Differential Equations and Modelling

A differential equation relates a quantity to its own rate of change. That is how nearly every physical law is stated, which is why solving them matters.

What a differential equation is

An equation containing a derivative. Its order is the highest derivative present:

$$\frac{dy}{dx} = 2x \qquad \text{(first order)} \qquad \frac{d^2y}{dx^2} + 4y = 0 \qquad \text{(second order)}$$

The simplest kind is solved by integrating directly:

$$\frac{dy}{dx} = 2x \quad \Rightarrow \quad y = x^2 + C$$

The $+C$ makes this the general solution — a whole family of curves. A boundary condition such as $y = 5$ when $x = 1$ picks one out: $5 = 1 + C$, so $C = 4$ and $y = x^2 + 4$. That is the particular solution.

Separation of variables

If the equation can be written as $\dfrac{dy}{dx} = f(x)g(y)$, gather the $y$'s on one side and the $x$'s on the other, then integrate both sides.

Solve $\dfrac{dy}{dx} = xy$.

$$\frac{1}{y}\,dy = x\,dx$$ $$\int \frac{1}{y}\,dy = \int x\,dx$$ $$\ln|y| = \frac{x^2}{2} + C$$ $$y = A e^{x^2/2}$$

where $A = e^C$ absorbs the constant.

The growth and decay equation. $\dfrac{dy}{dt} = ky$ separates the same way:

$$\ln|y| = kt + C \quad \Rightarrow \quad y = Ae^{kt}$$

with $A$ the value at $t = 0$. Positive $k$ gives growth; negative $k$ gives decay.

The integrating factor

For a first-order linear equation

$$\frac{dy}{dx} + P(x)\,y = Q(x)$$

multiply through by the integrating factor

$$\text{IF} = e^{\int P\,dx}$$

The left side then becomes an exact derivative:

$$\frac{d}{dx}\big(\text{IF} \times y\big) = \text{IF} \times Q$$

so integrating both sides finishes the job.

Solve $\dfrac{dy}{dx} + 2y = e^x$.

$$\text{IF} = e^{\int 2\,dx} = e^{2x}$$ $$\frac{d}{dx}\left(e^{2x}y\right) = e^{2x} \cdot e^x = e^{3x}$$ $$e^{2x}y = \frac{e^{3x}}{3} + C$$ $$y = \frac{e^x}{3} + Ce^{-2x}$$

Check: $y' + 2y = \left(\frac{e^x}{3} - 2Ce^{-2x}\right) + \left(\frac{2e^x}{3} + 2Ce^{-2x}\right) = e^x$ ✓

Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = x$.

$$\text{IF} = e^{\int \frac1x dx} = e^{\ln x} = x$$ $$\frac{d}{dx}(xy) = x^2 \quad \Rightarrow \quad xy = \frac{x^3}{3} + C \quad \Rightarrow \quad y = \frac{x^2}{3} + \frac{C}{x}$$

Second-order equations

For $a\dfrac{d^2y}{dx^2} + b\dfrac{dy}{dx} + cy = 0$, try $y = e^{mx}$. This produces the auxiliary equation

$$am^2 + bm + c = 0$$

and the shape of the solution depends on its roots:

Solve $y'' - 5y' + 6y = 0$.

$$m^2 - 5m + 6 = 0 \quad \Rightarrow \quad (m-2)(m-3) = 0$$ $$y = Ae^{2x} + Be^{3x}$$

Solve $y'' + 4y' + 4y = 0$.

$$m^2 + 4m + 4 = (m+2)^2 = 0$$

A repeated root $m = -2$, so

$$y = (A + Bx)e^{-2x}$$

Solve $y'' + 4y = 0$.

$$m^2 + 4 = 0 \quad \Rightarrow \quad m = \pm 2i$$

Here $\alpha = 0$ and $\beta = 2$:

$$y = A\cos 2x + B\sin 2x$$

— an oscillation, which is why complex roots describe vibrating systems.

Particular integrals

When the right-hand side is not zero,

$$\text{general solution} = \text{complementary function} + \text{particular integral}$$

The complementary function solves the equation with zero on the right; the particular integral is any single solution of the full equation. Guess its form from the right-hand side:

Solve $y'' - 5y' + 6y = 12$.

Complementary function: $Ae^{2x} + Be^{3x}$, from above.

Particular integral: try $y = k$, so $y' = y'' = 0$ and $6k = 12$, giving $k = 2$.

$$y = Ae^{2x} + Be^{3x} + 2$$

Modelling

Exponential growth and decay. "The rate of change is proportional to the amount present" is

$$\frac{dP}{dt} = kP \quad \Rightarrow \quad P = P_0e^{kt}$$

Population growth, compound interest and radioactive decay are all this equation, with $k$ negative for decay.

Newton's law of cooling. "The rate of cooling is proportional to the excess temperature over the surroundings":

$$\frac{d\theta}{dt} = -k(\theta - \theta_s)$$

Separating gives

$$\theta = \theta_s + Ae^{-kt}$$

so the temperature approaches $\theta_s$ but never quite reaches it — the object cools quickly at first, then ever more slowly, which matches what a cup of coffee actually does.