There're answered questions here about what I'm asking; however, I didn't study differential equations yet, and those answers use technics to solve differential equations. But my teacher, knowing that, asked me to do this:
Determine a function $u(x,y)$, such that, the differential $(x³ + x + y)u(x,y)dx – xu(x,y)dy$ be exact, and specify a potential for that differential. ($u$ is called an integrating factor for that differential)
I tried to use the assumption that
$$\begin{cases} \dfrac{\partial \varphi}{\partial x} = (x³ + x + y)u(x,y) \\ \dfrac{\partial \varphi}{\partial y} = – xu(x,y) \end{cases}$$
but I came to this differential equation:
$$\dfrac{\partial u(x,y)}{\partial y} (x³ + x + y) + \dfrac{\partial u(x,y)}{\partial x} x = -2u(x,y)$$
and I don't know how to proceed.
Best Answer
I tried $u(x,y)=u(x)$ Then got this
$$u(x)=-(u+xu') $$ $$\int \frac {du}{u}=-2\int \frac {dx}x$$ $$\implies u(x)=\frac 1 {x^2}$$
$$(x³ + x + y)u(x)dx - xu(x)dy$$ You have that : $$\partial_y \left (\frac {x³ + x + y}{x^2} \right )= \partial_x \left (-\frac 1x \right )$$