# What is a solution to the differential equation dy/dx=x/(y+3)?

Aug 9, 2016

$y = \pm \sqrt{{x}^{2} + C} - 3$

#### Explanation:

this is separable

$y ' = \frac{x}{y + 3}$

$\left(y + 3\right) y ' = x$

$\int \setminus \left(y + 3\right) y ' \setminus \mathrm{dx} = \int \setminus x \setminus \mathrm{dx}$

See ASIDE below
$\implies \int \setminus y + 3 \setminus \mathrm{dy} = \int \setminus x \setminus \mathrm{dx}$

${y}^{2} / 2 + 3 y = {x}^{2} / 2 + C$

${y}^{2} + 6 y = {x}^{2} + C$

completing the square
${\left(y + 3\right)}^{2} - 9 = {x}^{2} + C$

${\left(y + 3\right)}^{2} = {x}^{2} + C$

$y = \pm \sqrt{{x}^{2} + C} - 3$

ASIDE
Avoiding doing algebra with the expression $\frac{\mathrm{dy}}{\mathrm{dx}}$
Because $I = \int \setminus \left(y + 3\right) y ' \setminus \mathrm{dx}$
then $\frac{d}{\mathrm{dx}} \left(I\right) = \left(y + 3\right) y '$
And from the chain rule and the fact that $y = y \left(x\right)$ we can say that
$\frac{d}{\mathrm{dx}} \left(I\right) = I ' y '$
so $I ' = y + 3$
$\implies I = \int \setminus y + 3 \setminus \mathrm{dy}$