Problem 01 | Linear Equations

$(x^5 + 3y) \, dx - x \, dy = 0$

$\dfrac{(x^5 + 3y) \, dx}{x \, dx} - \dfrac{x \, dy}{x \, dx} = 0$

$\dfrac{x^5 + 3y}{x} - \dfrac{dy}{dx} = 0$

$x^4 + \dfrac{3y}{x} - \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} - \dfrac{3}{x}y = x^4$       → linear in y
 

Hence,
$P = -\dfrac{3}{x}$

$Q = x^4$
 

Integrating factor:
$e^{\int P \, dx} = e^{\int \left( -\frac{3}{x} \right)dx}$

$e^{\int P \, dx} = e^{-3\int \frac{dx}{x}}$

$e^{\int P \, dx} = e^{-3\ln x}$

$e^{\int P \, dx} = e^{\ln x^{-3}}$

$e^{\int P \, dx} = x^{-3}$
 

Thus,
$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx}\,dx + C$

$\displaystyle yx^{-3} = \int x^4(x^{-3})\,dx + \frac{c}{2}$

$\displaystyle yx^{-3} = \int x \,dx + \frac{c}{2}$

$\dfrac{y}{x^3} = \dfrac{x^2}{2} + \dfrac{c}{2}$
 

Multiply by 2x³
$2y = x^5 + cx^3$           answer