Question

Решить дифференциальное уравнение
(x-2y)y’=x+y
СРОЧНО!

Answer 1

Ответ:

$( x- 2y)y' = x + y \\ | \div x \\ (1 - \frac{2y}{x} )y'= 1 + \frac{y}{x} \\ \\ \frac{y}{x} = u \\ y' = u'x + u \\ \\ (1 - 2u)(u'x + u) = 1 + u \\ u'x + u = \frac{1 + u}{1 - 2u} \\ \frac{du}{dx} x = \frac{1 + u - u(1 - 2u)}{1 - 2u} \\ \frac{du}{dx} x = \frac{1 + 2 {u}^{2} }{1 - 2u} \\\int\limits \frac{1 - 2u}{1 + 2u {}^{2} } du = \int\limits \frac{dx}{x} \\ \int\limits \frac{du}{2u {}^{2} + 1} - \int\limits \frac{2udu}{1 + 2u {}^{2} } = ln( |x| ) + C \\ \frac{1}{ \sqrt{2} } \int\limits \frac{d( \sqrt{2} u)}{( \sqrt{2} u) {}^{2} + 1} - \frac{1}{2} \int\limits \frac{4udu}{2u {}^{2} + 1} = ln( |x| ) + C \\ - \frac{1}{ \sqrt{2} } arctg( \sqrt{2} u) - \frac{1}{2} \int\limits \frac{d(2u {}^{2} + 1) }{2u {}^{2} + 1} = ln( |x| ) + C\\ \frac{1}{ \sqrt{2} } arctg( \sqrt{2} u) - \frac{1}{2} ln( |2 {u}^{2} + 1 | ) = ln( |x| ) + C\\ \frac{1}{ \sqrt{2} } arctg( \sqrt{2} \frac{y}{x} ) - \frac{1}{2} ln( | \frac{2 {y}^{2} }{ {x}^{2} } + 1 | ) = ln( |x| ) + C$

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