Differential equation problems I need your help to finish my homework. Can somebody help me? 
I cannot finish all my homework especially in this problem.


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*$\dfrac{dy}{dx} = \dfrac{y^2 -1}{x}$; my answer is $-\dfrac{1}{y} -y = \ln x + x$, is it right?

*$\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$

*$\dfrac{dy}{dx} = \dfrac{2xy + 3y^2}{x^2 + 2xy}$


For 2. The equation can be formed as:$$\frac{dy}{dx}=\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(\frac{y}{x}\right)}$$
then if $x\neq0$ by taking $u=\frac{y}{x}$, we have $$\frac{dy}{dx}=\frac{\left(1+u^2\right)}{u}$$
but $u=\frac{y}{x}$ leads us to $xu=y$ and then $1+u'=y'$ where in $y'=\frac{dy}{dx}$. Now we have $$1+\frac{du}{dx}=\frac{\left(1+u^2\right)}{u}$$ This is a separable ode of first order. We have $$\frac{du}{dx}=\frac{\left(1+u^2\right)}{u}-1=\frac{1-u+u^2}{u}$$ so $$\frac{u\;du}{1-u+u^2}=dx$$ Now take an integral of both sides regarding to corresponding variable. We have $$\frac{1}{2}\ln(1-u+u^2)+\frac{\sqrt{3}}{3}\arctan\left(
\frac{\sqrt{3}}{3}(2u-1)\right)=x+C$$ put $u=\frac{y}{x}$.
