Question

A Bernoulli differential equation is one of the form dxdy+P(x)y=Q(x)yn Observe that, if n=0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u=y1−n transforms the Bernoulli equation into the linear equation dxdu+(1−n)P(x)u=(1−n)Q(x) Use an appropriate substitution to solve the equation xy+y=4xy2

Answer 1

It's unclear whether the ODE is

$xy'+y=4xy^2$

or

$xy+y'=4xy^2$

If the first case, then divide through both sides by $y^2$ to get

$xy^{-2}y'+y^{-1}=4x$

then substitute $z=y^{-1}$, so that $z'=-y^{-2}y'$. Then

$-xz'+z=4x\implies-\dfrac1xz'+\dfrac1{x^2}z=\left(-\dfrac1xz\right)'=\dfrac4x$

$\implies-\dfrac1xz=\displaystyle\int\frac4x\,\mathrm dx$

$\implies z=-4x\ln|x|+Cx$

$\implies\boxed{y=\dfrac1{Cx-4x\ln|x|}}$

If the second, then dividing through by $y^2$ gives

$xy^{-1}+y^{-2}y'=4x$

The same substitution as before gives

$xz-z'=4x\implies e^{-x^2/2}z'-xe^{-x^2/2}z=\left(e^{-x^2/2}z\right)'=-4xe^{-x^2/2}$

$\implies e^{-x^2/2}z=\displaystyle-4\int xe^{-x^2/2}\,\mathrm dx$

$\implies z=4+Ce^{x^2/2}$

$\implies\boxed{y=\dfrac1{4+Ce^{x^2/2}}}$