# What is a general solution to the differential equation y'=5x^(2/3)y^4?

Nov 14, 2016

$y = \sqrt[3]{\frac{1}{C - 9 {x}^{\frac{5}{3}}}}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{\frac{2}{3}} {y}^{4}$

This is a First Order separable DE, so collecting like terms and "separating the variables" we get;

$\int {y}^{-} 4 \mathrm{dy} = \int 5 {x}^{\frac{2}{3}} \mathrm{dx}$

Integrating we get:

${y}^{-} \frac{3}{-} 3 = 5 {x}^{\frac{5}{3}} / \left(\frac{5}{3}\right) + {C}_{1}$
${y}^{-} 3 = - 9 {x}^{\frac{5}{3}} - 3 {C}_{1}$
$\frac{1}{y} ^ 3 = - 9 {x}^{\frac{5}{3}} + C$
${y}^{3} = \frac{1}{C - 9 {x}^{\frac{5}{3}}}$
$y = \sqrt[3]{\frac{1}{C - 9 {x}^{\frac{5}{3}}}}$