Problem 18 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$\left(y e^{x y}+\cos x\right) d x+x e^{x y} d y=0, y(\pi / 2)=0$$

Step-by-Step Solution

Answer

The short answer for the given initial-value problem is: The particular solution is \(e^{x^2/2}(ye^x+\sin x) = 1\).

Step 1: Check for exactness

First, let's determine if the given differential equation is exact. An equation of the form \(M(x, y)dx + N(x, y)dy = 0\) is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Here, \(M(x, y) = ye^{xy} + \cos x\) and \(N(x, y) = x e^{xy}\). Compute the partial derivatives: \(\frac{\partial M}{\partial y} = xe^{xy} \) \(\frac{\partial N}{\partial x} = e^{xy} + xy e^{xy}\) Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the given equation is not exact as it is.

Step 2: Solve for potential functions

As the given equation is not exact, we won't be able to continue solving for potential functions directly. Instead, we need to rewrite it into an exact form. Divide both sides by \(xe^{xy}\), which gives: \(1 + \frac{\cos x}{xe^{xy}}dx + dy = 0\) Now, let the new equation as \(M'(x, y)dx + N'(x, y)dy = 0\), where \(M'(x, y) = 1 + \frac{\cos x}{xe^{xy}}\) and \(N'(x, y) = 1\). Calculate the partial derivatives: \(\frac{\partial M'}{\partial y} = -\frac{ye^{xy}\cos x}{(xe^{xy})^2} \) \(\frac{\partial N'}{\partial x} = 0\) The given equation is still not in its exact form, but since there's no straightforward way to make it exact, we move to an alternate method called integrating factors. We will find an integrating factor \(\mu(x)\) such that the product of \(\mu(x)\) and the differential equation becomes exact. This requires: \(\frac{d\mu}{dx} = \frac{\mu}{N}\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\) Using the obtained values of partial derivatives and plugging them into the equation: \(\frac{d\mu}{dx} = \frac{\mu}{x}\cdot xe^{xy} - 0\) Upon integrating, we get: \(\mu(x) = e^{x^2/2}\)

Step 3: Find the particular solution

Now, multiply the differential equation by the integrating factor \(\mu(x)\): \((ye^{xy}e^{x^2/2}+e^{x^2/2}\cos x)dx + x e^{xy}e^{x^2/2}dy = 0\) Now, we see that the equation is exact: \(\frac{\partial}{\partial y}(ye^{xy}e^{x^2/2}) = e^{xy}e^{x^2/2} + xye^{xy}e^{x^2/2} \) \(\frac{\partial}{\partial x}(xe^{xy}e^{x^2/2}) = e^{xy}e^{x^2/2} + xye^{xy}e^{x^2/2}\) Integrate \(M\) with respect to \(x\) to find the potential function \(F(x, y)\): \(F(x, y) = \int ye^{xy}e^{x^2/2} + e^{x^2/2}\cos x\,dx\) \(F(x, y) = e^{x^2/2}(ye^x + \sin x) + g(y)\) Next, integrate \(N\) with respect to \(y\): \(F(x, y) = \int xe^{xy}e^{x^2/2}\,dy\) \(F(x, y) = e^{x^2/2}(y e^x) + h(x)\) Compare the two expressions of \(F(x, y)\) and note that \(g(y) = 0\) and \( h(x) = e^{x^2/2}\sin x\). Thus, \(F(x, y) = e^{x^2/2}(ye^x + \sin x)\) Using the initial-value condition, substitute \(x = \pi/2\) and \(y = 0\): \(F(\pi/2, 0) = e^{(\pi/2)^2/2}(\sin(\pi/2))\) \(F(\pi/2, 0) = 1\) Finally, the particular solution is: \(e^{x^2/2}(ye^x+\sin x) = 1\)

Key Concepts

Integrating Factors

Sometimes, when dealing with differential equations, we come across equations that are not exact, meaning their partial derivatives do not match up correctly. To handle these, we use a technique called integrating factors. An integrating factor is a function, often denoted as \( \mu(x) \), that we multiply the entire differential equation by to turn it into an exact equation. This means it will then satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

Finding the right integrating factor can be thought of as tuning the equation to listen to one particular frequency that makes it solvable. In our context, the integrating factor \( \mu(x) = e^{x^2/2} \) was found by addressing the partially derived functions. The task is to solve for \( \mu(x) \) using the formula:

• \( \frac{d\mu}{dx} = \frac{\mu}{N}\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \)

Once it's applied, the differential equation becomes more straightforward to work with.

Partial Derivatives

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of multiple variables. They represent how a function changes as each variable changes independently.

In our differential equation form, we typically have \( M(x, y)dx + N(x, y)dy = 0 \). To check if this equation is exact, we need to compare the partial derivatives:

• \( \frac{\partial M}{\partial y} \)
• \( \frac{\partial N}{\partial x} \)

If these two derivatives are equal, the differential equation is considered exact. Otherwise, as in our exercise, further steps are needed, such as finding an integrating factor. It is crucial to calculate these derivatives accurately to determine the nature of the equation.

Initial-Value Problem

An initial-value problem involves finding a function that satisfies a differential equation and also meets specified conditions at a particular point. These conditions help uniquely determine the solution from potentially many possible solutions.

In the exercise we have:

• The differential equation \( (y e^{xy} + \cos x) dx + (x e^{xy}) dy = 0 \) and
• An initial condition \( y(\pi/2) = 0 \).

The initial condition \( y(\pi/2) = 0 \) is crucial in pinning down the exact solution after solving the differential equation. This requirement ensures that the function \( y(x) \) goes through the point \( (\pi/2, 0) \), which we then use to find constants when integrating. It gives us the particular solution that satisfies both the differential equation and this specified condition.

Potential Function

A potential function, in the context of differential equations, is essentially the solution we arrive at. When an equation is exact, there's a potential function \( F(x, y) \) such that \( \frac{\partial F}{\partial x} = M \) and \( \frac{\partial F}{\partial y} = N \). This potential function often holds the key to finding the solution to the differential equation.

In our specific problem, after using the integrating factor, we integrated both \( M \) and \( N \) to find:

• \( F(x, y) = e^{x^2/2}(ye^x + \sin x) \)

Here, this potential function is what describes the relationship between \( x \) and \( y \) that satisfies our equation. When combined with the initial condition, it provides us the complete particular solution. Integrating the derivatives gives us clues about the form of this potential function, leading to the exact relation that defines the function \( y(x) \).