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

The solution for (y+1)y^'-x^2y-y=0,y(3)=1 is y+ln(y)=(x^3)/3+x-11

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

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

y + ln (y) = \frac{x ^{3}}{3} + x - 11

Solution steps

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

Solve separable ODE:

y + ln (y) = \frac{x ^{3}}{3} + x - 11

y + ln (y) = \frac{x ^{3}}{3} + x - 11

\frac{d y}{d x} = - \frac{7}{8} x^{3} y^{4}

y^{^{'}} + \frac{1}{50} y = 0.2

\frac{d y}{d x} = - \frac{7}{3} x^{5} y^{6}

\frac{d y}{d x} = - k y

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

What is the solution for (y+1)y^'-x^2y-y=0,y(3)=1 ?
The solution for (y+1)y^'-x^2y-y=0,y(3)=1 is y+ln(y)=(x^3)/3+x-11