Answer to Question #270993 in Differential Equations for Aysu

Question

Solve the equation in exact differentials: (y²-1)dx+(2xy+3y)dy=0

Answer 1

$(y^2-1)dx+(2xy+3y)dy=0$

$Mdx+Ndy=0$

M=y²-1

N=2xy+3

$\frac{\partial{M}}{\partial{y}}=2y$

$\frac{\partial{N}}{\partial{x}}=2y$

$\therefore \frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}$ equation is exact

$\int Mdx=\int(y^2-1)dx=y^2x-x,(y=constant)$

$\int Ndy=\int(2xy+3y)dy=xy^2+\frac{3}{2}y^2,(x=constant)$

Hence the required solution is;

$xy^2-x+\frac{3}{2}y^2=C$

(rejecting the repeated term)