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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

d er i v a t i v e \frac{e ^{2 x}}{x}

d er i v a t i v e f (x) = \frac{x ^{2} - 2}{2 x + 1}

\frac{d y}{d x} = x^{8} y^{- 5}

(3 x y^{2} - 3) d x + (3 x^{2} y + 4) d y = 0

\int \frac{( x + 7 ) ^{3}}{2 x} d x

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