An equation about rates of change can be pulled apart, integrated piece by piece, and reassembled into the function it was describing all along.
10 min read4 worked examples
Prerequisite
Connected Rates of Change in the Differentiation module covers how a differential equation like $\dfrac{dV}{dt}=kV$ gets set up from a real-world description, often via the chain rule linking several changing quantities. This lesson picks up from an already-formed differential equation and focuses on solving it.
Separating the variables
A first-order differential equation of the form $\dfrac{dy}{dx}=f(x)g(y)$ is called separable, because all the $y$-terms can be gathered on one side and all the $x$-terms on the other. Treating $\dfrac{dy}{dx}$ algebraically as a fraction (which is valid here, since it falls out of the chain rule when integrating both sides),
Both sides are then integrated independently — with respect to $y$ on the left and $x$ on the right. Since two separate integrations are involved, only a single constant of integration is needed overall (the two constants can be combined into one, conventionally written on the right-hand side).
General vs. particular solutions
Integrating both sides produces a general solution: a whole family of curves, one for each value of the arbitrary constant $c$. A boundary condition — a known point $(x_0,y_0)$ that the solution must pass through — pins down the exact value of $c$, giving a single particular solution.
Working order
1) Separate the variables. 2) Integrate both sides, adding one constant $c$. 3) Substitute the boundary condition to find $c$. 4) Rearrange to express $y$ explicitly in terms of $x$, if required.
Interpreting the solution's limitations
A differential equation is usually a model of some real situation — population growth, cooling, chemical concentration — so the algebraic solution should always be checked against the context it came from. Solutions are often only valid for a restricted domain (e.g. $t\ge0$ for a time variable), and predictions far outside the range where the model was fitted (extrapolation) should be treated with caution, since the underlying assumptions may no longer hold.
Worked example 1
Finding a general solution
Find the general solution of $\dfrac{dy}{dx}=\dfrac{x}{y}$, for $y>0$.
MethodSeparate the variables, then integrate each side
1
Separate: $y\,dy = x\,dx$.
Multiply both sides by y and by dx to gather like terms.
2
$\displaystyle\int y\,dy = \int x\,dx$.
Integrate both sides independently.
3
$\dfrac{y^2}{2} = \dfrac{x^2}{2}+c$.
A single constant suffices for both integrations combined.
4
$y^2 = x^2+2c$, or writing $A=2c$: $y^2=x^2+A$.
Multiply through by 2 and rename the constant for a tidier form.
5
Since $y>0$: $y=\sqrt{x^2+A}$.
Take the positive square root, consistent with the given restriction.
▪ solved
A family of curves
Each value of $A$ traces out a different curve $y=\sqrt{x^2+A}$ — the general solution describes an entire family, all satisfying the same differential equation, differing only by a vertical shift-like parameter.
Worked example 2
Finding a particular solution
Given that $\dfrac{dy}{dx}=6x^2y$ and $y=4$ when $x=0$, find $y$ in terms of $x$.
MethodSeparate, integrate, then apply the boundary condition
For the model in Example 3 (where $P$ represents a population in thousands, $t$ in years), state what happens to $P$ as $t\to\infty$, and comment on a limitation of the model.
MethodAnalyse the long-term behaviour of the solution, then evaluate the model's realism
1
As $t\to\infty$, $e^{10t}\to\infty$, so $P=\dfrac{10e^{10t}}{4+e^{10t}} \to \dfrac{10e^{10t}}{e^{10t}} = 10$.
Divide numerator and denominator by e^(10t) to see the dominant behaviour.
2
So the population approaches a long-term limit of $P=10$ (i.e. 10,000), never quite reaching or exceeding it.
This matches P(10 − P) = 0 at P = 10, a natural "capacity" where growth stops.
3
This behaviour is consistent with the differential equation itself: when $P$ is close to $10$, $\dfrac{dP}{dt}=P(10-P)$ is close to zero, so growth slows to a halt near this ceiling.
A sanity check that the algebraic solution matches the original equation's structure.
4
Limitation: the model assumes continuous, smooth growth with a fixed carrying capacity of exactly 10 — real populations are affected by discrete events (disease, migration, resource shocks) that this simple model cannot capture, especially over long time horizons.
Solutions to real-world differential equations should always be interpreted with their modelling assumptions in mind.
▪ solved
Look deeper — the shape of the family
Different boundary conditions (different starting populations) all produce curves approaching the same limit $P=10$ as $t\to\infty$, but starting populations above $10$ would decrease towards it instead — the general solution describes this entire family of possible behaviours, of which the particular solution is just one member.
Common mistake
Adding a constant of integration to both sides after separating the variables. Since the two constants can always be combined into one (by subtracting one from the other), only a single constant $c$ should appear in the working — usually written on the side with the "simpler" variable, by convention.
Check yourself
Given $\dfrac{dy}{dx}=\dfrac{2y}{x}$ for $x>0,\,y>0$, and $y=3$ when $x=1$, find $y$ in terms of $x$. (Hint: separate the variables, integrate, then use the boundary condition to find c.)