Question

use the result of Problem 27 to determine whether the given equation is exact. If so, solve the equation. $$ x^{2} y^{\prime \prime}+x y^{\prime}-y=0, \quad x>0 $$

Short answer

Given equation: \(x^2y'' + xy' - y = 0\) Answer: Yes, the given equation is exact. The solution is any function, y(x).

Step-by-step solution

Check if the equation is exact

To check if the given equation is exact, we need to find a function that satisfies the following conditions: $$ \frac{\partial F}{\partial x} = xy', \quad \frac{\partial F}{\partial y} = x^2y'' $$ Since \(y'\) appears only in the first condition, we can integrate the first equation with respect to \(x\): $$ F(x, y) = \int xy' \, dx $$ Let's call the integration constant \(G(y)\) since it depends only on \(y\). So, we have: $$ F(x, y) = \frac{1}{2}x^2y' + G(y) $$ Now, we can differentiate \(F(x, y)\) with respect to \(y\): $$ \frac{\partial F}{\partial y} = x^2y' + G'(y) $$ We want this to be equal to \(x^2y''\). Comparing the two equations, we have: $$ x^2y' = x^2y'', \quad G'(y) = -y $$ Integrating the G'(y) equation with respect to \(y\), we obtain the function G(y): $$ G(y) = -\frac{1}{2}y^2 + C $$ So, the function F(x, y) is given by: $$ F(x, y) = \frac{1}{2}x^2y' - \frac{1}{2}y^2 + C $$ As we found a function F(x, y) that satisfies both conditions, the given equation is exact.

Solve the exact equation

From Step 1, we have \(F(x, y) = \frac{1}{2}x^2y' - \frac{1}{2}y^2 + C = 0\). Rearranging the terms, we get: $$ y' = \frac{y}{x} $$ This is a first order linear differential equation on y. To solve this equation, we can first identify the integrating factor: $$ IF = e^{\int \frac{1}{x} \, dx} = e^{\ln{x}} = x $$ Now, we multiply the whole equation by the integrating factor and use product rule: $$ x y' + \frac{1}{x} xy = x({\frac{y}{x}}) $$ Integrating both sides of the equation with respect to \(x\), we get: $$ (xy)' = \int x\cdot \frac{y}{x} \, dx \Rightarrow xy = \int y \, dx \Rightarrow xy=y(x) \int dx $$ This states that there is a function \(y(x)=y\) whose derivative multiplied by \(x\) is equal to y. Given this, we can find out that the original equation is an identity and will be true for all functions y(x). So the solution for the given equation is any function y(x).

Key concepts

Exactness Condition

When dealing with differential equations, the concept of exactness is crucial. It mainly revolves around determining whether a given differential equation can be expressed as the derivative of some function.
An equation of the form \(M(x,y) + N(x,y)\frac{dy}{dx} = 0\) is said to be exact if there exists a function \(F(x, y)\) such that:

• \(\frac{\partial F}{\partial x} = M(x, y)\)
• \(\frac{\partial F}{\partial y} = N(x, y)\).

The system is similar to a gradient of some potential function, which changes the equation into an exact differential.
In the given exercise, we expressed the equation as a system and managed to find such a function \(F(x, y)\), proving exactness and subsequently finding a simplified solution. Understanding exactness is essential for breaking down more complex differential equations into solvable parts.

First Order Linear Differential Equation

First order linear differential equations are among the simplest forms of differential equations that mathematics students encounter. They take the form \(\frac{dy}{dx} + P(x)y = Q(x)\).
In the original problem, once the equation \(x^2y'' + xy' - y = 0\) was reduced to \( y' = \frac{y}{x} \), it became a first order linear differential equation.
Solving such equations often involves utilizing an integrating factor or directly integrating if it's already separable. Recognizing that it can be solved by integrating directly is part of mastering the technique for these types of equations.
First order linear differential equations are powerful tools in modeling exponential growth, decay, cooling, and more.

Integrating Factor

An integrating factor is a special function used to simplify solving first order linear differential equations. This method aims to transform the equation into an easily integrable form. To find the integrating factor, \(IF\), for equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), we use:

• \(IF = e^{\int P(x) \, dx}\).

Multiplying the entire differential equation by this integrating factor allows the left side to become a derivative of the product \(IF \cdot y\).
In the given solution, we identified \(IF = e^{\ln{x}} = x\) to multiply throughout, turning the derivative \((xy)'\) into a much simpler form to integrate. By making the left-hand side an exact derivative, the solution becomes straightforward.
Mastery of applying integrating factors is invaluable in handling a wide array of differential equations.