Recipe

Solve a separable first-order equation

Use when: the equation can be written as \dfrac{dy}{dx}=f(x)g(y) — every y-term collects on one side, every x-term on the other

Goal: an implicit or explicit solution for y, plus any equilibrium solutions the separation step would otherwise lose

Taught in Separable equations

1. 1

   Rewrite \dfrac{dy}{dx}=f(x)g(y) as \dfrac{dy}{g(y)}=f(x)\,dx.

   Example

   Take \dfrac{dy}{dx}=xy(2-y), so f(x)=x and g(y)=y(2-y). Separating: \dfrac{dy}{y(2-y)}=x\,dx

   Why: this is legal exactly because the original equation already factors into a pure function of x times a pure function of y — nothing besides regrouping happens here.

2. 2

   Dividing by g(y) assumes g(y)\ne0; find every root of g(y)=0 before moving on.

   Example

   Here g(y)=y(2-y), so g(y)=0 exactly when y=0 or y=2. Both roots have to be checked against the original equation once the rest of the recipe is done.

   Why: each root of g(y)=0 gives a constant solution y=c that the separation step silently excludes — it has to be checked separately, not discovered by accident.

3. 3

   Integrate \displaystyle\int\dfrac{dy}{g(y)}=\int f(x)\,dx independently.

   Example

   \dfrac1{y(2-y)} splits by partial fractions into \dfrac12\left(\dfrac1y+\dfrac1{2-y}\right), so \int\dfrac12\left(\dfrac1y+\dfrac1{2-y}\right)dy=\int x\,dx \;\Longrightarrow\; \dfrac12\ln|y|-\dfrac12\ln|2-y|=\dfrac{x^2}{2}+C_1 Multiplying by 2 and combining the logs: \ln\left|\dfrac{y}{2-y}\right|=x^2+C_2, so \dfrac{y}{2-y}=Ae^{x^2} for a renamed constant A.

   Why: once the variables are apart, each side is an ordinary single-variable integral.

4. 4

   Isolate y only if the algebra stays clean; an implicit equation in x and y is a complete, legitimate answer on its own.

   Example

   Here the algebra does stay clean: cross-multiplying \dfrac{y}{2-y}=Ae^{x^2} gives y=Ae^{x^2}(2-y), so y+Ae^{x^2}y=2Ae^{x^2}, and y=\dfrac{2Ae^{x^2}}{1+Ae^{x^2}} If y(0)=1 is given, substituting x=0,\,y=1 gives 1=\dfrac{2A}{1+A}, so A=1, and \boxed{y=\dfrac{2e^{x^2}}{1+e^{x^2}}}

   Why: forcing an implicit relation into y=f(x) form is a separate task from solving the equation, not a required last step.

5. 5

   Substitute each root of g(y)=0 from Step 2 back into the ORIGINAL, unseparated equation.

   Example

   Substituting y=0: left side \dfrac{dy}{dx}=0, right side x\cdot0\cdot(2-0)=0 — solves it. Substituting y=2: left side 0, right side x\cdot2\cdot(2-2)=0 — also solves it. So y=0 and y=2 are genuine equilibrium solutions, alongside the family found in Step 4.

   Why: this is what confirms whether a constant solution is genuinely new, rather than already hiding inside the general family for some value of C.