2007-11-02 09:25:38 · 3 answers · asked by jarynth5 1 in Science & Mathematics ➔ Mathematics
Hint: product and quotient rules.
2007-11-02 15:09:35 · update #1
The LHS is pretty close to being the derivative of xyy', but it's not quite there. So the next try would be to look at the derivative of zy', where z = x/y.
Hah. Unless I'm doing something wrong, your LHS is y^2 * (zy')'. So your equation is tantamount to (zy')' = 0. zy' = c. xy' = cy. y = x^c.
What I've just suggested is that y = x^c satisfies that equation for any constant c. Trying to check that, the LHS applied to x^c gives x^(2c-1) * (c(c-1) -c^2 + c). And, lo and behold, that sure looks a lot like 0.
Any constant multiple would factor out -- i.e., ax^c would be a solution for any constants a and c.
Nice one!
2007-11-02 17:51:22 · answer #1 · answered by Curt Monash 7 · 0⤊ 0⤋
It's nonlinear, which makes it harder right off. By solve, do you mean get a closed form solution? Or are solutions involving infinite series or special functions allowable?
As I don't know how to solve this equation, I would be much obliged if you posted a solution. If you don't want to publish it, you can email it to me. Incidentally, does this problem arise in some application?
2007-11-02 16:59:02 · answer #2 · answered by acafrao341 5 · 0⤊ 0⤋
what the heck is that, I'm taking college algebra and that doesn't exist
2007-11-02 16:33:04 · answer #3 · answered by pinkNzebra 2 · 0⤊ 2⤋