Answer to Question #284296 in Differential Equations for Aysu

Question

Solve the equation in exact differentials:

(x²+y)dx+(x-2y)dy=0

Answer 1

$\begin{aligned} &(x-2 y) \mathrm{d} y+\left(y+x^{2}\right) \mathrm{d} x=0 \\ &\text { Comparing with } M(x, y) \mathrm{d} y+N(x, y) \mathrm{d} x=0 \\ &\text { where } M(x, y)=x-2 y \text { and } N(x, y)=y+x^{2} \\ &\text { Check for full differential: } M(x, y)_{x}^{\prime}=N(x, y)_{y}^{\prime}=1 \\ &\text { Find } F(x, y): \mathrm{d} F(x, y)=F_{y}^{\prime} \mathrm{d} y+F_{x}^{\prime} \mathrm{d} x \\ &F(x, y)=\int N(x, y) \mathrm{d} x=\int y+x^{2} \mathrm{~d} x=\frac{x^{3}}{3}+y x+C_{y} \\ &\left(\frac{x^{3}}{3}+y x\right)_{y}^{\prime}=x \\ &C_{y}=\int M(x, y)-\left(\frac{x^{3}}{3}+y x\right)_{y}^{\prime} \mathrm{d} y=\int-2 y \mathrm{~d} y=-y^{2} \\ &\quad F(x, y)=\frac{x^{3}}{3}+y x+C_{y}=-y^{2}+\frac{x^{3}}{3}+y x \\ &-y^{2}+\frac{x^{3}}{3}+y x=C \end{aligned}$