Find solutions of a differential equation of a given form

Let $f(x)$ be a function such that
$2f'(x) = f \left( \frac{1}{x} \right) \qquad \text{for } x>0, \quad \text{and} \quad f(1) = 2.$

Let $y = f(x)$ and show that $y$ satisfies

$x^2 y'' + axy' + by = 0,$

for constants $a,b$ . Determine the values of the constants.
Find a solution $f(x) = Cx^n$ .

Incomplete.

April 20, 2019 at 12:03 pm

Evangelos says:

a.) From our givens, we have

$\begin{align*} y &= f(x) \\ y' &= f'(x) \\ &= \frac{1}{2}f(\frac{1}{x}) \\ y'' &= [\frac{1}{2}f(\frac{1}{x})]' \\ &= \frac{1}{2}f'(\frac{1}{x})(\frac{-1}{x^{2}}) \\ & = \frac{-1}{4x^{2}}f(x) \end{align*}$

We are to find constants a and b such that

$\begin{align*} x^{2}y'' + axy' + by &= 0 \end{align*}$

Written another way, simplifying terms:

$\begin{align*} -\frac{1}{4}f(x) + \frac{ax}{2}f(\frac{1}{x}) + bf(x) &= 0 \end{align*}$

We can satisfy this equation with the following constants

$\begin{align*} a = 0 \quad b = \frac{1}{4} \end{align*}$

b.) Now, to find a solution to the above second order equation of the form Cx^n

From part (a.) we found the values of constants a and b. We can plug in these values to give our second-order equation, and since the solution y is of the form Cx^n, we have

$\begin{align*} y &= Cx^n y' &= nCx^{n-1} y'' &= n(n-1)Cx^{n-2} \end{align*}$