Question

Найти общее решение уравнения.

Answer 1

Ответ:

$(y - {x}^{2} y)dy -(x + x {y}^{2}) dx = 0 \\ y(1 - {x}^{2} )dy = x(1 + {y}^{2}) dx \\ \int\limits \frac{ydy}{1 + {y}^{2} } = \int\limits \frac{xdx}{1 - {x}^{2} } \\ \frac{1}{2} \int\limits \frac{2ydy}{1 + {y}^{2} } = - \frac{1}{2} \int\limits \frac{( - 2x)dx}{1 - {x}^{2} } \\ \frac{1}{2} \int\limits \frac{d(1 + {y}^{2} )}{1 + {y}^{2} } = - \frac{1}{2} \int\limits \frac{d(1 - {x}^{2}) }{1 - {x}^{2} } \\ \frac{1}{2} ln(1 + {y}^{2} ) = - \frac{1}{2} ln(1 - {x}^{2} ) + ln(C) \\ ln(1 + {y}^{2} ) = - ln(1 - {x}^{2} ) + ln(C) \\ ln(1 + {y}^{2} ) = ln( \frac{C}{1 - {x}^{2} } ) \\ {y}^{2} + 1 = \frac{C}{1 - {x}^{2} } \\ {y}^{2} = \frac{C}{1 - {x}^{2} } - 1 \\ {y}^{2} = \frac{C- (1 - {x}^{2}) }{1 - {x}^{2} } \\ {y}^{2} = \frac{C - 1 + {x}^{2} }{1 - {x}^{2} } \\ {y}^{2} = \frac{ {x}^{2} + C }{1 - {x}^{2} }$