Find the solution to the initial value problem
$$y'=(1-11t)y^2, $$ $$y(0)=-1/9$$
First I separate the variables
$dy/y^2=(1-11t)dt$
Integrate both sides
$-\frac 1y = t-11t^2+c$
$y(t)=-\frac 1{(t-11t^2+c)}$
Using the initial condition, I find c=9
$y(t)=- \frac 1 {(t-11t^2+9)}$
Is this correct? My textbook gives me a different solution