I want to solve the following differential equation:
$$ y' + xy = y^4,\ \ \ \ y(0)=1.$$
I don't know if this is a bernoulli equation, or a homogenous equation, but when I attempted it as a bernoulli equation, I got stuck.
I ended up with:
$$e^{\frac{-3x^{2}}{2}}v = -3\int e^{\frac{-3x^{2}}{2}}dx$$
But the book says answers does not need to be in closed form, but we have an initial condition..
Considering the equation $$y' + xy = y^4$$ let $$y=\frac{1}{{z^{1/3}}}\implies y'=-\frac{z'}{3 z^{4/3}}$$ which makes $$z'-3x z+3=0$$ which is separable and the general solution is $$z=c_1 e^{\frac{3 x^2}{2}}-\sqrt{\frac{3 \pi }{2}} e^{\frac{3 x^2}{2}} \text{erf}\left(\sqrt{\frac{3}{2}} x\right)$$ Using the initial condition $z(0)=1$, you get $c_1=1$ making $$z=\frac{1}{2} e^{\frac{3 x^2}{2}} \left(2-\sqrt{6 \pi } \text{erf}\left(\sqrt{\frac{3}{2}} x\right)\right)$$ Back to $y$ $$y=\frac{e^{-\frac{x^2}{2}}}{\sqrt[3]{1-\sqrt{\frac{3 \pi }{2}} \text{erf}\left(\sqrt{\frac{3}{2}} x\right)}}$$