Question

Find the solution to the differential equation dx/dy+y/3=0 subject to the initial conditions y(0)=12.

Answer 1

$\dfrac{\mathrm dx}{\mathrm dy}+\dfrac y3=0\iff\mathrm dx=-\dfrac y3\,\mathrm dy$
$\implies\displaystyle\int\mathrm dx=-\frac13\int\mathrm y\,dy$
$\implies x=-\dfrac16y^2+C$

Given that $y(0)=12$, we have

$0=-\dfrac16(12)^2+C\implies C=24$

So the particular solution to the initial value problem is

$x=-\dfrac16y^2+24\iff y^2=144-6x$