What is a solution to the differential equation sqrtx+sqrtyy'=0 with y=4 when x=1?

Sep 7, 2016

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

Explanation:

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

this can be separated

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

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

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

by the power rule

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

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

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

applying the IV:$y \left(1\right) = 4$

$4 = {\left(C - {1}^{\frac{3}{2}}\right)}^{\frac{2}{3}}$

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

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