Answer to Question #237493 in Differential Equations for James

Question

Solve the differential equation by separation of variables.

(x^3 + 2)y = x(y^4 + 3)y’

Answer 1

$({x^3} + 2)y = x\left( {{y^4} + 3} \right)y' \Rightarrow ({x^3} + 2)y = x\left( {{y^4} + 3} \right)\frac{{dy}}{{dx}} \Rightarrow \frac{{{x^3} + 2}}{x}dx = \frac{{{y^4} + 3}}{y}dy \Rightarrow \left( {{y^3} + \frac{3}{y}} \right)dy = \left( {{x^2} + \frac{2}{x}} \right)dx \Rightarrow \frac{{{y^4}}}{4} + 3\ln |y| = \frac{{{x^3}}}{3} + 2\ln |x| + C$

Answer: $\frac{{{y^4}}}{4} + 3\ln |y| - \frac{{{x^3}}}{3} - 2\ln |x| = C$