Answer to Question #130229 in Differential Equations for Mehmet

Question

X^2 ydx + (y +x^3)dy=0 Find the general solution of differential equations.

Answer 1

$$Solution$$

$x^2 ydx + (y +x^3)dy=0 \implies x^2ydx=-(y +x^3)dy\\ \frac{dx}{dy}=\frac{-y-x^3}{x^2y}\\ \frac{dx}{dy}=-\frac{1}{x^2}-\frac{x}{y}\implies \frac{dx}{dy}+ \frac{x}{y}= -\frac{1}{x^2}\\ \frac{dx}{dy}+ Rx= S\\ R=\frac{1}{y},\ S=-\frac{1}{x^2}\\ IF=\epsilon^{\int Rdy}=\epsilon^{\int \frac{1}{y}dy}=\epsilon^{log\ y}=y\\ x \cdot IF = \int S \cdot IF\ dy+C\\ x \cdot y = \int -\frac{1}{x^2} \cdot y\ dy+C= \int -\frac{y}{x^2}\ dy+C= -\frac{y^2}{2x^2}+C\\ \therefore xy=-\frac{y^2}{2x^2}+C-------->Answer$