## anonymous one year ago Hello all, I'm having a hard time with this: y'-y/2x=xsin(x/y) ... it seems an homogeneus with z=x/y sostitutions but it don't works. Any idea?

1. anonymous

$y'-\frac{y}{2x}=x\sin\frac{x}{y}~~?$

2. anonymous

There's a nice general way to check if a given ODE is homogeneous. Given an ODE of the form $$y'=f(x,y)$$, the ODE is homogeneous if for some real $$\alpha$$, $$f(t x,t y)=t^\alpha f(x,y)$$. Let's check: $y'-\frac{y}{2x}=x\sin\frac{x}{y}~~\implies~~f(x,y)=\frac{y}{2x}+x\sin\frac{x}{y}$ So you have \begin{align*} f(tx,ty)&=\frac{ty}{2tx}+tx\sin\frac{tx}{ty}\\[1ex] &=\frac{y}{2x}+tx\sin\frac{x}{y}\\[1ex] &\neq t^\alpha f(x,y)&\text{for any }\alpha\in\mathbb{R} \end{align*} and so the substitution you mentioned would not work.

3. anonymous

Actually, your substitution isn't the one one usually uses for homogeneous equations. I think you meant $$y=zx$$ i.e. $$z=\dfrac{y}{x}$$?

4. anonymous

Thank you for your checking method! I've tryed both $y/x$ and that unhortodox $x/y$ for the substitution but nothing. Can we say we have not elementary solution for this ODE?

5. anonymous

I would say so (though that doesn't mean I'm not wrong). Like IrishBoy mentioned, the $$\sin$$ term is the main source of trouble. Had its argument been just $$x$$ instead of $$\dfrac{x}{y}$$, there would have been a chance. Numerical methods may be the best bet, but even then you would need an initial condition which you don't seem to have.

6. anonymous

No, I havent initial contition, this is the text "nude and crude" ... I'm suspecting there is a mistake in the text even if it is an official exam at my University. I will make some investigations again and let you know :-). Thank you by now for you time (And sorry for my non perfect English)

7. anonymous

It's sure: was a mistake in the text :-)