How do you find the particular solution to sqrtx+sqrtyy'=0 that satisfies y(1)=4?

Jan 24, 2017

$\implies y = {\left(9 - {x}^{\frac{3}{2}}\right)}^{\frac{2}{3}}$

Explanation:

$\sqrt{x} + \sqrt{y} \setminus y ' = 0$

This is separable:

$\sqrt{y} \setminus y ' = - \sqrt{x}$

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

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

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

$y \left(1\right) = 4 \implies {\left(y \left(1\right)\right)}^{\frac{3}{2}} = 8 = - 1 + C \implies C = 9$

$\implies y = {\left(9 - {x}^{\frac{3}{2}}\right)}^{\frac{2}{3}}$