What is the solution for y^'= 2/(3xy+y) ?

The solution for y^'= 2/(3xy+y) is y=sqrt((4ln(3x+1)+c_{1))/3},y=-sqrt((4ln(3x+1)+c_{1))/3}

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y^'= 2/(3xy+y)

y^{^{'}} = \frac{2}{3 x y + y}

y = \frac{4 ln ( 3 x + 1 ) + c _{1}}{3}, y = - \frac{4 ln ( 3 x + 1 ) + c _{1}}{3}

Solution steps

y^{'} (x) = \frac{2}{3 x y + y}

Solve separable ODE:

y = \frac{4 ln ( 3 x + 1 ) + c _{1}}{3}, y = - \frac{4 ln ( 3 x + 1 ) + c _{1}}{3}

y = \frac{4 ln ( 3 x + 1 ) + c _{1}}{3}, y = - \frac{4 ln ( 3 x + 1 ) + c _{1}}{3}

x y \cdot y^{^{'}} = y - 1

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

\frac{d y}{d x} = - \frac{0.8}{cos ( y )}, y (0) = \frac{π}{4}

\frac{1}{1 - 2 y} d y = x d x

\frac{d x}{y ^{2}} = \frac{d y}{x ^{2}}

What is the solution for y^'= 2/(3xy+y) ?
The solution for y^'= 2/(3xy+y) is y=sqrt((4ln(3x+1)+c_{1))/3},y=-sqrt((4ln(3x+1)+c_{1))/3}