What is the solution for (dy)/(dx)=((x^2y-y))/((y+1)),y(0)=1 ?

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

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(dy)/(dx)=((x^2y-y))/((y+1)),y(0)=1

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

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

Solution steps

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

Solve separable ODE:

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

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

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