First order non linear ODE with Bernoulli I have a problem with this equation: $ y'(x)-xy(x)=-xy^4(x) $ with initial condition $ y(x_{0})=y_{0}$.
I'm arrived to prove that $ y_{0}= (Ce^{-\frac{3}{2}x_{0}^{2}}+1)^{-3} $ but now i can't move on. Moreover, WolframAlpha's solution is next to impossible…
Thanks for any help!
 A: Hint:  Write this as $$\frac{y'(x)}{y(x)-y(x)^4}=x$$
A: Let us solve the Bernoulli differential equation $$y'(x)-xy(x)=-xy(x)^4$$ with initial condition $y (x_0)=y_0$. Dividing by $y^4$ and setting $u=y^{-3}$, we have the linear ODE
$$
\frac{1}{3} u'(x)+xu(x) = x 
$$
with initial condition $u (x_0)={y_0}^{-3}$. The solution obtained by integrating factor reads
\begin{aligned}
u(x) &= e^{-3 (x^2-{x_0}^2)/2} \left({y_0}^{-3} + 3 \int_{x_0}^x t e^{3t^2/2} \,\text d t \right) \\
&= e^{-3 (x^2-{x_0}^2)/2} \left({y_0}^{-3} + e^{3x^2/2} - e^{3{x_0}^2/2} \right) ,
\end{aligned}
from which one deduces $y=u^{-1/3}$.
A: We have $$C=e^{\frac{3{x_0}^2}{2}}({y_0}^{-1/3}-1)$$
A: Substitute $z=1/y^3$
$$z'+3xz=3x$$
Tis equation is separable
$$z'=3x(1-z)$$
$$\int \frac {dz}{1-z}=\frac 32{x^2}+C$$
$$-\ln ({z-1})=\frac 32{x^2}+C$$
$$\implies y^3(x)=\frac 1 {Ke^{-3x^2/2}+1}$$

Therefore
$$y_0^3=\frac 1 {Ke^{-3x_0^2/2}+1}$$
$${Ke^{-3x_0^2/2}}=\frac 1 {y_0^3}-1$$
$$K=\left (\frac 1 {y_0^3}-1 \right)e^{3x_0^2/2}$$
A: To reach a better understanding of my problem, i will write all the passages.
$\frac{y^{'}}{y^{4}}=\frac{xy}{y^{3}}-x\rightarrow \frac{y^{'}}{y^{4}}=xy^{-3}-x$
Now i put $z=y^{-3}\rightarrow z^{'}=-3xz+3x$.
$y_{0}(x)=Ce^{A(x)}\rightarrow A(x)=\int -3xdx=-\frac{3}{2}x^{2}\rightarrow y_{0}(x)=Ce^{-\frac{3}{2}x^{2}}$
$y_{p}(x)=e^{A(x)}B(x)\rightarrow B(x)=\int -3x\cdot e^{A(x)}dx=\int -3x\cdot e^{\frac{3}{2}x^{2}}dx$
Now i put $\frac{3}{2}x^{2}=t\rightarrow dt=3xdx\rightarrow dx=\frac{dx}{3x}$
Since $\int -3xe^{t}\frac{dt}{3x}=-e^{\frac{3}{2}x^{2}}\rightarrow y_p{x}=1$, i obtain $y(x)=Ce^{-\frac{3}{2}x^{2}}+1$. But since $y^{-3}=z$...the result that i wrote.
