Question

solve the equation in exact differentials

(y²-1)dx+(2xy+3y)dy=0

Answer 1

Let us solve the equation in exact differentials $(y^2-1)dx+(2xy+3y)dy=0,$

Since $\frac{\partial (y^2-1)}{\partial y}=2y=\frac{\partial (2xy+3y)}{\partial x}$, the differential equation is indeed exact.

It follows that there exists the function $u=u(x,y)$ such that

$\frac{\partial u}{\partial x}=y^2-1,\ \frac{\partial u}{\partial y}=2xy+3y.$ Therefore, $u(x,y)=xy^2-x+c(y),$ and hence

$\frac{\partial u}{\partial y}=2xy+c'(y)=2xy+3y.$ It follows that $c'(y)=3y,$ and thus $c(y)=\frac{3}2y^2+C.$

We conclude that the general solution is

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