Solutions

Exponential solutions arise naturally when solving linear homogeneous differential equations. They reflect the system's growth or decay processes and are expressed in the form \( e^{rt} \), where \( r \) is the root of the characteristic equation.

For the solution to approach zero as \( t \) approaches infinity, the real parts of the roots must be negative. This ensures that the exponent in the exponential expression gets more negative over time leading to a rapidly diminishing value of \( e^{rt} \). In our exercise, we require the solutions to be multiples of \( e^{-t} \), which means that our characteristic roots should be negative to achieve a decaying exponential as time progresses.

Homogeneous ODE

A differential equation is considered a homogeneous ODE when all terms are a product of the function \( y \) or its derivatives, and the equation is set equal to zero. The term 'homogeneous' refers to the fact that the equation remains the same if we multiply the function by a scalar.

In our example, once the characteristic equation is applied, and its roots determined, we are left with a homogeneous ODE of the form \( y'' + (\text{sum of roots})y' + (\text{product of roots})y = 0 \). This is significant because the general solution to a homogeneous second-order ODE with constant coefficients can be expressed based on the nature of the roots (real and distinct, real and repeated, or complex conjugates). The task was to tailor the equation such that the long-term behavior of the solution is aligned with an exponential decay.