Question #157063

dy/dx + y/x = y^{3 solve in Bernoulli equation}

Expert's answer

"\\frac{dx}{dy} + \\frac{y}{x} = y"

"\\frac{1}{y^3} . \\frac{dy}{dx} + \\frac{1}{y^2}. \\frac{1}{x} = 1"

supposed

"\\frac{1}{y^2} = v" "\\frac{-2}{y^3}.\\frac{dy}{dx}=\\frac{dv}{dx}"

"\\frac{-1}{2}.\\frac{dv}{dx}+ \\frac{v}{x}=1" "\\frac{dv}{dx} \\frac{-2}{x} = -2"

"\\frac{dv}{dx} +" "ev =Q" for Bernoulli equation , "p=" "f(x) ; Q= f (x)"

Integrating factors I.F = "e^{{\\intop p dx}}"

I.F = "e^{{\\intop \\frac{-2}{x}dx}}" = "e^{{-2\\ln x}}" = "e^{{\\ln x (\\frac{1}{x^2}) }}" = "\\frac{1}{x^2}"

"\\frac{1}{x^2}" "\\frac{dv}{dx}" "\\frac{-2}{x^3}.v= \\frac{-2}{x^2}"

= "\\frac{dv}{x^2}- \\frac{2dx}{x^3}.v =\\frac{dx}{x^2}"

"\\intop d(\\frac{v}{x^2}) = \\intop \\frac{dv}{x^2} = \\frac{v}{x^2} = x ^{{-1}} + c"

"\\frac{1}{y^2}. \\frac{1}{x^2} = \\frac{-1}{x} + c"

"\\frac{1}{x^2y^2} + \\frac{1}{x} = c"

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