Question

(x^2+y^2)dx+2xydy=0
срооочно

Answer 1

Ответ:

$( {x}^{2} + {y}^{2} )dx + 2xydy = 0 \: \: \: | \div {x}^{2} \\ (1 + \frac{ {y}^{2} }{ {x}^{2} } )dx + \frac{2y}{x} dy = 0 \\ - \frac{2y}{x} dy = (1 + \frac{ {y}^{2} }{ {x}^{2} } )dx \\ \frac{2y}{x} y' = - (1 + \frac{ {y}^{2} }{ {x}^{2} } ) \\ \\ \frac{y}{x} = u \\ y '= u'x + u \\ \\ 2u(u'x + u) = - (1 + {u}^{2} ) \\ u'x + u = - \frac{1 + {u}^{2} }{2u} \\ \frac{du}{dx} x = \frac{ - 1 - {u}^{2} - 2 {u}^{2} }{2u} \\ \frac{du}{dx} x = - \frac{1 + 3 {u}^{2} }{2u} \\ \int\limits \frac{2u}{ 3u {}^{2} + 1 } du = - \int\limits \: \frac{dx}{x} \\ \frac{1}{3} \int\limits \frac{3 \times 2u}{3u {}^{2} + 1 } du = ln |x| + ln |C| \\ \frac{1}{3} \int\limits \frac{d(3u {}^{2} + 1)}{3u {}^{2} + 1 } = ln |Cx| \\ \frac{1}{3} ln |3 {u}^{2} + 1| = ln |Cx| \\ ln |3 {u}^{2} + 1 | = 3ln |Cx| \\ 3 {u}^{2} + 1 = C {x}^{3} \\ \frac{3 {y}^{2} }{ {x}^{2} } + 1 = C {x}^{3} \\ 3 {y}^{2} + {x}^{2} = C {x}^{5} \\ 3 {y}^{2} = C {x}^{5} - {x}^{2}$

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