How to find all solutions of $y'=c-y/x$?

Question

How to show that every solution of the ODE $f'(x)=c-f(x)/x$ is of the form

$f(x) = cx/2 + k_1/x$ for some $k_1 \in \mathbb{R}$?

(Let's say that we limit the domain of $f$ to $x>0$ so $1/x$ is a valid solution).

You might find this Wikipedia page helpful. – K.defaoite May 04 '20 at 10:32 — K.defaoite, May 04 '20 at 10:32

score 2 · Accepted Answer · answered May 04 '20 at 07:49

2

$$y' + \frac{1}{x}y = c$$

Multiplying both sides by $x$

$$xy' + y = cx$$ $$(xy)' = cx$$ $$xy = c\frac{x^2}{2} + d$$ $$y = \frac{cx}{2} + \frac{d}{x}$$

All assuming $x\gt 0$

answered May 04 '20 at 07:49

score -1 · Answer 2 · answered May 04 '20 at 07:51

-1

Let $z(x)=\frac{y(x)}{x}, $ then $c-z=y'=z+xz'.$

Can you proceed ?

answered May 04 '20 at 07:51

Fred

Hi. I am not sure how did you mean to proceed from here. Can you please elaborate? (I did not downvote). – Asaf Shachar May 04 '20 at 08:03
The equation $xz'=c-2z$ is a linear, first order differential equation. – Fred May 04 '20 at 08:04

2 Answers2