Question

Show that if \(y=\phi(t)\) is a solution of \(y^{\prime}+p(t) y=0,\) then \(y=c \phi(t)\) is also a solution for any value of the constant \(c .\)

Short answer

Question: Show that if \(y=\phi(t)\) is a solution of the first-order linear homogeneous ordinary differential equation (ODE) \(y^{\prime}+p(t)y=0\), then \(y=c\phi(t)\), where \(c\) is any constant, is also a solution of the ODE. Answer: To show that \(y = c\phi(t)\) is also a solution of the ODE, we substituted it into the ODE, found the derivative with respect to \(t\), and simplified the equation. We eventually obtained \(c(\phi^{\prime}(t)+p(t)\phi(t)) = 0\), which is true for any value of the constant \(c\). Thus, \(y=c\phi(t)\) is also a solution of the given ODE.

Step-by-step solution

Substitute the solution into the ODE

We are given that \(y=\phi(t)\) is a solution of the ODE. Let's now consider a new function \(y=c\phi(t)\). To show that this new function is also a solution of the ODE, we will substitute it into the given ODE and verify that it is satisfied: \begin{align*} y^{\prime}+p(t)y &= (c\phi(t))^{\prime}+p(t)(c\phi(t)). \end{align*}

Find the derivative

For the left side of the equation, we need to find the derivative of \(y\) with respect to \(t\). Since \(y=c\phi(t)\), we have: \begin{align*} y^{\prime} &= (c\phi(t))^{\prime} = c\phi^{\prime}(t). \end{align*}

Substitute derivative back into the ODE

Now we will substitute \(y^{\prime}=c\phi^{\prime}(t)\) back into the ODE: \begin{align*} y^{\prime}+p(t)y &= c\phi^{\prime}(t)+p(t)(c\phi(t)). \end{align*}

Simplify the equation

Next, we will simplify the equation by factoring out the constant \(c\): \begin{align*} c\phi^{\prime}(t)+p(t)(c\phi(t)) &= c(\phi^{\prime}(t)+p(t)\phi(t)). \end{align*}

Verify the solution

Recall that \(y=\phi(t)\) is a solution of the ODE \(y^{\prime}+p(t)y=0\). This means that \(\phi^{\prime}(t)+p(t)\phi(t) = 0\). Now we can substitute this into our simplified equation: \begin{align*} c(\phi^{\prime}(t)+p(t)\phi(t)) &= c(0) = 0. \end{align*} Since the ODE is satisfied for any value of the constant \(c\), we conclude that \(y=c\phi(t)\) is also a solution of the given ODE.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are mathematical equations that describe the relationship between a function and its derivatives. These equations play a critical role in various fields such as physics, engineering, and economics, offering insights into the behavior of dynamic systems over time.

Typically, an ODE is expressed in the form \(f(t, y, y^{\text{'}}) = 0\), where \(t\) is the independent variable, \(y\) is the dependent variable we wish to solve for, and \(y^{\text{'}}\) represents the derivative of \(y\) with respect to \(t\). The order of an ODE is determined by the highest derivative present, with first-order ODEs being the simplest.

For students to better understand ODEs, it is essential to grasp the notion of a derivative, which measures how a function changes as its input changes. The process of finding the function \(y\) that satisfies the ODE is known as solving the ODE, and the function \(y\) is referred to as the solution.

Solutions to ODE

Solutions to Ordinary Differential Equations (ODEs) are functions that, when substituted into an ODE, satisfy the equation completely. Finding these solutions is one of the primary goals when working with ODEs. They can be represented explicitly, where \(y\) is given as a function of \(t\), or implicitly, where the relationship between \(y\) and \(t\) is expressed without giving a direct formula for \(y\).

For a first-order ODE, the general solution often involves an integration process and may include an arbitrary constant, reflecting the fact that there may be an infinite family of solutions. The particular solution is obtained by specifying an initial condition or boundary condition that allows the arbitrary constant to be determined.

In our exercise, we see this concept illustrated: \(y=\phi(t)\) is a known solution, and it is shown that \(y=c\phi(t)\) is also a solution for any constant \(c\), exemplifying that the constants play a crucial role in defining the specific solution out of a family of potential solutions.

Homogeneous Equations

Homogeneous equations in the context of ODEs are a special type of differential equation where every term is dependent on the function \(y\) or its derivatives. When faced with the equation \(y^{\text{'}} + p(t)y = 0\), we're looking at a homogeneous, linear, first-order ODE, indicating that every term except for zero involves the variable function \(y\) or its derivative.

The term 'homogeneous' refers to the zero term on one side of the equation. It's important to observe that for homogeneous equations, if \(y=\phi(t)\) is a solution, then any multiple of that solution, \(y=c\phi(t)\) where \(c\) is a constant, is also a solution. This is because the homogeneous equation's structure allows us to factor out \(c\) as seen in the exercise example, resulting in the original solution which we know equates to zero.

The principle exploited here is the superposition principle, which is a characteristic of linear systems. Understanding this concept helps students recognize patterns in ODEs and build a more intuitive grasp of linear homogeneous equations and their solutions.