Differential Equations and Linear Algebra (4th Edition)

$y = \frac{\ln(x)}{x}$
$e^{xy} + x = c$ Implicit differentiation: differentiating the equation in terms of $x$: $e^{xy}(x\frac{dy}{dx} + y) - 1 = 0$ $e^{xy}(x\frac{dy}{dx} + y) = 1$ Rearranging the equation, we get: $\frac{1 - ye^{xy}}{xe^{xy}}$ = 1 Since $y(1) = 0$, therefore $c = 0$. As such, $e^{xy} - x = 0$ $e^{xy} = x$ $xy = \ln(x)$ $y = \frac{\ln(x)}{x}$