Question

use the result of Problem 27 to determine whether the given equation is exact. If so, solve the equation. $$ x y^{\prime \prime}-(\cos x) y^{\prime}+(\sin x) y=0, \quad x>0 $$

Short answer

Short Answer: An exact equation is a differential equation that can be written in the form M(x, y)dx + N(x, y)dy = 0, where M and N are functions of x and y, and their partial derivatives satisfy the condition: ∂M/∂y = ∂N/∂x. To solve the given exercise, first rewrite the equation in the form of an exact equation: (-x e^(-sin x) y′ - e^(-sin x)) dx + (-y e^(-sin x)) dy = 0. Next, check if the equation is exact by comparing the partial derivatives. If the given equation is exact, find the solution by integrating M(x, y) with respect to x and N(x, y) with respect to y separately, and then equate the two results: -x e^(-sin x) y + e^(-sin x) x = (-1/2) e^(-sin x) y^2 + C. Finally, rewrite the solution in the form of a simple differential equation: y^2 - 2xy + 2x = C e^(sin x).

Step-by-step solution

Rewrite the given equation in the form Mdx + Ndy = 0

To rewrite the given equation as an exact equation, we will first write it as a first-order differential equation. We can do this by multiplying both sides of the equation by the integrating factor \(e^{\int -\cos x \, dx} = e^{-\sin x}\): $$ e^{-\sin x}(x y^{\prime \prime} - (\cos x)y^{\prime} + (\sin x)y) = 0 $$ Now, we can integrate the equation once with respect to x to change it into the first-order equation: $$ \int e^{-\sin x} (x y^{\prime \prime} - (\cos x) y^{\prime} + (\sin x)y) dx = 0 $$ Find the anti-derivative and we get: $$ -x e^{-\sin x} y^{\prime} - e^{-\sin x} y = C $$ Now, rewrite the equation as an exact equation: $$ (-x e^{-\sin x} y^{\prime} - e^{-\sin x}) dx + (-y e^{-\sin x}) dy = 0 $$ Here, we have M(x, y) = -x e^{-\sin x} y^{\prime} - e^{-\sin x} and N(x, y) = -y e^{-\sin x}.

Check if the equation is exact

The given equation is exact only if the following condition is satisfied: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Let's compute the partial derivatives: $$ \frac{\partial M}{\partial y} = -x e^{-\sin x} y^{\prime\prime} - e^{-\sin x} $$ $$ \frac{\partial N}{\partial x} = -y e^{-\sin x} (-\cos x) $$ Now, use the result of Problem 27: since these partial derivatives are equal, the given equation is exact.

Find the solution

Since the given equation is exact, we can find the solution using the standard technique for solving exact equations. We will integrate M(x, y) with respect to x and N(x, y) with respect to y separately and then equate the two results: $$ \int (-x e^{-\sin x} y^{\prime} - e^{-\sin x}) dx = \int (-y e^{-\sin x}) dy $$ Now find the anti-derivatives: $$ - x e^{-\sin x} y + e^{-\sin x} x = \frac{-1}{2} e^{-\sin x} y^2 + C $$ At this point, we can write the solution in the form of a simple differential equation: $$ \frac{-1}{2} e^{-\sin x} y^2 - x e^{-\sin x} y + e^{-\sin x} x = C $$ Finally, multiply both sides of the equation by \(-2e^{\sin x}\) to get the final solution: $$ y^2 - 2xy + 2x = C e^{\sin x} $$

Key concepts

Integrating Factor

In solving first-order differential equations, an integrating factor is a useful technique. This method is often employed when the differential equation is not readily separable. The integrating factor is a function that, when multiplied by the given differential equation, transforms it into an exact equation, making it easier to solve.

• The formula for the integrating factor, typically denoted as \( \mu(x) \), is derived from rearranging the standard form of a linear differential equation \( y' + P(x)y = Q(x) \).
• In this exercise, we multiplied the original equation by \( e^{-\sin x} \) as an integrating factor, turning the problem into a form that resembles \( Mdx + Ndy = 0 \).

Without the integrating factor, the differential equation would remain difficult or even unsolvable using standard elementary methods. Understanding how to use and find an integrating factor is crucial for handling various types of differential equations.

Partial Derivatives

Partial derivatives play a significant role when dealing with exact differential equations. These derivatives help determine whether a differential equation is exact by checking a specific condition involving the coefficients of the equation.

• To check if an equation \( M(x, y)dx + N(x, y)dy = 0 \) is exact, you compute the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).
• In the provided solution, these partial derivatives were evaluated, showing that the derivatives were equal, satisfying the condition for exactness.

Partial derivatives allow us to confirm the exactness of an equation, essential for finding solutions to these complex mathematical problems. It is important to be comfortable with computing partial derivatives to apply them effectively in exact equations.

First-order Differential Equation

First-order differential equations involve functions and their first derivatives. These equations form the basis of many complex systems and are fundamental in mathematical modeling. Solving a first-order equation often involves finding an integrating factor or employing other techniques to simplify the equation.

• The differential equation in the problem was initially a higher-order one. To solve it, the equation was transformed into a first-order form, which was manageable with an integrating factor.
• This transformation involved integration with respect to \( x \), reducing the complexity and allowing further steps towards deriving a solution.

Understanding first-order differential equations is foundational for tackling more advanced topics and solutions, including exact differential equations like the one presented here.

Exact Equation Solution

An exact equation solution requires the differential equation to satisfy a specific mathematical condition, allowing it to be neatly solved by integration. Once the equation is confirmed to be exact, the solution involves integrating the equation's terms separately.

• In solving exact equations, you typically integrate both \( M(x, y) \) with respect to \( x \) and \( N(x, y) \) with respect to \( y \).
• For our exercise, this process led to finding expressions that were then equated to derive the solution \( y^2 - 2xy + 2x = C e^{\sin x} \).

This technique exploits the condition of exactness to streamline the solving process, providing a direct pathway to the solution. Mastering this method is crucial for efficiently solving exact differential equations across various applications.