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2xyy'-y^2+x^2=0 differential equation please?
Could you tell me how to solve as well? I have an exam tomorrow :(
2 Answers
- SteinerLv 77 years agoFavorite Answer
Put y = ux, so that y' = u + u'x. Then,
2 ux^2 (u + u'x) - u^2 x^2 + x^2 = 0
2 u^2 + 2 u u' x - u^2 + 1= 0 (x ≠ 0)
2 u u' x = -u^2 - 1
2 u du/dx x = -u^2 - 1
2u/(u^2 + 1) du = -dx/x
Integrating,
ln(u^2 + 1) = -ln(x) + C, x > 0
u^2 + 1 = K/x, K = e^C a constant
u^2 = (y/x)^2 = K/x - 1
y^2 = Kx - x^2
y = sqrt(K/x - x^2), for K/x - x^2 ≥ 0, in the real domain
- Anonymous7 years ago
Yes
