Differential equation $(1-x)(y')^2+(y-x)y'+y=0.$

Question

I have the following problem.

Find all differentiable on $[0,1)$ non-negative functions $f$, for which for any $u\in(0,1)$ the tangent line to graph of $f$ in the point $(u,f(u))$ intersects the lines $y=0$ and $x=1$ in the points $A$ and $B$ respectively and $x_{A}=y_{B}$. Also, $f'(0)=0$.

Easy to show that our function it's a root of the equation $$y=xy'-\frac{y'^2}{1+y'}$$ or $$(1-x)(y')^2+(y-x)y'+y=0.$$ My question: is there some way to solve this equation directly?

Also, I assumed that the graph of $f$ is parabola with focus $F\left(\frac{1}{2},\frac{1}{2}\right)$ and directrix $y=x-1,$ which gives an equation of the parabola: $$(x+y)^2=4y,$$ which gives a solution of our equation.

My second question: is there some better way to get this parabola?

Thank you!

Answer 1

"Is there some better way to get this ellipse?" I don't know if the following methods are better or not, anyway here they are.

The focus of a parabola lies on the circumcircle of the triangle formed by any three of its tangents. The circumcircles of the right triangles formed by lines $y=0$, $x=1$ and $AB$ all pass through point $F=(1/2,1/2)$ (see here for more details), hence line $AB$ is tangent to the parabola with focus $F$ and also tangent to $y=0$, $x=1$. By symmetry, the axis of that parabola must be the line passing through $F$ and $T=(1,0)$ and its directrix passes through $T$ (because perpendicular tangents of a parabola meet on the directrix). This proves that an arc of that parabola is a solution of your equation.

Another method: the curve you are looking for is the envelope of the lines passing through $A=(t,0)$ and $B=(1,t)$, whose equations are $$ y(1-t)=tx-t^2. $$ We can then follow the usual procedure to find the equation of the envelope: differentiate the above equation with respect to $t$, solve the resulting equation for $t$ and substitute back the result into the above equation: the final result is exactly the cartesian equation of the parabola found above.

Answer 2

$$(1-x)(y')^2+(y-x)y'+y=0.$$ $$y=xy'-\frac{y'^2}{1+y'}$$ This is Clairaut's differential equation. $$y=xy'+f(y')$$