Question

Bernoulli differential equation... y'+xy=xy^2

Answer 1

$y'+xy=xy^2\implies y^{-2}y'+xy^{-1}=x$

Let $z=y^{-1}$, so that $z'=-y^{-2}y'$. Then the ODE becomes linear in $z$ with

$-z'+xz=x\implies z'-xz=-x$

Find an integrating factor:

$\mu(x)=\exp\left(\displaystyle\int-x\,\mathrm dx\right)=e^{-x^2/2}$

Multiply both sides of the ODE by $\mu$:

$e^{-x^2/2}z'-xe^{-x^2/2}z=-xe^{-x^2/2}$

The left side can be consolidated as a derivative:

$\left(e^{-x^2/2}z\right)'=-xe^{-x^2/2}$

Integrate both sides with respect to $x$ to get

$e^{-x^2/2}z=e^{x^2/2}+C$

where the right side can be computed with a simple substitution. Then

$z=1+Ce^{x^2/2}$

Back-substitute to solve for $y$.

$y^{-1}=1+Ce^{x^2/2}\implies y=\dfrac1{1+Ce^{x^2/2}}$