Question

ПОМОГИТЕ НУЖНО РЕШЕНИЕ!!!
Решите уравнение y'+2x^2y'+2xy-2x=0
ОТВЕТ: (1-y)√1+2x^2

Answer 1

$y'+2x^2y'+2xy-2x=0 \\ y'(1 + 2 {x}^{2} ) = 2x - 2xy \\ \\ \frac{dy}{dx} (1 + 2 {x}^{2} ) = 2x(1 - y) \\ \\ \int \frac{dy}{1 - y} = \int \frac{ 2x}{1 + 2 {x}^{2} }dx \\ \\ - \int\frac{d(1 - y)}{1 - y} = \frac{1}{2} \int \frac{ d(1 + 2 {x}^{2} )}{1 + 2 {x}^{2} } \\ \\ - ln |1 - y| = \frac{1}{2} ln |1 + 2 {x}^{2} | + \frac{1}{2} ln |C| \\ \\ ln |1 - y| = - \frac{1}{2} ln |C(1 + 2 {x}^{2} )| \\ \\ 1 - y =( C(1 + 2 {x}^{2} )) ^{ - \frac{1}{2} } \\ \\ y = 1 - \frac{1}{ \sqrt{C(1 + 2 {x}^{2} )}} = 1 - \frac{C}{ \sqrt{(1 + 2 {x}^{2} )}} \\ \\ OTBET: \ y=1 - \frac{C}{ \sqrt{(1 + 2 {x}^{2} )}}$