What is the solution for 3(x+1)y^2y^'=ln(x+1),y(0)=2 ?

The solution for 3(x+1)y^2y^'=ln(x+1),y(0)=2 is y=\sqrt[3]{(ln^2(1+x)+16)/2}

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3(x+1)y^2y^'=ln(x+1),y(0)=2

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

Solution steps

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

Solve separable ODE:

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

\frac{d y}{d x} = \frac{5 x ^{4}}{3 y ^{2} e ^{y^{3}}}

u^{^{'}} = 6 e^{7 x - H (x)}

\frac{d y}{d x} = 9.8 y

y^{^{'}} = - 3 y + 2

What is the solution for 3(x+1)y^2y^'=ln(x+1),y(0)=2 ?
The solution for 3(x+1)y^2y^'=ln(x+1),y(0)=2 is y=\sqrt[3]{(ln^2(1+x)+16)/2}