IB Math AA: Differential Equations

Differential equations are the mathematical language of change, allowing you to model everything from population dynamics to the cooling of your morning coffee. In IB Math Analysis and Approaches HL, mastering first-order differential equations is not just an exam requirement—it's about developing a powerful toolkit for translating real-world processes into solvable mathematics, moving from a general equation to a specific, predictive function.

What is a Differential Equation?

A differential equation is an equation that relates a function to its derivatives. In essence, it describes how a quantity changes rather than stating what the quantity is directly. Your goal is to "solve" it, which means finding the original function, often called the general solution. The order of a differential equation is determined by the highest derivative present; this article focuses exclusively on first-order equations, where the highest derivative is the first derivative ($dy/dx$).

These equations are typically presented in forms like $dy/dx=f(x,y)$. The solution will involve a constant of integration, usually denoted $C$. To find a particular solution—a single, specific function—you must use an initial condition (e.g., $y(0)=5$) to solve for the exact value of $C$. This process transforms a family of possible curves into the one curve that fits the observed data.

Solving Separable Differential Equations

The first and most straightforward technique you must master is for separable differential equations. An equation is separable if you can algebraically manipulate it to get all terms involving $y$ (and $dy$) on one side and all terms involving $x$ (and $dx$) on the other. The standard form is: $$\frac{dy}{dx}=f(x)g(y)$$

The solution method is a direct application of integration:

1. Separate the variables: $\frac{1}{g(y)}dy=f(x)dx$. Assume $g(y)\ne 0$.
2. Integrate both sides: $\int \frac{1}{g(y)}dy=\int f(x)dx$.
3. Perform the integrations, which yields an equation involving $y$, $x$, and the constant $C$.
4. If possible, rearrange the result to express $y$ explicitly as a function of $x$.

Example: Solve $\frac{dy}{dx}=2xy$, given $y(0)=3$.

1. Separate: $\frac{1}{y}dy=2xdx$ (assuming $y\ne 0$).
2. Integrate: $\int \frac{1}{y}dy=\int 2xdx\Rightarrow \ln |y|=x^2+C$.
3. Solve for $y$: $|y|=e^{x^2+C}=e^C{e}^{x^2}$. Let $A=\pm e^C$, so the general solution is $y=A{e}^{x^2}$.
4. Apply the initial condition $y(0)=3$: $3=A{e}^0\Rightarrow A=3$.
5. The particular solution is $y=3e^{x^2}$.

Solving First-Order Linear Differential Equations

The next key type is the first-order linear differential equation. Its standard form is: $$\frac{dy}{dx}+P(x)y=Q(x)$$ Here, $P(x)$ and $Q(x)$ are functions of $x$ only. The equation is "linear" because $y$ and its derivative appear to the first power and are not multiplied together. The powerful, standard method for solving these involves an integrating factor, $I(x)$.

The algorithm is systematic:

1. Ensure the equation is in standard form: $dy/dx+P(x)y=Q(x)$.
2. Compute the integrating factor: $I(x)=e^{\int P(x)dx}$. You do not need a constant of integration here.
3. Multiply every term in the equation by $I(x)$: $I(x)\frac{dy}{dx}+I(x)P(x)y=I(x)Q(x)$.
4. The left-hand side is now the derivative of a product: $\frac{d}{dx}[I(x)y]=I(x)Q(x)$.
5. Integrate both sides with respect to $x$: $I(x)y=\int I(x)Q(x)dx+C$.
6. Finally, solve for $y$ by dividing by $I(x)$.

Example: Solve $x\frac{dy}{dx}+2y=x^3$, for $x>0$.

1. Convert to standard form: $\frac{dy}{dx}+\frac{2}{x}y=x^2$.
2. Here, $P(x)=2/x$. The integrating factor is $I(x)=e^{\int (2/x)dx}=e^{2\ln x}=x^2$.
3. Multiply through: $x^2\frac{dy}{dx}+2xy=x^4$.
4. The left side is $\frac{d}{dx}(x^{2}y)=x^4$.
5. Integrate: $x^{2}y=\int x^{4}dx=\frac{x^5}{5}+C$.
6. Solve for $y$: $y=\frac{x^3}{5}+\frac{C}{x^2}$.

