# dy/dx = sqrt(x/y) ?

Apr 26, 2018

$y = \pm \sqrt{\frac{4}{3} {x}^{3 / 2} + C}$

#### Explanation:

We have the differential equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{x}}{y}$

Use separation of variables to get $x$ and $y$ on opposite sides of the equation:

$y \textcolor{w h i t e}{.} \mathrm{dy} = \sqrt{x} \textcolor{w h i t e}{.} \mathrm{dx}$

Now integrate both sides:

$\int y \textcolor{w h i t e}{.} \mathrm{dy} = \int {x}^{1 / 2} \textcolor{w h i t e}{.} \mathrm{dx}$

$\frac{1}{2} {y}^{2} = \frac{1}{3 / 2} {x}^{3 / 2} + C$

$\frac{1}{2} {y}^{2} = \frac{2}{3} {x}^{3 / 2} + C$

${y}^{2} = \frac{4}{3} {x}^{3 / 2} + C$

(I'll just keep writing $C$ to represent any arbitrary constant, it doesn't matter that we've multiplied it by two.)

$y = \pm \sqrt{\frac{4}{3} {x}^{3 / 2} + C}$

Apr 27, 2018

${y}^{\frac{3}{2}} = {x}^{\frac{3}{2}} + C$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{\frac{x}{y}}$

We can collect terms:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{x}}{\sqrt{y}}$

$\sqrt{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{x}$

Which is a Separable DE, so we can "separate the variables" to get

$\int \setminus \sqrt{y} \setminus \mathrm{dy} = \int \setminus \sqrt{x} \setminus \mathrm{dx}$

Now we can integrate:

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

${y}^{\frac{3}{2}} = {x}^{\frac{3}{2}} + C$