Question

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

Answer 1

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