Modeling Real-World Phenomena

The true power of differential equations lies in their ability to model dynamic systems. You will encounter several canonical models in the IB syllabus.

Exponential Growth and Decay (Population & Radioactivity): This is the simplest model, governed by the law that the rate of change is proportional to the current amount. For a population $P$, the model is $dP/dt=kP$, where $k>0$ for growth and $k<0$ for decay (as in radioactive decay). This is a separable equation whose solution is the familiar exponential function $P(t)=P_0{e}^{kt}$, where $P_0$ is the initial population.

Newton's Law of Cooling: This states that the rate of change of an object's temperature ($T$) is proportional to the difference between its temperature and the ambient temperature ($T_s$). The model is $dT/dt=-k(T-T_s)$, where $k>0$. This is also separable (or can be treated as linear), leading to a solution of the form $T(t)=T_s+(T_0-T_s)e^{-kt}$, showing exponential decay towards $T_s$.

The Logistic Model (Limited Population Growth): The exponential model is unrealistic long-term. The logistic differential equation introduces a carrying capacity $K$: $dP/dt=rP(1-P/K)$. This separable equation models S-shaped growth, starting exponentially but slowing as $P$ approaches $K$.

Mixing Problems: A classic application involves a tank with a liquid of a certain concentration, where a second liquid flows in and the mixture flows out. You typically model the rate of change of the amount of a substance (e.g., salt) in the tank ($A(t)$). The general principle is: $$\frac{dA}{dt}=\text{Rate In}-\text{Rate Out}$$ "Rate In" is (concentration of inflow) × (flow rate in). "Rate Out" is (current concentration in the tank, which is $A(t)/\text{Volume}(t)$) × (flow rate out). This often sets up a first-order linear differential equation.

Common Pitfalls

1. Misapplying Separation of Variables: The most common error is trying to separate variables when the equation is not separable. For example, $dy/dx=x+y$ cannot have all $y$ terms moved to one side. You must recognize this and consider if it is linear instead.
2. Forgetting the Constant of Integration (or Misplacing It): When integrating, the $+C$ is essential. A frequent mistake is adding it only on one side or forgetting it entirely, which makes finding a particular solution impossible. Remember: an indefinite integral on each side requires one constant; combine them into a single $C$ on one side.
3. Incorrect Integrating Factor: For first-order linear equations, the integrating factor must be $e^{\int P(x)dx}$ precisely. Errors arise from using the wrong $P(x)$ (ensure the equation is in standard form first) or incorrectly integrating $P(x)$.
4. Neglecting Domain and Initial Conditions: The solution $y=A{e}^{x^2}$ is valid, but if the initial condition is $y(0)=-3$, then $A=-3$. Many students incorrectly assume $A>0$. Also, in separation of variables, if you divide by $g(y)$, you must consider the case where $g(y)=0$ as a possible constant solution.

Summary

• A differential equation relates a function to its derivatives. Solving it involves integration, producing a general solution with a constant $C$. An initial condition allows you to find a particular solution.
• Separable equations of the form $dy/dx=f(x)g(y)$ are solved by separating variables and integrating: $\int \frac{1}{g(y)}dy=\int f(x)dx$.
• First-order linear equations in the form $dy/dx+P(x)y=Q(x)$ are solved using an integrating factor $I(x)=e^{\int P(x)dx}$. The solution is found from $I(x)y=\int I(x)Q(x)dx+C$.
• These equations model key real-world phenomena: unrestricted growth/decay ($dP/dt=kP$), Newton's Law of Cooling ($dT/dt=-k(T-T_s)$), limited logistic growth ($dP/dt=rP(1-P/K)$), and mixing problems built on the balance law $dA/dt=\text{Rate In}-\text{Rate Out}$.
• Always check if an equation is separable first. If not, check if it is linear. Carefully track constants of integration and apply initial conditions precisely to find the correct particular solution.