What is the solution for y^'=(x^2y-y)/(y+1) ?

The solution for y^'=(x^2y-y)/(y+1) is y=\W(1/(e^{(-x^3+3x-c_{1))/3}})

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y^'=(x^2y-y)/(y+1)

y^{^{'}} = \frac{x ^{2} y - y}{y + 1}

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x - c _{1}}{3}}})

Solution steps

y^{'} (x) = \frac{x ^{2} y - y}{y + 1}

Solve separable ODE:

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x - c _{1}}{3}}})

y = W (\frac{1}{e ^{\frac{- x ^{3} + 3 x - c _{1}}{3}}})

\int 2 x - 5 d x

x \to 2 lim ((3 - x)^{2})

\int (3 x^{2} - 7 x)^{5} (6 x - 7) d x

\int x (e^{(3 x - 2)} + 5) d x

tan (θ)

What is the solution for y^'=(x^2y-y)/(y+1) ?
The solution for y^'=(x^2y-y)/(y+1) is y=\W(1/(e^{(-x^3+3x-c_{1))/3}})