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.

Writing your equation in the form
$$y(x)=-y'(x)-\frac{x}{y'(x)}$$ differentiating this equation with respect to $x$
$$\frac{d}{dx}(y'(x))=\frac{y'(x)^3+y'(x)}{x-y'(x)}$$
substituting
$$v(x)=y'(x)$$
$$x'(v)=-\frac{v^2}{v^3+v}+\frac{x}{v^3+v}$$
Calculating
$$\mu(x)=\int e^{1/(v^3+v)}dx=\frac{\sqrt{v^2+1}}{v}$$ so we get
$$\frac{\sqrt{v^2+v}}{v}v'(x)-\frac{\sqrt{v^2+1}x(v)}{v(v^3+v)}=-\frac{v\sqrt{v^2+1}}{v^3+v}$$ and now integrate
$$\int\frac{d}{dv}\left(\frac{\sqrt{v^2+1}x(v)}{v}\right)dv=\int-\frac{v\sqrt{v^2+1}}{v^3+v}dv$$
