Question

Find a formula for the solution of the differential equation
dy/dt=1/(y+2)^2 , y(0)=1

I got C=6.333 but need the domain of the definition of the solution. I am not sure how to get this from
y^3/3+2y^2+4y=t+6.333

Answer 1

\[\int\limits \left( y+2 \right)^2 dy =\int\limits dt+c\]
\[\frac{ \left( y+2 \right)^3 }{ 3 }=t+c\]
when t=0,y=1
\[\frac{ 3^3 }{ 3 }=0+c,c=3^2=9\]
\[\frac{ \left( y+2 \right)^3 }{ 3 }=t+9\]