**Problem 04**

$y' = y - xy^3e^{-2x}$

**Solution 04**

$y' = y - xy^3e^{-2x}$
From which

$P = -1$

$\dfrac{dy}{dx} - y = -xe^{-2x}y^3$

$dy - y~dx = -xe^{-2x}y^3~dx$ → Bernoulli's equation

$dy + Py~dx = Qy^n~dx$

$P = -1$

$Q = -xe^{-2x}$

$n = 3$

$(1 - n) = -2$

$z = y^{1 - n} = y^{-2}$

Integrating factor,

$u = e^{(1 - n)\int P~dx} = e^{-2\int (-1)~dx}$

$u = e^{2\int dx} = e^{2x}$

Thus,

$\displaystyle zu = (1 - n)\int Qu~dx + C$

$\displaystyle y^{-2}(e^{2x}) = -2\int (-xe^{-2x})(e^{2x})~dx + C$

$\displaystyle e^{2x}y^{-2} = 2\int x~dx + C$

$\dfrac{e^{2x}}{y^2} = x^2 + c$

$e^{2x} = y^2(x^2 + c)$ *answer*