Question

Show that 𝑦 = ln (𝑥) is a solution of 𝑥𝑦′′ + 𝑦′ = 0 

Answer 1

$xy" + y' = 0\\ so: y = ln(x) y' = \frac{1}{x y"} = \frac{-1}{x^2}$

Plug this into the differential equation:

This shows that there is a solution, but not necessarily on what interval.

Now to show whic interval this is valid in, we need to use

the existence and uniqueness theorem.

Given a differential equation: $y" + p(x)y' = q(x) y(x0) = y_0$

A unique solution is guaranteed where p(x)0 and q(x) are continuous.

Put this into proper form by dividing through by x to get:

Thus:

These functions are both continuous on :