Homogeneous and Reducible Differential Equations

A function $f(x,y)$ is homogeneous of degree n if $f(\lambda x, \lambda y) = \lambda^n f(x,y)$ for every $\lambda$. The equation $\dfrac{dy}{dx} = \dfrac{P(x,y)}{Q(x,y)}$ is homogeneous when:

• $P$ and $Q$ are homogeneous of the same degree (so the ratio has degree 0), OR equivalently
• the right side can be rewritten as a function of $\dfrac{y}{x}$ alone: $\dfrac{dy}{dx} = g\!\left(\dfrac{y}{x}\right)$.

Quick tests: $y^2 + 2xy$ is homogeneous of degree 2; $2x - 3y$ of degree 1; $\sin\!\left(\tfrac{y}{x}\right)$ of degree 0. But $\cos x + \sin y$ is not homogeneous — mixing $x$ and $y$ separately (not as a ratio) breaks the scaling test.

For a homogeneous equation $\dfrac{dy}{dx} = g\!\left(\dfrac{y}{x}\right)$:

• Put $y = vx$, so by the product rule $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
• Substitute: $v + x\dfrac{dv}{dx} = g(v)$, hence $x\dfrac{dv}{dx} = g(v) - v$.
• Separate: $\dfrac{dv}{g(v) - v} = \dfrac{dx}{x}$, then integrate both sides.
• **Substitute $v = \dfrac{y}{x}$ back at the end** to return to $x, y$, and fit any initial condition to find the constant.

The full procedure on a worked homogeneous DE:

• Set up: rewrite the right side as a function of $v = y/x$, put $y = vx$, and use $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
• Separate and integrate the resulting $v$-$x$ equation.
• Back-substitute $v = y/x$.
• Fit the initial condition: substitute the given point to evaluate the constant of integration. For a *particular* solution the constant is a specific number, not $c$.

Watch the algebra of $g(v) - v$: for $(y^2 - x^2)\,dx = xy\,dy$ this becomes $\dfrac{v^2 - 1}{v} - v = -\dfrac{1}{v}$, giving $-v\,dv = \dfrac{dx}{x}$.

The slope is $\dfrac{dy}{dx} = \dfrac{y}{x} + h\!\left(\dfrac{y}{x}\right)$ where $h$ is a trig function of the ratio. Put $y = vx$:

• $v + x\dfrac{dv}{dx} = v + h(v)$, and the $v$ on both sides cancels, leaving $x\dfrac{dv}{dx} = h(v)$.
• Separate: $\dfrac{dv}{h(v)} = \dfrac{dx}{x}$. The integral is standard:
• slope $\dfrac{y}{x} + \sec\dfrac{y}{x}$: $\dfrac{dv}{\sec v} = \cos v\,dv \Rightarrow \sin v = \log x + c$.
• slope $\dfrac{y}{x} - \cos^2\dfrac{y}{x}$: $\dfrac{dv}{-\cos^2 v} = -\sec^2 v\,dv \Rightarrow -\tan v = \log x + c$, i.e. $\tan v = -\log x + c$.

Then substitute $v = y/x$ and use the given point to find $c$.

For $x\dfrac{dy}{dx} = y\!\left(\log\dfrac{y}{x} + 1\right)$:

• Divide by $x$: $\dfrac{dy}{dx} = \dfrac{y}{x}\!\left(\log\dfrac{y}{x} + 1\right)$ — homogeneous.
• Put $v = \dfrac{y}{x}$, $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$: $v + x\dfrac{dv}{dx} = v(\log v + 1) = v\log v + v$.
• Cancel $v$: $x\dfrac{dv}{dx} = v\log v$, so $\dfrac{dv}{v\log v} = \dfrac{dx}{x}$.
• The key integral: $\displaystyle\int\dfrac{dv}{v\log v} = \log(\log v)$ (put $t = \log v$). Hence $\log(\log v) = \log x + \log c$, giving $\log v = c x$, i.e. $\log\dfrac{y}{x} = c x$.

For $\dfrac{dy}{dx} = f(x + y)$ put $v = x + y$:

• $\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac{dv}{dx} - 1$.
• The equation becomes $\dfrac{dv}{dx} - 1 = f(v)$, i.e. $\dfrac{dv}{1 + f(v)} = dx$ — separable.
• Worked forms:
• $\dfrac{dy}{dx} = (x + y)^2$: $\dfrac{dv}{1 + v^2} = dx \Rightarrow \tan^{-1}v = x + c \Rightarrow \tan^{-1}(x + y) = x + c$.
• $\cos(x + y)\,dy = dx$: $\dfrac{du}{1 + \cos u} = dy \Rightarrow y = \tan\dfrac{x + y}{2} + c$.
• Scaled pair: for $\dfrac{dy}{dx} = (x + 9y)^2$ put $u = x + 9y$, $\dfrac{du}{dx} = 1 + 9\dfrac{dy}{dx}$, giving $\dfrac{du}{1 + 9u^2} = dx$.