How to solve $y'^2 +yy'+x=0$? I encountered this ODE while looking for a curve which is orthogonal to the family of lines given by $$ y = mx + \frac{1}{m} \ \ \   m \in \Re  \ \ ...[1] \\ $$
I setup an ODE for [1] by puting $m = y'$ in [1],
$$ \ \ y = xy' + \frac{1}{y'}  \ \ ...[2]. \\ $$ Next, to get a family of orthogonal curves to [1], we change $y' \rightarrow - 1/y'$ in Eq. [2], we get
$$y'~^2 +yy'+x = 0 \ \ ...  [3] $$
The handbook by G.M.Murphy gives the solution of equation [3] as 
$$x = -t\left(\frac{C+\sinh^{-1}t}{\sqrt{1+t^2}}\right)~ \mbox{and}~ y=-t-\frac{x}{t}.$$
Can some one help me to get  to this solution. The interesting point is that even for a simple family of lines [1] the form of family of orthogonal curves is unfamiliar and involved.
This family of lines [1] may also cut or touch the required curve at some point other than that of normalcy.

Let $y'=p$, then the ODE can be written as 
$$p^2+yp+x=0,  \tag1$$ 
d.w.t. y we get 
$$2p\frac{dp}{dy}+y\frac{dp}{dy}+p+\frac{1}{p}=0 \Rightarrow (p^2+1)dy+(2p^2+py)dp=0  \tag2$$
It is of the type $M dy +N dp=0$, and it in-exact ODE. However, by the integrating factor $\mu=\int e^{h(p)} dp$, where $h(p) = \frac{2p-p}{1+p^2} \Rightarrow \mu(p)=\frac{1}{\sqrt{1+p^2}}$. Multiplying Eq. (2) by $\mu(p)$, we get the exact ODE as 
$$\sqrt{1+p^2}~ dy + \frac{2p^2+py}{\sqrt{1+p^2}}~ dp=0.   \tag3$$
it's solution is:
$$\int \sqrt{1+p^2}~ dy ~~(p-\mbox{const})~~ + \int \frac{2p^2}{\sqrt{1+p^1}}~ dp =C.   \tag4$$
We get
$$y=\frac{C+\sinh^{-1} p}{\sqrt{1+p^2}}-p ~\mbox{and using (1)}, x=- p\frac{C+\sinh^{-1}p}{\sqrt{1+p^2}},   \tag5$$ 
where $p$ merely acts like a real parameter and Eq. (5) defines a family of curves which are solution of (1).
