What is the solution for derivative of x^2y^2+xy=0 ?

The solution for derivative of x^2y^2+xy=0 is y=(-1+c_{1})/(2x),y=-(c_{1}+1)/(2x)

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derivative of x^2y^2+xy=0

\frac{d}{d x} (x^{2} y^{2} + x y) = 0

y = \frac{- 1 + c _{1}}{2 x}, y = - \frac{c _{1} + 1}{2 x}

Solution steps

\frac{d}{d x} (x^{2} y^{2} + x y) = 0

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

x^{2} y^{2} + x y = \int 0 d x

\int 0 d x = c_{1}

x^{2} y^{2} + x y = c_{1}

Isolate

y : y = \frac{- 1 + 4 c _{1} + 1}{2 x}, y = - \frac{4 c _{1} + 1 + 1}{2 x}

y = \frac{- 1 + 4 c _{1} + 1}{2 x}, y = - \frac{4 c _{1} + 1 + 1}{2 x}

Simplify

y = \frac{- 1 + c _{1}}{2 x}, y = - \frac{c _{1} + 1}{2 x}

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

(x^{2} + 1) y^{^{'}} + 4 x y = 0

\frac{d y}{d x} = \frac{4 y}{x}, y = 2, x = 1

y^{2} sec (3 x) \frac{d y}{d x} = 1, y (0) = 0

y^{^{'}} = sin (3 x) y^{2}

What is the solution for derivative of x^2y^2+xy=0 ?
The solution for derivative of x^2y^2+xy=0 is y=(-1+c_{1})/(2x),y=-(c_{1}+1)/(2x)