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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^-frac32x_0^2+1)^-3 $ but now i can't move on. Moreover, WolframAlpha's solution is next to impossible…

Thanks for any help!

Hint: Write this as $$fracy'(x)y(x)-y(x)^4=x$$

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
$$
frac13 u'(x)+xu(x) = x
$$
with initial condition $u (x_0)=y_0^-3$. The solution obtained by integrating factor reads
beginaligned
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^3x_0^2/2 right) ,
endaligned
from which one deduces $y=u^-1/3$.

Substitute $z=1/y^3$
$$z'+3xz=3x$$
Tis equation is separable
$$z'=3x(1-z)$$
$$int frac dz1-z=frac 32x^2+C$$
$$-ln (z-1)=frac 32x^2+C$$
$$implies y^3(x)=frac 1 Ke^-3x^2/2+1$$

Therefore
$$y_0^3=frac 1 Ke^-3x_0^2/2+1$$
$$Ke^-3x_0^2/2=frac 1 y_0^3-1$$
$$K=left (frac 1 y_0^3-1 right)e^3x_0^2/2$$

We have $$C=e^frac3x_0^22(y_0^-1/3-1)$$

To reach a better understanding of my problem, i will write all the passages.

$fracy^'y^4=fracxyy^3-xrightarrow fracy^'y^4=xy^-3-x$

Now i put $z=y^-3rightarrow z^'=-3xz+3x$.

$y_0(x)=Ce^A(x)rightarrow A(x)=int -3xdx=-frac32x^2rightarrow y_0(x)=Ce^-frac32x^2$

$y_p(x)=e^A(x)B(x)rightarrow B(x)=int -3xcdot e^A(x)dx=int -3xcdot e^frac32x^2dx$

Now i put $frac32x^2=trightarrow dt=3xdxrightarrow dx=fracdx3x$

Since $int -3xe^tfracdt3x=-e^frac32x^2rightarrow y_px=1$, i obtain $y(x)=Ce^-frac32x^2+1$. But since $y^-3=z$...the result that i wrote.