Answer to Question #247050 in Differential Equations for Sarah

Question

(x³-y)dx+xdy=0

Answer 1

Let us solve the differential equation $(x³-y)dx+xdy=0,$ which is equivalent to $xy'-y=-x^3$ and to $\frac{y'}x-\frac{y}{x^2}=-x$ after dividing by $x^2.$ It follows that $(\frac{y}x)'=-x,$ and hence $\frac{y}x=-\frac{x^2}2+C.$ We conclude that $y=-\frac{x^3}2+Cx$ is the general solution of the differential equation.