Question

Solve the initial-value problems. Show that if the equation $$ M(x, y) d x+N(x, y) d y=0 $$ is homogeneous and \(M(x, y) x+N(x, y) y \neq 0\), then \(1 /[M(x, y) x+N(x, y) y]\) is an integrating factor of (A).

Short answer

Given the homogeneous equation \(M(x, y) dx + N(x, y) dy = 0\), we first verify its homogeneity by checking if \(M(tx, ty) = t^nM(x, y)\) and \(N(tx, ty) = t^nN(x, y)\) for some integer n. Then, we confirm the condition \(M(x, y) x + N(x, y) y \neq 0\). Finally, we prove that \(1 / [M(x, y) x + N(x, y) y]\) is an integrating factor by showing that the product of this factor and the original equation is an exact differential equation, i.e., \(\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}\), where F_x and F_y are the x and y components of the vector field F, respectively.

Step-by-step solution

Check Homogeneous Condition

We need to check the homogeneity of the equation first. To do that, we will plug tx and ty into M and N and compare it with t^nM(x, y) and t^nN(x, y): 1. \(M(tx, ty) = t^nM(x, y)\) 2. \(N(tx, ty) = t^nN(x, y)\) If the above two conditions are satisfied for some n, then the equation is homogeneous.

Verify Condition

Next, we must verify the given condition: \(M(x, y) x + N(x, y) y \neq 0\) This condition ensures that the denominator of the integrating factor always has a non-zero value.

Prove Integrating Factor

Now, let's show that \(1 / [M(x, y) x + N(x, y) y]\) is an integrating factor. To do this, we must prove that the product of the integrating factor and the original equation is an exact differential equation: \(dF = [1 / (M(x, y) x + N(x, y) y)](M(x, y) dx + N(x, y) dy)\) We need to show that: 1. \(\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}\) 2. \(\frac{\partial}{\partial x}(\frac{N(x,y)}{M(x, y) x + N(x, y) y}) = \frac{\partial}{\partial y}(\frac{M(x,y)}{M(x, y) x + N(x, y) y})\) where F_x and F_y are the x and y components of the vector field F, respectively. If we can show that the above partial derivative conditions are satisfied, then the integrating factor given by \(1 / [M(x, y) x + N(x, y) y]\) would indeed be an integrating factor for our homogeneous equation.

Key concepts

Initial-Value Problems

In the realm of differential equations, initial-value problems are essential for determining a specific solution amongst the infinite sea of possible solutions. These problems not only include the differential equation itself, but also an accompanying condition which states the exact value of the unknown function at a certain point, typically known as 'initial conditions'. For instance, when we're dealing with a first-order ordinary differential equation (ODE), the initial condition will give us the value of the dependent variable at a specific point. This enables us to not just solve the equation but to pin down the particular solution curve that passes through the given point.

Imagine launching a rocket and knowing its acceleration pattern but not knowing where to start—it's like having the differential equation without the initial condition. In providing an initial velocity and starting location, you set the trajectory. This is analogous to solving an initial-value problem: pinpointing the exact path out of myriad possibilities that satisfies both the mathematical rules of motion and your initial setup. This is crucial in real-world applications where specific solutions, rather than a family of solutions, are necessary for accurate predictions and designs.

Integrating Factor

An integrating factor is a mathematical lifesaver that can turn a confusing differential equation into a more manageable format. It's a function, often denoted as \(\mu(x)\), which, when multiplied by a non-exact differential equation, transforms it into an exact one. This means that once we identify the proper integrating factor, we can rewrite our troublesome differential equation to then easily find a potential function, often called \(F(x, y)\), whose differential is equal to our equation.

Think of an integrating factor as a pair of glasses that make blurry text clear. Applying this 'lens' allows us to view and solve our problem in a new light. To find this 'magical' factor, we usually look for a function that depends solely on \(x\) or solely on \(y\), and that, through multiplication, can make the mixed partial derivatives of our potential function coincide—which is what we need for an exact differential equation.

Exact Differential Equations

An exact differential equation is a particular type of equation that is music to mathematicians' ears because it signifies that there is a direct path to the solution. To identify one, we rely on a condition involving partial derivatives, which dictate that if we have an equation in the form \(M(x, y) dx + N(x, y) dy = 0\), we need the partial derivative of \(M\) with respect to \(y\) to be identical to the partial derivative of \(N\) with respect to \(x\). When this condition is met, it tells us that there exists a function \(F(x,y)\) whose gradient vector coincides with the vector field defined by \(M\) and \(N\), and the path to solving begins by finding this function.

To solve an exact differential equation, the strategy lies in recovering this elusive function \(F(x,y)\), which is tactfully hidden and encapsulated by the differential equation itself. Imagine you're given a description of a landscape — the exact differential equation is like having a map that tells you precisely how to navigate through it. By piecing together these descriptions (integrating them), you can uncover the landscape's topography—comparable to finding the potential function \(F\).

Partial Derivatives

In the exploration of multi-variable functions, partial derivatives are indispensable tools, representing how a function changes as you tweak just one of many variables, keeping the others constant. It’s like examining the impact of changing water temperature on the brewing of tea, disregarding all other variables such as steeping time or the type of tea. Partial derivatives provide us with a microscopic view of the function's behavior along each axis.

Applying this concept in the context of differential equations, partial derivatives are critical in the discussion of exact differential equations because they are part and parcel of satisfying the exactness criterion. When a differential equation is multiplied by an integrating factor, we’re looking at the partial derivatives to align nicely—this is how we prove that our transformed equation is indeed exact. Without aligning partial derivatives, we cannot obtain the function \(F(x,y)\) whose gradient constitutes our differential equation and, as a consequence, cannot proceed to find the solution of the initial-value problem with confidence and ease.