How to solve $y'^2 +yy'+x=0$?

Question

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.

Answer 1

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).

Answer 2

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 $$\int\frac{d}{dv}\left(\frac{\sqrt{v^2+1}x(v)}{v}\right)dv=\int-\frac{v\sqrt{v^2+1}}{v^3+v}dv$$