Question

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

Answer 1

`(\mathrm dx)/(\mathrm dy)+\frac y3=0\iff\mathrm dx=-\frac y3\,\mathrm dy`

`\implies\displaystyle\int\mathrm dx=-\frac13\int\mathrm y\,dy`

`\implies x=-\frac16y^2+C`

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


`0=-\frac16(12)^2+C\implies C=24`

So the particular solution to the initial value problem is


`x=-\frac16y^2+24\iff y^2=144-6x`