91Ó°ÊÓ

Rewrite the given differential equation as a first-order differential equation, dividing both sides by \(dx\): \[\frac{dy}{dx} + \frac{y}{x} = 1\]

The equation is in the form \(M(x, y) + N(x, y)\frac{dy}{dx} = 0\). Identify \(M(x, y)\) and \(N(x, y)\): \[M(x, y) = x + y\] \[N(x, y) = -x\] Now, find the partial derivatives of \(M\) and \(N\) with respect to \(y\) and \(x\), respectively: \[\frac{\partial M}{\partial y} = 1\] \[\frac{\partial N}{\partial x} = -1\] Since the partial derivatives are not equal, the equation is not exact.

To make the differential equation exact, let's find an integrating factor \(\mu(x)\). Divide both sides by \(x\): \[\frac{1}{x}\left(1 + y\right) + \left(-1\right)\frac{dy}{dx} = 0\] Now, we can use the formula for integrating factors: \[\mu(x) = e^{\int \frac{N_x - M_y}{M} dx}\] Substitute the values: \[\mu(x) = e^{\int \frac{-1 - 1}{x + y} dx}\] Simplify the equation: \[\mu(x) = e^{\int \frac{-2}{x + y} dx}\]

Multiply both sides of the equation by the integrating factor \(\mu(x)\): \[\left(e^{\int \frac{-2}{x + y} dx}\right)\left(\frac{1}{x}\left(1 + y\right) + \left(-1\right)\frac{dy}{dx}\right) = 0\]

Define new functions \(\tilde{M}(x, y)\) and \(\tilde{N}(x, y)\): \[\tilde{M}(x, y) = \mu(x) M(x, y)\] \[\tilde{N}(x, y) = \mu(x) N(x, y)\] Compute the partial derivatives: \[\frac{\partial \tilde{M}}{\partial y} = e^{\int \frac{-2}{x + y} dx}\] \[\frac{\partial \tilde{N}}{\partial x} = e^{\int \frac{-2}{x + y} dx}\] Since the partial derivatives of the new functions are equal, the new equation is exact.

The potential function \(F(x, y)\) satisfies the following relations: \[\frac{\partial F}{\partial x} = \tilde{M}(x, y)\] \[\frac{\partial F}{\partial y} = \tilde{N}(x, y)\] Integrate \(\frac{\partial F}{\partial x}=\tilde{M}(x, y)\) with respect to \(x\): \[F(x, y) = \int \tilde{M}(x, y) dx\] Integral of \(\tilde{M}(x, y)\) with respect to \(x\): \[F(x, y) = xe^{\int \frac{-2}{x + y} dx} + g(y)\] Now differentiate \(F(x, y)\) with respect to \(y\) and set it equal to \(\tilde{N}(x, y)\): \[\frac{\partial F}{\partial y} = \tilde{N}(x, y)\] \[g'(y) - 2x e^{\int \frac{-2}{x + y} dx} = -x e^{\int \frac{-2}{x + y} dx}\] \[g'(y) = xe^{\int \frac{-2}{x + y} dx}\] Integrate \(g'(y)\) with respect to \(y\): \[g(y) = \int xe^{\int \frac{-2}{x + y} dx} dy\] As the integral is devoid of a closed-form solution, we leave the answer in the form of an implicit solution: \[F(x, y) = xe^{\int \frac{-2}{x + y} dx} + \int xe^{\int \frac{-2}{x + y} dx} dy = C\] Where \(C\) is the constant of integration.