MISSION · Calculus
Differential Equations
When you know how fast something changes but not what it is
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The idea in one sentence
A differential equation relates a function to its own derivatives, and solving it means finding the function — usually a whole family of them, narrowed down by one extra fact.
Why they’re everywhere
Almost no law of nature says what a quantity is. They say how it changes.
- Population grows in proportion to its size: \(\dfrac{dP}{dt} = kP\)
- A hot drink cools in proportion to the temperature difference: \(\dfrac{dT}{dt} = -k(T - T_{room})\)
- Force is mass times acceleration, and acceleration is a second derivative.
Every one of these is a differential equation, and in each case what you want — the population, the temperature, the position — is the thing you have to solve for.
The simplest kind
\[\frac{dy}{dx} = 2x\]Just integrate: \(y = x^2 + C\). The \(C\) isn’t a technicality — it says that infinitely many curves have this gradient everywhere, all vertical shifts of each other. Knowing the gradient fixes the shape but not the height.
An initial condition picks one out. If \(y = 5\) when \(x = 0\), then \(C = 5\). The general solution is the family; the particular solution is the member you wanted.
Separable equations
When the derivative is a product of a function of \(x\) and a function of \(y\), you can separate them:
\[\frac{dy}{dx} = xy \quad \Rightarrow \quad \frac{1}{y}dy = x\,dx\]Integrate both sides:
\[\ln|y| = \frac{x^2}{2} + C \quad \Rightarrow \quad y = Ae^{x^2/2}\]where \(A = e^C\) absorbs the constant. Treating \(dy\) and \(dx\) as things you can move around is informal, and it can be justified properly — the result is right, and it’s the standard working.
The growth equation
\[\frac{dy}{dx} = ky \quad \Rightarrow \quad y = y_0 e^{kx}\]This is the most important differential equation in the subject, and it deserves reading as a sentence: a quantity that grows in proportion to its own size grows exponentially. Compound interest, bacterial cultures, radioactive decay (\(k < 0\)), and the initial phase of an epidemic all run on this one equation.
For \(\frac{dy}{dx} = 2y\) with \(y(0) = 5\), the solution is \(y = 5e^{2x}\).
Slope fields
You can see the solutions without solving anything. At a grid of points, draw a short segment with gradient \(\frac{dy}{dx}\) as given by the equation. The result is a field of tick marks, and a solution curve is any curve that follows them.
The picture makes two things obvious: the solutions form a family of non-crossing curves, and an initial condition selects one by naming a point it must pass through. It also works for equations with no formula solution at all, which is most of them.
When there’s no formula
\(\frac{dy}{dx} = x^2 + y^2\) has no solution in elementary functions. This is the normal case. Real work uses numerical methods — Euler’s method steps along the slope field in small increments, and better methods refine the same idea. That’s what weather models and orbit calculations are doing.
Common mix-ups
- Dropping the constant of integration, or adding it after taking exponentials.
- Applying the initial condition too late, after already discarding the constant.
- Separating what can’t be separated. \(\frac{dy}{dx} = x + y\) won’t separate; it needs an integrating factor.
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Forgetting the modulus in $$\ln y $$, which loses the negative solutions. - Reading \(k < 0\) as impossible. Negative \(k\) is decay, and it’s the same equation.
Check yourself
Solve \(\frac{dy}{dx} = 2y\) with \(y(0) = 5\), and separate \(\frac{dy}{dx} = xy\).
Proportional growth gives \(y = 5e^{2x}\), so k = 2. Separating the second gives \(\int \frac{1}{y}dy = \int x\,dx\), so \(\ln|y| = x^2/2 + C\).
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