> [!theorem] Strategy > > The first order [[Ordinary Differential Equation|ordinary differential equation]] > $ > M(x, y)dx + N(x, y)dy = 0 > $ > or > $ > M(x, y) + N(x, y)\frac{dy}{dx} = 0 > $ > is homogeneous if $M$ and $N$ are both homogeneous of the same degree. > > Make a substitution on $x$ or $y$ with $y = xu(x)$. Then > $ > \frac{dy}{dx} = x\frac{du}{dx} + u > $ > Replacing $dy$ with $x du + u dx$, then the equation becomes > $ > M(x, xu) dx + N(x, xu)(xdu + udx) = 0 > $ > Since $M$ and $N$ are both homogeneous of the same degree, we can take out the $x$ factor > $ > x^dM(1, u)dx + x^dN(1, u)(xdu + udx) = 0 > $ > and > $ > M(1, u)dx + N(1, u)(xdu + udx) = 0 > $ > to obtain > $ > (M(1, u) + uN(1, u))dx + N(1, u)xdu = 0 > $ > The resulting equation should always be separable. > $ > \frac{dx}{x} + \frac{N(1, u)}{M(1, u) + uN(1, u)}du = 0 > $ > Simply integrate to solve.