$\begingroup$

I am stuck trying to solve for the below ODE,

$$ \dfrac{d y}{dx}=\dfrac{y}{x}+1 $$

it would be trivial to solve if it did not have the one at the end since I could use separation of variables. I tried to use a change of variables $ y = \xi -x$ but that did not get me anywhere. Does a simple solution to this ODE exist?

$\endgroup$ 1

2 Answers

$\begingroup$

HINT:

Set $\displaystyle\frac yx=v$

$\displaystyle\implies y=vx\implies\frac{dy}{dx}=v+x\frac{dv}{dx}$

Reference : Homogeneous Ordinary Differential Equation

$\endgroup$ 3 $\begingroup$

$$ \dfrac{d y}{dx}=\dfrac{y}{x}+1\quad\Longrightarrow\quad \frac{1}{x}\dfrac{d y}{dx}-\frac{1}{x^2}y=\frac{1}{x}\quad\Longrightarrow\quad \left(\frac{y}{x}\right)'-(\ln x)'=0\quad\Longrightarrow\quad \frac{y}{x}=\ln x+c, $$ for some $c$ constant.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy