please give me a hint how to start this@
x(dy/dx) = y + 2xsin^2(y/x)
basically i dont know what the first step is because the way i learnt this, i can't see how to apply that method to this example?
$\displaystyle \displaystyle \frac{-1}{2\tan\frac{y}{x}} = \ln x + c$
$\displaystyle \displaystyle \frac{-1}{\ln x+ c} = 2\tan\frac{y}{x}$
$\displaystyle \displaystyle \frac{-1}{2\ln x+ c} = \tan\frac{y}{x}$
$\displaystyle \displaystyle \tan^{-1}\left(\frac{-1}{2\ln x+ c}\right) = \frac{y}{x}$
$\displaystyle \displaystyle x\tan^{-1}\left(\frac{-1}{2\ln x+ c}\right) = y$
Different people learn this method different ways. I learned it this way:
$\displaystyle \displaystyle x\,\frac{dy}{dx} = y + 2x\sin^{2}\!\left(\dfrac{y}{x}\right)$
$\displaystyle \displaystyle x\,dy-\left(y+2x\sin^{2}\!\left(\dfrac{y}{x}\right)\righ t)dx=0.$
If we follow your substitution of $\displaystyle z=y/x,$ then $\displaystyle y=zx,$ and $\displaystyle dy=z\,dx+x\,dz.$ (This is how I "find" $\displaystyle dy/dx$. If you wish, you can write $\displaystyle dy/dx=z+x(dz/dx)$.) Plugging into the DE yields
$\displaystyle x(z\,dx+x\,dz)-(zx+2x\sin^{2}(z))\,dx=0$
$\displaystyle x^{2}\,dz-2x\sin^{2}(z)\,dx=0$
$\displaystyle x\,dz-2\sin^{2}(z)\,dx=0.$ Separating out yields
$\displaystyle \csc^{2}(z)\,dz-\dfrac{2\,dx}{x}=0$
$\displaystyle -\cot(z)-2\ln|x|=C$
$\displaystyle \cot(z)+2\ln|x|=C$ (absorbing the minus sign into the constant).
$\displaystyle \cot\left(\dfrac{y}{x}\right)=-2\ln|x|+C.$
At this point, as with all solutions to any differential equation, you should verify that this implicitly defined function satisfies the original DE. The reason to differentiate now is that if you do, the constant will go away, which is what you want when you're verifying the solution to a DE. The solution I've got here is, incidentally, the same as yours, I believe. So far, so good. If you do want to isolate $\displaystyle y,$ then take the $\displaystyle \cot^{-1}$ of both sides, and then multiply the resulting equation by $\displaystyle x$ to obtain
$\displaystyle y=x\cot^{-1}\left(-2\ln|x|+C\right).$
On what interval(s) is this solution valid?