Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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?

Answer 5

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.

Answer 6

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)

Answer 7

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