Answer to Question #231864 in Differential Equations for Meep

Question

show that y=ln (x) is a solution of xy" + y'=0.

Answer 1

$y=\ln x$ , find

$y'=\frac{1}{x}, y''=-\frac{1}{x^2}$

Substitute in equation

$x\cdot (-\frac{1}{x^2})+\frac{1}{x}=-\frac{1}{x}+\frac{1}{x}=0$

so $y=\ln x$ is a solution of $xy''+y'=0$.