Recipe

Solve a homogeneous first-order equation

Use when: \dfrac{dy}{dx}=f(x,y), and dividing the numerator and denominator by the highest power of x present leaves an expression depending only on the ratio y/x

Goal: an implicit or explicit solution for y, converted back from the substituted variable v

Taught in Homogeneous first-order equations

1. 1

   Divide every term of f(x,y) by the appropriate power of x and check that only the ratio y/x remains.

   Example

   For \dfrac{dy}{dx}=\dfrac{x^2+3y^2}{2xy}: dividing numerator and denominator by x^2 gives \dfrac{1+3(y/x)^2}{2(y/x)} — depends only on y/x, so the equation is homogeneous.

   Why: only when f depends on y/x alone does the substitution in Step 2 turn the equation separable — this check is what confirms the technique applies before starting it.

2. 2

   Replace y by vx everywhere in f(x,y), and replace \dfrac{dy}{dx} by v+x\dfrac{dv}{dx} (the product rule applied to y=vx).

   Example

   With y=vx: \dfrac{x^2+3y^2}{2xy}=\dfrac{x^2+3v^2x^2}{2x\cdot vx}=\dfrac{1+3v^2}{2v}, so the equation becomes v+x\dfrac{dv}{dx}=\dfrac{1+3v^2}{2v}

   Why: this substitution is chosen for exactly one reason — an f(x,y) depending only on y/x becomes a function of v alone once y is replaced by vx, and the product rule is what \dfrac{dy}{dx} turns into under that same substitution.

3. 3

   Subtract v from both sides and simplify the right side into a single fraction.

   Example

   x\dfrac{dv}{dx}=\dfrac{1+3v^2}{2v}-v=\dfrac{1+3v^2-2v^2}{2v}=\dfrac{1+v^2}{2v}

   Why: this is guaranteed to be separable — the substitution was chosen precisely so that x and v always end up factoring apart at this step.

4. 4

   Separate the v and x variables, then integrate both sides.

   Example

   \dfrac{2v}{1+v^2}\,dv=\dfrac{dx}{x}. Integrating (the left side's numerator is exactly the derivative of its denominator): \ln(1+v^2)=\ln|x|+C

   Why: once separated, this is an ordinary single-variable integral on each side, the same integration Section 1 already covers.

5. 5

   Replace v by y/x everywhere in the solved equation and simplify.

   Example

   Exponentiating and renaming the constant: 1+v^2=A|x|. Substituting v=y/x: 1+\dfrac{y^2}{x^2}=A|x|, so, multiplying by x^2, \boxed{x^2+y^2=A|x|^3}

   Why: v was never the actual unknown — leaving the final answer in terms of v is an unfinished substitution, the same discipline the Bernoulli recipe's last step already asks for.