Answer to Question #157063 in Civil and Environmental Engineering for zahid

Question #157063

dy/dx + y/x = y3 solve in Bernoulli equation

1
Expert's answer
2021-01-26T02:58:34-0500

"\\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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