Need Help with Heterogeneous Differential Equation

[Kvad]

Hi Everyone!

I am really stuck at one differential equation:

$(p-y)^{2}y'(x)+(x-y)(x+y-2p)=0$

where $p<x<1; \ 0<y<p; \ p - parameter, 0<p<1$

I have a suspicion that it does not have a solution, but is there a way to prove it mathematically in this case? I would appreciate any hints on tackling this equation. Thank you!

 

[JJacquelin]

Hi !

The solutions of the ODE cannot be expressed on the form y(x), but on a parametric form :

 

[Kvad]

Thanks, JJacquelin!

I am not strong in differential mathematics and I am not sure what parametric solution stands for, so sorry for a naive question, but is it possible to somehow plot the parametric form solution or at least verify that if it could have been expressed on a y(x) form its slope would be downward?

 

[JJacquelin]

at least verify that if it could have been expressed on a y(x) form its slope would be downward?

There is no need to solve the equation and no need to know explicitly y(x) to answer the question. Just show that y'(x) < 0 from the equation.

 

[Kvad]

There is no need to solve the equation and no need to know explicitly y(x) to answer the question. Just show that y'(x) < 0 from the equation.

Do you mean the initial equation, where we are given that basically y<p<x and therefore the slope is evidently negative, or do we know for sure the signs of $X(\theta) \ \ and \ \ \theta$? In this case from the parametric solution I believe it would be a more formal proof...

 

[JJacquelin]

Do you mean the initial equation, where we are given that basically y<p<x and therefore the slope is evidently negative..

Yes, only the initial equation with all the specified conditions. Forget the analytic resolution.

 

[Kvad]

Ok, thanks again! One last naive question, could you, please, tell me if the analytical solution also shows the downward slope of the function, because I am not sure I completely understand it...

 

[JJacquelin]

Forget the analytical solution. It should be a thousand times more complicated than using directly the differential equation with the associated conditions in order to prove that y'(x)<0.

 

[Kvad]

Forget the analytical solution. It should be a thousand times more complicated than using directly the differential equation with the associated conditions in order to prove that y'(x)<0.

Yes, this I understand, but I wonder does the analytical way you solved this equation allows to come to some kind of explicit solution that would map x to y, so that when, say, p=0,5 or any other number withing given range, we would always be able to identify the value of function y, given the value of argument x?

 

[JJacquelin]

Yes, this I understand, but I wonder does the analytical way you solved this equation allows to come to some kind of explicit solution that would map x to y, so that when, say, p=0,5 or any other number withing given range, we would always be able to identify the value of function y, given the value of argument x?

What you expect suppose to express analytically the inverse (reciprocal) function of X($\theta$) which is not possible in the general case. If it was possible, no need to a prametric form of solution : I would have written y(x) instead of.

 

[Kvad]

Ok, thanks again for all your answers!