> [!theorem] Strategy > > Let $(L)$ be a first order [[Ordinary Differential Equation|ODE]] with > $ > M(x, y)dx + N(x, y)dy = 0 > $ > but not exact, then there may exist $u(x, y)$ such that > $ > uMdx + uNdy = 0 > $ > is exact. Suppose that > $ > \frac{\partial}{\partial y}uM = \frac{\partial}{\partial x}uN > $ > then > $ > M\frac{\partial u}{\partial y} + u \frac{\partial M}{\partial y} = N\frac{\partial u}{\partial x} + u\frac{\partial N}{\partial x} > $ > This partial differential equation would be easier to deal with if we restrict $u$ as a function of one variable. Suppose that $u$ is a function of $x$ *only*, then we have > $ > u\frac{\partial M}{\partial y} = N\frac{d u}{d x} + u\frac{\partial N}{\partial x} > $ > and > $ > u\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{du}{dx} > $ > If the ugly fraction is a function of $y$ only, then the problem has been converted into a linear first order ODE.