$(y + 1) \, dx + (4x - y) \, dy = 0$
$\dfrac{(y + 1) \, dx}{(y + 1) \, dy} + \dfrac{(4x - y) \, dy}{(y + 1) \, dy} = 0$
$\dfrac{dx}{dy} + \dfrac{4x - y}{y + 1} = 0$
$\dfrac{dx}{dy} + \dfrac{4x}{y + 1} - \dfrac{y}{y + 1} = 0$
$\dfrac{dx}{dy} + \left( \dfrac{4}{y + 1} \right)x = \dfrac{y}{y + 1}$ → linear in x
Hence,
$P = \dfrac{4}{y + 1}$
$Q = \dfrac{y}{y + 1}$
Integrating factor:
$\displaystyle e^{\int P \, dy} = e^{\int \frac{4}{y + 1} \, dy}$
$\displaystyle e^{\int P \, dy} = e^{4\int \frac{dy}{y + 1}}$
$e^{\int P \, dy} = e^{4\ln (y + 1)}$
$e^{\int P \, dy} = e^{\ln (y + 1)^4}$
$e^{\int P \, dy} = (y + 1)^4$
Thus,
$\displaystyle xe^{\int P\,dy} = \int Qe^{\int P\,dy} \, dy + C$
$\displaystyle x(y + 1)^4 = \int \left( \dfrac{y}{y + 1} \right) (y + 1)^4 \, dy + c$
$\displaystyle x(y + 1)^4 = \int y(y + 1)^3 \, dy + c$
Using integration by parts
$u = y$, $du = dy$
$dv = (y + 1)^3 \, dy$, $v = \frac{1}{4}(y + 1)^4$
$\displaystyle x(y + 1)^4 = \frac{1}{4}y(y + 1)^4 - \frac{1}{4}\int (y + 1)^4 \, dy + c$
$x(y + 1)^4 = \frac{1}{4}y(y + 1)^4 - \frac{1}{20}(y + 1)^5 + \frac{1}{20}c$
Multiply 20(y + 1)-4
$20x = 5y - (y + 1) + c(y + 1)^{-4}$
$20x = 4y - 1 + c(y + 1)^{-4}$ answer
