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).