Find family of curves for which lenght of the line segment between origin and the point in which line normal to the curve intersects with X axis is equal to $\frac{y^2}x$
The equation should be $yy'+x=\frac{y^2}x$ but I have no idea where did this come from.
$\endgroup$ 42 Answers
$\begingroup$One little picture says more than a long speech !
The ODE leads to $y(x)=x\sqrt{c-2*\ln(x)}$ or $y(x)=-x\sqrt{c-2*\ln(x)}$
The curve drawn on the figure is symbolic : it is not supposed to be a true representation of the function.
$\endgroup$ $\begingroup$Your approach is right when picking an arbitrary point $(X,Y)$ on the curve and writing down the line perpendicular to the tangent in that point: $$y-Y=\frac{-1}{y'(X)}(x-X).$$ Now the assumptions is that $(\frac{Y^2}{X},0)$ lies on this line and plugging this into the eqaution yields $$-Y=\frac{-1}{y'(X)}(\frac{Y^2}{X}-X)$$ which is equivalent $$Yy'(X)+X=\frac{Y^2}{X}.$$ Since $(X,Y)$ was arbitrary this gives you your differential equation. Now how to solve the differential equation? Note that the equation is equivalent to $$\frac{dy}{dx}=\frac{y^2-x^2}{xy}$$ and this is a homogeneous differential equation. Put $y=ux$ and use the product rule : $$\frac{dy}{dx}=u+x\frac{du}{dx}.$$ This gives you a separable differential equation that can easily be solved. Can you take it from here?
$\endgroup$Solution: $y^2=cx^2-x^2ln(x^2)$, with $c \in \mathbb{R}$
