Solution to a wierd differential equation I solved this problem: find the area of the triangle formed by the $x$ and $y$ axis and the tangent line to the curve of $f(x) = \frac 1x$ at any point $(x_0,f(x_0))$.
I found out that the area was equal to $2$ for every point.
I then wondered what kind of function can have that particular area being independent of the point we choose.
And I saw that for the area to be equal to a constant $c$, the function $y(x) $must verify this differential equation:
$$y'x^2 + \frac 1{y'}y^2 - 2xy = 2c$$
Taking $C = 2$ Area gives $C = 2c$ to make it simpler.
But I can't find the solutions. I really need a help.
 A: We first correct what appears to be a sign mistake in the question, then we show the only curves in the first quadrant that yield triangles of constant area $a$, are the lines

$$y(x)=-c^2x+c\sqrt{2a}$$ 

for every $c>0$, and the hyperbola

$$y(x)=\frac ax$$


So, one is considering curves of equation $$y=u(x)$$ in the first quadrant, with $u$ decreasing. The tangent line at $(x_0,y_0)=(x_0,u(x_0))$ has equation $$y=u'(x_0)(x-x_0)+y_0$$ hence it meets the $x$-axis at the point of abscissa $$x=x_0-\frac{y_0}{u'(x_0)}$$ and it meets the $y$-axis at the point of ordinate $$y=-u'(x_0)x_0+y_0$$ The area $a$ of the triangle formed by the tangent line and the two axes is half the product of this abscissa and this ordinate, that is, $$a=\frac12\left(x_0-\frac{y_0}{u'(x_0)}\right)(-u'(x_0)x_0+y_0)$$ Reordering, one sees that the solutions correspond to functions $u$ such that

$$(xu'(x)-u(x))^2=-2au'(x)\tag{$\ast$}$$


Now, to the resolution of this differential equation $(\ast)$. First, let us note that $(\ast)$ involves a second degree polynomial in $u'(x)$, namely, $(\ast)$ is equivalent to $$x^2u'(x)^2+2(a-xu(x))u'(x)+u(x)^2=0$$
The discriminant of this polynomial should be nonnegative for a solution $u'(x)$ to exist, thus one can choose $v(x)\geqslant0$ such that $$v^2(x)=(a-xu(x))^2-x^2u(x)^2=a^2-2axu(x)$$ then the formula for the roots of a quadratics with nonnegative discriminant shows that $$x^2u'(x)=xu(x)-a\pm v(x)$$ One can express $u(x)$ and $u'(x)$ in terms of $v(x)$ and $v'(x)$, since $$2axu(x)=a^2-v^2(x)\qquad x^2u'(x)=xu(x)-a\pm v(x)$$ But now, 
$$v(x)v'(x)=-a(xu(x))'=-axu'(x)-au(x)$$ hence $$2xv(x)v'(x)=-2a(x^2u'(x))-(2axu(x))=-(a^2-v^2(x)-2a^2\pm2av(x))-(a^2-v^2(x))$$ that is,
$$2xv(x)v'(x)=2v^2(x)\mp2av(x)$$ and finally,
$$v(x)\cdot(xv'(x)-v(x)\pm a)=0$$ The solution $v(x)=0$ yields $$2xu(x)=a$$ which describes the hyperbola $$y=\frac a{2x}$$ The solution $xv'(x)-v(x)\pm a=0$ yields $$|v(x)\mp a|=bx$$ for some nonnegative $b$. Since $v(x)$ is always between $0$ and $a$, this yields $v(x)=a-bx$, thus, $v^2(x)=a^2-2abx+b^2x^2$, that is, $$u(x)=b-\frac{b^2}{2a}x$$ This last equation indeed describes the lines given as solutions at the beginning of this post.
