> [!theorem] Strategy > > Let $(L)$ be a first order [[Ordinary Differential Equation|ODE]] with > $ > M(x, y)dx + N(x, y)dy = 0 > $ > Then $(L)$ is **exact** if > $ > \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} > $ > Resist the urge to cancel terms between $M$ and $N$. Suppose that the equation has an implicit solution $f(x, y(x)) = C$, then > $ > \frac{df}{dx} = C \Rightarrow \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} = 0 > $ > and with some witchcraft we get > $ > \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = 0 > $ > which matches the form of the exact ODE. > > Over more witchcraft we can obtain > $ > \begin{align*} > f(x, y) &= \int_{x_0}^{x}M(s, y)ds + > \int_{y_0}^{y}N(x, t)dt + > \int_{y_0}^{y}\int_{x_0}^{x}\frac{\partial M}{\partial y}(s, t)dsdt \\ > &= \int_{x_0}^{x}M(s, y)ds + > \int_{y_0}^{y}N(x, t)dt > \end{align*} > $