# Help with first order second degree diff. eq.

Hi, can anybody help me with this differential equation? (px-y)(py-x)=a^2*p, where p=y'. I tried to solve it by expliciting the two solutions for y as functions of p and x and then derive, in order to obtain two first order differential equations in p but these are impossible to solve. I also tried to find the singular solution of the equation but it turned out that there is not a singular solution! I really don't know what to do, can anybody help me? Thank you in advance :)

Mark44
Mentor
Hi, can anybody help me with this differential equation? (px-y)(py-x)=a^2*p, where p=y'. I tried to solve it by expliciting the two solutions for y as functions of p and x
"Explicit" is not a verb, so I don't understand what "expliciting" is supposed to mean. The fact that you have a product on the left side is useful only if the right side happens to be zero. For example, if (x + 2)(y + 3) = 0, then I can say something about x and y, but if (x + 2)(y + 3) = 1, then I really can't say anything useful about x and y.
dam said:
and then derive
Do you mean "differentiate"? If so, I'm not following what you're trying to do. Or do you mean "integrate"?
dam said:
, in order to obtain two first order differential equations in p but these are impossible to solve. I also tried to find the singular solution of the equation but it turned out that there is not a singular solution! I really don't know what to do, can anybody help me? Thank you in advance :)
If you expand the product on the left side of your equation above, you get a quadratic in y'. You can use the Quadratic formula to get two first-order differential equations. I haven't gone any further than this, as what you get is pretty messy.

Yeah I'm sorry if I wrote it so terribly, I'm not English so I did My best. As for the form of the equation, I didn't change it, that was the way it was written, so it should supposedly help to solve it. Actually written in that form it is quite clear that the equation is solved by two families of curves, and each curve of a family has an inverse function in the other family. You can also derive some other properties of the solutions, such as that the two families don't have any envelope, except when a=0. My real problem is how to find the expression for the curves of these two families. As for the word deriving, yeah, I mean differentiating. If you solve it for y and then differentiate, you find a first order differential equation in p=y', which however is still horrible. Thank for your interest :)

Svein
So your equation is: $(y'x - y)(y'y - x) = a^{2}y'$?

If you multiply out, you get a 2. degree equation in y'. Solve for y' and then solve the resulting two differential equations.

Mark44
Mentor
If you multiply out, you get a 2. degree equation in y'. Solve for y' and then solve the resulting two differential equations.
That's pretty much what I said in post #2.

Svein