Question

In each exercise, the general solution of the differential equation \(y^{\prime}+p(t) y=g(t)\) is given, where \(C\) is an arbitrary constant. Determine the functions \(p(t)\) and \(g(t)\). \(y(t)=C t^{-1}+1, \quad t>0\)

Short answer

Question: Determine the functions \(p(t)\) and \(g(t)\) for the given differential equation \(y' + p(t)y = g(t)\), with the given general solution \(y(t) = Ct^{-1} + 1\). Answer: The functions \(p(t)\) and \(g(t)\) for the given differential equation are both \(p(t) = -t^{-1}\) and \(g(t) = -t^{-1}\).

Step-by-step solution

Differentiate the given general solution

Differentiate \(y(t) = Ct^{-1} + 1\) with respect to \(t\): \(y'(t) = \frac{d}{dt}(Ct^{-1} + 1) = -Ct^{-2}\).

Substitute \(y(t)\) and \(y'(t)\) into the given differential equation

We have \(y'(t) = -Ct^{-2}\) and \(y(t) = Ct^{-1} + 1\). Now, substitute these expressions into the given differential equation \(y'+p(t)y=g(t)\): \(-Ct^{-2} + p(t)(Ct^{-1} + 1) = g(t)\).

Solve the resulting equation for \(p(t)\) and \(g(t)\)

Rewrite the equation from Step 2 with the original expressions for \(y(t)\) and \(y'(t)\): \(-Ct^{-2} + p(t)(Ct^{-1} + 1) = g(t)\). Collect the terms containing the arbitrary constant \(C\) on the left side of the equation and the remaining terms on the right side to isolate the functions \(p(t)\) and \(g(t)\): \(C(-t^{-2} + p(t)t^{-1}) = g(t) - p(t)\). Since \(C\) is an arbitrary constant, we can identify the corresponding expressions for \(p(t)\) and \(g(t)\) as follows: \(p(t)t^{-1} = -t^{-2}\), so \(p(t) = -t^{-1}\). \(g(t) - p(t) = 0\), so \(g(t) = p(t) = -t^{-1}\). Therefore, we have determined the functions \(p(t) = -t^{-1}\) and \(g(t) = -t^{-1}\).

Key concepts

Differential Equation

A differential equation is an equation that relates a function with its derivatives. In simplest terms, it describes how a particular quantity changes with respect to another, often representing physical phenomena such as motion, heat, or wave propagation.

In the exercise provided, we are looking at a first-order linear differential equation of the form \(y' + p(t)y = g(t)\), where \(y'\) represents the derivative of the function y with respect to \(t\), \(p(t)\) is a function of \(t\), and \(g(t)\) is another function of \(t\). The variable \(t\) typically stands for time, but it can represent any independent variable. This type of equation is particularly useful because it occurs frequently in various fields of science and engineering.

General Solution

The general solution of a differential equation is an expression that contains all possible solutions to the equation. It is 'general' because it includes an arbitrary constant (or constants) which can be tailored to meet specific initial conditions or boundary values.

In our exercise, the general solution given is \(y(t) = Ct^{-1} + 1\), where \(C\) is an arbitrary constant. This implies that for any value of \(C\), the function \(y(t)\) will satisfy the differential equation for \(t > 0\). A particular solution can be found by determining a specific value for \(C\) that conforms to additional information or initial conditions not provided in this exercise.

Arbitrary Constant

An arbitrary constant in the context of differential equations is a symbol, typically \(C\), that represents an undetermined constant value. It reflects the general nature of solutions in differential equations where multiple solutions may exist.

When deriving the general solution to a differential equation, we integrate functions, which introduces the arbitrary constant. This constant can take on any real number value and gives rise to a family of functions that represent the general solution. In the provided exercise, \(C\) is used as the arbitrary constant in the solution \(y(t) = Ct^{-1} + 1\), indicating that there is a whole set of solutions depending on the value of \(C\).

Function Differentiation

Differentiating a function involves computing its derivative, which measures the rate at which the function's value changes with respect to changes in its input value. Essentially, differentiation provides the slope of the function at any point.

For the solution \(y(t) = Ct^{-1} + 1\), we differentiate with respect to \(t\) to find \(y'(t) = -Ct^{-2}\). This step is crucial in solving the differential equation because it allows us to substitute the derivative \(y'(t)\) back into the original equation and match terms involving \(t\) and \(C\) to find \(p(t)\) and \(g(t)\). By differentiating correctly, we can proceed with confidence knowing we have accurately represented the behavior of \(y(t)\) as it relates to \(t\).