# How to you find the general solution of dy/dx=xsqrt(5-x)?

Nov 16, 2016

$y = \frac{2}{5} {\left(5 - x\right)}^{\frac{5}{2}} - \frac{10}{3} {\left(5 - x\right)}^{\frac{3}{2}} + C$, Provided $x \le 5$

#### Explanation:

We have $\frac{\mathrm{dy}}{\mathrm{dx}} = x \sqrt{5 - x}$ which is a First Order Separable Differential Equation. We can therefore e separable the variables, as follows;

$\int \mathrm{dy} = \int x \sqrt{5 - x} \mathrm{dx}$

And so $y = \int x \sqrt{5 - x} \mathrm{dx} + C$

For the RHS integral, Let $u = 5 - x \implies x = 5 - u$
And, $\frac{\mathrm{du}}{\mathrm{dx}} = - 1 \implies \int \ldots \mathrm{du} = - \int \ldots \mathrm{dx}$

So $\int x \sqrt{5 - x} \mathrm{dx} = \int \left(5 - u\right) \sqrt{u} \left(- 1\right) \mathrm{du}$
$\therefore \int x \sqrt{5 - x} \mathrm{dx} = \int \left(5 {u}^{\frac{1}{2}} - {u}^{\frac{3}{2}}\right) \left(- 1\right) \mathrm{du}$
$\therefore \int x \sqrt{5 - x} \mathrm{dx} = \int \left({u}^{\frac{3}{2}} - 5 {u}^{\frac{1}{2}}\right) \mathrm{du}$
$\therefore \int x \sqrt{5 - x} \mathrm{dx} = {u}^{\frac{5}{2}} / \left(\frac{5}{2}\right) - 5 {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)$
$\therefore \int x \sqrt{5 - x} \mathrm{dx} = \frac{2}{5} {u}^{\frac{5}{2}} - 5 \cdot \frac{2}{3} {u}^{\frac{3}{2}}$
$\therefore \int x \sqrt{5 - x} \mathrm{dx} = \frac{2}{5} {u}^{\frac{5}{2}} - \frac{10}{3} {u}^{\frac{3}{2}}$

So the DE solution is;

$y = \frac{2}{5} {\left(5 - x\right)}^{\frac{5}{2}} - \frac{10}{3} {\left(5 - x\right)}^{\frac{3}{2}} + C$