UK A-Level: Differential Equations

Differential equations are the mathematical language of change, describing how quantities evolve over time or space. Mastering them is crucial for A-Level success and provides the foundation for understanding everything from population dynamics to the cooling of a cup of tea. This guide focuses on the essential technique of separation of variables, empowering you to solve key first-order equations and construct models from worded scenarios.

1. What is a First-Order Separable ODE?

A differential equation is an equation that involves an unknown function and its derivatives. A first-order equation involves only the first derivative (e.g., $dy/dx$). It is termed separable if you can algebraically manipulate it to get all terms involving the dependent variable (usually $y$) on one side and all terms involving the independent variable (usually $x$) on the other.

The general form of a separable differential equation is: $$\frac{dy}{dx}=f(x)g(y).$$ Here, the rate of change of $y$ with respect to $x$ is expressed as a product of a function of $x$ and a function of $y$. The core idea of the separation of variables technique is to "separate" these parts to allow integration. We treat $dy/dx$ as a fraction (a useful heuristic at this level) and rearrange: $$\frac{1}{g(y)}dy=f(x)dx.$$ We then integrate both sides: $$\int \frac{1}{g(y)}dy=\int f(x)dx.$$ This process yields a general solution, which will include a constant of integration, $C$.

Example: Solve $\frac{dy}{dx}=xy$.

1. Separate: $\frac{1}{y}dy=xdx$ (assuming $y\ne 0$).
2. Integrate: $\int \frac{1}{y}dy=\int xdx$.
3. Solve: $\ln |y|=\frac{1}{2}{x}^2+C$.
4. Solve for $y$: $|y|=e^{\frac{1}{2}{x}^2+C}=e^C{e}^{\frac{1}{2}{x}^2}$. Let $A=\pm e^C$, giving the general solution $y=A{e}^{\frac{1}{2}{x}^2}$.

2. Formulating Equations from Word Problems

A major application is formulating differential equations from word problems. The challenge is translating a textual description of a rate into a precise mathematical statement. Look for phrases like "the rate of change of [quantity] is proportional to..." or "...decreases at a rate proportional to itself."

Key Steps:

1. Define your variables. (e.g., Let $P$ be the population at time $t$).
2. Translate the rate. The phrase "the rate of change of $P$ with respect to $t$" is $dP/dt$.
3. Express the proportionality. "Is proportional to $P$" means $=kP$, where $k$ is a constant of proportionality.
4. Determine the sign of $k$. If a quantity is increasing, $k>0$; if it is decreasing, $k<0$.
5. State the initial condition. (e.g., When $t=0$, $P=P_0$).

For instance: "The number of bacteria in a culture grows at a rate proportional to the number present." This translates directly to the differential equation $dN/dt=kN$, where $N(t)$ is the number of bacteria and $k>0$.

3. Exponential Growth and Decay Models

The differential equation $dP/dt=kP$ is the fundamental exponential growth and decay model. Its solution, found via separation of variables, is $P(t)=P_0{e}^{kt}$, where $P_0$ is the initial value at $t=0$.

• Exponential Growth: Occurs when $k>0$. Examples include unrestricted population growth and compound interest.
• Exponential Decay: Occurs when $k<0$ (often written as $k=-\lambda$ where $\lambda >0$). Examples include radioactive decay and Newton's Law of Cooling.

Worked Example (Radioactive Decay): A radioactive substance decays at a rate proportional to its mass. Initially, there are 100g. After 10 years, 80g remain. Find the mass after 25 years.

1. Formulate: Let $m$ be mass at time $t$. $dm/dt=-km$ (decay implies negative rate).
2. Solve: Separate and integrate: $\int (1/m)dm=\int -kdt$. This gives $\ln m=-kt+C$, so $m=A{e}^{-kt}$. Using $m(0)=100$, we get $A=100$. Thus, $m(t)=100e^{-kt}$.
3. Find $k$: Use $m(10)=80$: $80=100e^{-10k}$. So $e^{-10k}=0.8$. Taking natural logs: $-10k=\ln (0.8)$, so $k=-\frac{1}{10}\ln (0.8)$.
4. Find $m(25)$: $m(25)=100e^{-25k}=100e^{2.5\ln (0.8)}=100(e^{\ln (0.8)}{)}^{2.5}=100\times (0.8{)}^{2.5}$. Evaluating gives $m(25)\approx 57.2$ grams.

4. Verifying Solutions by Substitution

A critical skill is verifying solutions by substitution into original equations. This checks your algebraic and integral work. To verify a proposed solution to a differential equation, you must:

1. Differentiate your solution to find $dy/dx$.
2. Substitute both your original solution for $y$ and your calculated derivative for $dy/dx$ into the original differential equation.
3. Confirm that the left-hand side (LHS) equals the right-hand side (RHS) for all $x$, or that the equation simplifies to an identity (like $0=0$).

Example: Verify that $y=\sqrt{x^2+9}$ is a solution to $xdx-ydy=0$.

1. Differentiate $y$: $dy/dx=x/\sqrt{x^2+9}=x/y$.
2. Substitute into $x-y(dy/dx)=0$: $x-y(x/y)=x-x=0$. The equation holds, so the solution is verified.

Common Pitfalls

1. Forgetting the Constant of Integration (and Its Domain): Always add $+C$ after integrating. In the final solution, $C$ or the constant $A=\pm e^C$ can often be positive or negative, or sometimes only positive, depending on the physical context (e.g., a population cannot be negative). Failing to consider this can lose you marks.

1. Incomplete Separation Before Integrating: A common error is attempting to integrate before fully separating the variables. For $\frac{dy}{dx}=x+xy$, you must factor to $\frac{dy}{dx}=x(1+y)$ first, then separate to $\frac{1}{1+y}dy=xdx$. Integrating $\int (x+xy)dx$ is incorrect.

1. Misapplying Initial Conditions: The constant of integration must be found by substituting the initial condition into the general solution, not into an intermediate step. For example, use the form $\ln |y|=\frac{1}{2}{x}^2+C$ with the initial condition to find $C$, not the later form $y=A{e}^{\frac{1}{2}{x}^2}$—though both are valid, the logarithmic form is often more straightforward.

1. Neglecting the Modulus in Logarithms: When integrating $\int (1/y)dy$, the result is $\ln |y|$, not just $\ln y$. For A-Level purposes, you can often drop the modulus later by absorbing the $\pm$ into the constant $A$, but you should show the step with the modulus to demonstrate understanding, especially when an initial condition might determine the sign.

Summary

• The separation of variables technique is the primary method for solving first-order separable ODEs of the form $dy/dx=f(x)g(y)$, involving rearranging and integrating both sides.
• Formulating differential equations from word problems hinges on correctly interpreting phrases describing rates and proportionalities to write an expression for $dy/dx$.
• The equation $dP/dt=kP$ models exponential growth ($k>0$) and decay ($k<0$), with the general solution $P(t)=P_0{e}^{kt}$.
• Always verify your solution by differentiating it and substituting both $y$ and $dy/dx$ back into the original differential equation to confirm it holds true.
• Avoid common algebraic and calculus errors by separating completely before integrating, carefully handling the constant, and applying initial conditions at the correct stage.