Question

3xy'+y=12x find general solution

Answer 1

The equation is linear, as you can write

$3xy'+y=12x\implies y'+\dfrac1{3x}y=4$

An integrating factor would be

$\mu(x)=\exp\left(\displaystyle\int\frac{\mathrm dx}{3x}\right)=x^{1/3}$

Multiplying both sides of the ODE by this yields

$x^{1/3}y'+\dfrac1{3x^{2/3}}y=4x^{1/3}$

where the LHS is a derivative:

$\dfrac{\mathrm d}{\mathrm dx}\left[x^{1/3}y\right]=4x^{1/3}$

Integrating both sides, we get

$x^{1/3}y=4\displaystyle\int x^{1/3}\,\mathrm dx$
$x^{1/3}y=3x^{4/3}+C$
$y=3x^{3/3}+Cx^{-1/3}=3x+\dfrac C{x^{1/3}}$