the solution of a Cauchy problem

Question

I have $y'+\frac{y}{(x+1)^2}=x^2y^2$ with $y(0)=1$. I know exist local solution because $f(x,y)=-\frac{1}{(x+1)^2}y+x^2y^2$ is $C^1(R\setminus\{-1\} \times R)$. It's a Bernoulli equation but when i put $z=y^{-1}$ as substitution I found an expression in which the integrable is not calculable.It's right?

Answer 1

With the substitution $v = 1/y$, you get the linear d.e. $$ - v' + \frac{v}{(1+x)^2} = x^2$$ whose general solution is $$ v = e^{-1/(x+1)} \left( -\int x^2 e^{1/(x+1)} \; dx + C \right) $$ That integral is not elementary, but it can be expressed in terms of the Exponential integral function as $$ \frac{2 x^3 + x^2 - 3 x - 2 }{6} e^{1/(x+1)} - \frac{\text{Ei}(1/(x+1))}{6}$$