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

The solution for (dx)/(dy)=((x^2y^2)/((x+1))) is y=\sqrt[3]{(3xln(x)-3+3c_{1}x)/x}

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(dx)/(dy)=((x^2y^2)/((x+1)))

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

y = 3 \frac{3 x ln ( x ) - 3 + 3 c _{1} x}{x}

Solution steps

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

Solve separable ODE:

y = 3 \frac{3 x ln ( x ) - 3 + 3 c _{1} x}{x}

y = 3 \frac{3 x ln ( x ) - 3 + 3 c _{1} x}{x}

v^{^{'}} = \frac{( v + \frac{4}{v} )}{t}

\frac{d y}{d x} 6 x y + 1 = 0

(x^{2} + 6)^{\frac{1}{2}} d x + \frac{1}{x csc ( 8 y )} d y = 0

\frac{d y}{d x} = 72 \cdot y \cdot x^{7}

(x^{2} + 1) \frac{d y}{d x} + x y = 3 x, y (0) = 4

What is the solution for (dx)/(dy)=((x^2y^2)/((x+1))) ?
The solution for (dx)/(dy)=((x^2y^2)/((x+1))) is y=\sqrt[3]{(3xln(x)-3+3c_{1}x)/x}