Question

Solve the initial-value problems. Show that if \(\mu(x, y)\) and \(v(x, y)\) are integrating factors of $$ M(x, y) d x+N(x, y) d y=0 $$ such that \(\mu(x, y) / v(x, y)\) is not constant, then $$ \mu(x, y)=c v(x, y) $$ is a solution of Equation (A) for every constant \(c\).

Short answer

To show that if the ratio \(\frac{\mu(x, y)}{v(x, y)}\) is not constant, then \(\mu(x, y) = c v(x, y)\) is a solution of Equation (A) for every constant \(c\), we first multiply the given differential equation by integrating factors \(\mu(x, y)\) and \(v(x, y)\). Then, we consider the case when \(\mu(x, y) = c v(x, y)\) and show that it satisfies the differential equation for any constant \(c\). Thus, \(\mu(x, y) = c v(x, y)\) is a solution of Equation (A).

Step-by-step solution

Define the concepts of integrating factor and Equation (A)

An integrating factor is a function that, when multiplied by a differential equation, transforms it into an exact differential equation. Equation (A) refers to the equation that defines the condition for a function to be an integrating factor. For a given differential equation \(M(x, y) dx + N(x, y) dy = 0\), Equation (A) is given by: \[ \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = -Mf_y + Nf_x \] where \(f(x, y)\) is the integrating factor.

Rewrite the given differential equation using integrating factors

We are given that \(\mu(x, y)\) and \(v(x, y)\) are integrating factors of the differential equation \(M(x, y) dx + N(x, y) dy = 0\). Let's rewrite the given differential equation with integrating factors: Multiplying by \(\mu(x, y)\): \[ \mu(x, y) M(x, y) dx + \mu(x, y) N(x, y) dy = 0 \] Multiplying by \(v(x, y)\): \[ v(x, y) M(x, y) dx + v(x, y) N(x, y) dy = 0 \]

Show that if \(\frac{\mu(x, y)}{v(x, y)}\) is not constant, then \(\mu(x, y) = c v(x, y)\) is a solution, for every constant \(c\)

Suppose the ratio \(\frac{\mu(x, y)}{v(x, y)}\) is not constant. We want to show that \(\mu(x, y) = c v(x, y)\) is a solution of Equation (A) for every constant \(c\). The given condition implies that \(\frac{\mu M + \mu N}{v M + v N} \neq\) constant. Now, let's consider \(\mu(x, y) = c v(x, y)\), where \(c\) is a constant. Then, the differential equations in Step 2 become: \(c v(x, y) M(x, y) dx + c v(x, y) N(x, y) dy = 0\) Dividing both sides by \(c\) and equating the above equation with the equation obtained by multiplying by \(v(x, y)\): \(v(x, y) M(x, y) dx + v(x, y) N(x, y) dy = 0\) Since the left-hand sides of the equations are equal, we can say that \(\mu(x, y) = c v(x, y)\) satisfies the differential equation for any constant \(c\). Therefore, it is a solution of Equation (A).

Key concepts

Initial-Value Problems

In the context of differential equations, an initial-value problem (IVP) refers to finding a function that solves a differential equation and meets specific conditions, known as initial values, at a certain point. These problems are essential because they allow us to find unique solutions to differential equations that describe real-world phenomena.

To illustrate, consider a simple IVP where we have a differential equation along with an initial condition, such as at a point \( (x_0, y_0) \). The solution to this problem is not just any solution to the differential equation but the one that passes through the point \( (x_0, y_0) \). This specific requirement drives the uniqueness of the solution, making sure that the behavior of the dynamic system at that point is accurately captured.

Working with IVPs typically involves integrating a function and using initial conditions to determine the constants of integration. Through this process, one can assure that the solution reflects the initial state of the system, which is critical for accurately modeling physical situations, such as the motion of an object under certain forces.

Exact Differential Equations

Exact differential equations hold a special place in the study of ordinary differential equations (ODEs). An exact differential equation is one that can be written in the form \( M(x, y) dx + N(x, y) dy = 0 \) where \( M \) and \( N \) have continuous partial derivatives that satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

When an equation is exact, it suggests the existence of a function \( F(x, y) \) such that \( dF = M dx + N dy \) is its total differential. To solve such an equation, we look for this potential function \( F \) that simplifies the solution process, essentially reducing the problem to integrating partial derivatives.

However, not all differential equations are born exact, which leads us to the concept of integrating factors. Integrating factors are functions that, when multiplied with the differential equation, turn a non-exact equation into an exact one. This transformation is crucial because it opens the door to solving a broader range of differential equations by leveraging the concept of exactness.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) form a fundamental category of differential equations characterized by functions of a single variable and their derivatives. They are termed 'ordinary' to distinguish them from 'partial' differential equations, which involve partial derivatives with respect to multiple variables.

ODEs are used extensively to describe the dynamics of various systems, from simple motion to complex biological processes. The order of an ODE is given by the highest derivative present in the equation, with linear ODEs being a subset where the equation can be written as a linear combination of the function and its derivatives.

An important aspect of understanding ODEs is recognizing that solutions can be expressed as functions, which means we're often looking for an unknown function rather than a number. This brings an unequivocal richness and complexity to the study of ODEs, as it involves intricate methods of integration and a deep understanding of the behavior of functions and their derivatives.

From separation of variables to transformation through integrating factors like \(\mu(x, y)\) and \(v(x, y)\), solving ODEs can be an engaging challenge that requires both analytical and creative problem-solving skills.