solve the homogenous differential equation? y'= xy/x^2-y^2?

Answer 1

Note that Paul D's answer is incorrect: he gives
x*ln(y) - (y^2/2*x) = x + C.
Differentiating implicity gives
ln y + x (1/y) y' - (y^2/2)(-1/x^2) - 2yy'/(2x) = 1
<=> y' (x/y - y/x) = 1 - ln y - y^2/2x^2
<=> y' (x^2 - y^2) = xy - xy ln y - y^3/2x
<=> y' = [xy (1 - ln y) - y^3/2x] / (x^2 - y^2).
His error is that Int(x/y) dy, for example, is not equal to x ln y when x and y are related. Similarly Int (y/x) dy is not equal to y^2 / 2x.

Now, to solve the equation
y' = xy / (x^2 - y^2)
Let y = xz. Then y' = z + xz', so we get
z + xz' = x^2z / (x^2 - x^2z^2)
<=> z + xz' = z / (1 - z^2)
<=> xz' = (z^2 + z - 1) / (1 - z^2)
<=> (1-z^2) z' / (z^2 + z - 1) = 1/x
<=> ∫(-1 + z/(z^2 + z - 1)) dz = ∫1/x dx
<=> -z + ∫(1/2)(2z+1)/(z^2+z-1) dz - ∫(1/2)(1/(z^2+z-1)) dz = ln |x| + c.
<=> -z + ln |z^2 + z - 1| / 2 - (1/2)∫du/(u^2 - 5/4) = ln |x| + c, where u = z + 1/2.
<=> -z + ln |z^2 + z - 1| / 2 - (1/2)∫[(1/√5 (1/(u-√5/2) - 1/(u+√5/2)] du = ln |x| + c
<=> -z + ln |z^2 + z - 1| / 2 - (1/(2√5)) (ln |u - √5/2| - ln |u + √5/2|) = ln |x| + c
<=> -z + ln |z^2 + z - 1| / 2 - (1/(2√5)) (ln |z + (1-√5)/2| - ln |z + (1+√5/2)|) = ln |x| + c
<=> -y/x + ln |y^2/x^2 + y/x - 1| / 2 - 1/(2√5) . (ln |(y/x + 1/2 - √5/2) / (y/x + 1/2 + √5/2)|) = ln |x| + c.

So there you have it; an extremely nasty expression which you can play around with if you really want to. I wouldn't give you much chance of getting it into a nice form, and certainly not an explicit expression for y...

Answer 2

We have dy/dx = (xy)/(x^2-y^2), so separating, we get (x^2-y^2)/(xy) dy = dx, or (x^2/xy - y^2/xy) dy = (x/y - y/x) dy = dx, so integrating gives us Int x/y-y/x dy = Int dx, so x*ln(y) - (y^2/2*x) = x + C.


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