Question

y^3(y-1)dx+3xy^2(y-1)dy=(y+2)dy, y(1/4)=2
Помогите решить

Answer 1

Ответ:

∂/∂y(e^(-y^3x - 3y^2x + g(y))(y^3(y-1))) = (3y^2(y-1)e^(-y^3x - 3y^2x + g(y)) + e^(-y^3x - 3y^2x + g(y))(3y^3 - 6y^2)) = 3y^2(y - 1)e^( -y^3x - 3y^2x + g(y)) + 3y^3e^(-y^3x - 3y^2x + g(y)) - 6y^2e^(-y^3x - 3y^2x + g (у)),

∂/∂x(e^(-y^3x - 3y^2x + g(y))(y + 2)) = -(y^3 + 3y^2)e^(-y^3x - 3y^2x + g(y)) + e^(-y^3x - 3y^2x + g(y))(y + 2)(-y^3 - 3y^2).

3y^2(y - 1)e^(-y^3x - 3y^2x + g(y)) + 3y^3e^(-y^3x - 3y^2x + g(y)) - 6y^2e^ (-y^3x - 3y^2x + g(y)) = -(y^3 + 3y^2)e^(-y^3x - 3y^2x + g(y)) + e^(-y^ 3x - 3y^2x + g(y))(y + 2)(-y^3 - 3y^2).

Пошаговое объяснение:

y = 0 и y = 1.