Separable differential equations

When separation is needed

• $\dfrac{dy}{dx} = f(x)$ can be integrated directly, because the right-hand side contains only $x$.
• $\dfrac{dy}{dx} = ky$ cannot — the right-hand side contains $y$, and you cannot integrate $y$ with respect to $x$ without knowing what $y$ is.
• Separation of variables rearranges the equation so that each side contains only one variable.

The method

• Separate — get all the $y$ terms with $dy$ on one side, and all the $x$ terms with $dx$ on the other.
• Integrate both sides, each with respect to its own variable.
• Include one constant, conventionally on the $x$ side.
• Apply the initial condition to find the constant.
• Rearrange to make $y$ the subject if the question asks for it.

$$\frac{dy}{dx} = \frac{g(x)}{h(y)} \quad \Longrightarrow \quad \int h(y)\,dy = \int g(x)\,dx$$

The standard case: $\dfrac{dy}{dx} = ky$

• This is by far the most common form, and it deserves to be known by heart.

• Separate:

$$\frac{1}{y}\,dy = k\,dx$$

• Integrate both sides:

$$\ln|y| = kx + c$$

• Exponentiate both sides to free $y$:

$$|y| = e^{kx + c} = e^{c}e^{kx}$$

• $e^c$ is just another positive constant. Absorbing the sign as well and calling it $A$:

$$y = Ae^{kx}$$

• Each part:

  • $A$ — the value of $y$ when $x = 0$, since $e^0 = 1$
  • $k$ — the rate constant; positive means growth, negative means decay

• This result is worth memorising. Any equation of the form "rate of change is proportional to the amount present" has an exponential solution.

Handling the constant carefully

• The constant can be carried in two equivalent ways:
  • Leave it as $+c$ inside the logarithm stage, then exponentiate: $y = e^c e^{kx}$
  • Convert to a multiplicative constant $A$ immediately
• Be consistent. The 2025 report warns: "care must be taken to use the constant of integration consistently throughout a solution."
• Do not write $y = e^{kx} + c$. Exponentiating a sum gives a product, not a sum — that is the single most common error in this topic.

Separating when both sides are more complicated

• Worked through — solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y = 3$ when $x = 0$:
  • Separate: $y\,dy = x\,dx$
  • Integrate: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + c$
  • Apply the condition: $\dfrac{9}{2} = 0 + c$, so $c = \dfrac{9}{2}$
  • Rearrange: $y^2 = x^2 + 9$, so $y = \sqrt{x^2 + 9}$
  • Note that the positive root is chosen because $y = 3 > 0$ at the initial condition, and $y$ is continuous.

Choosing the sign of a square root

• Separation often produces $y^2 = \ldots$, and the square root has two branches.
• Use the initial condition to decide which branch applies, and say why.
• The solution cannot jump between branches, because it would have to pass through $y = 0$, where the original equation is usually undefined.

Two conditions: finding $k$ as well

• Applied questions often give two pieces of information, because both $A$ and $k$ are unknown:
  • The first (usually at $t = 0$) gives $A$.
  • The second (a later measurement) gives $k$, and finding it needs logarithms.
• The 2026 specification notes that "solving differential equations may require manipulation of exponents and logarithms".