A differential equation (DE) is an equation that involves an unknown function and one or more of its derivatives. In VCE Specialist Mathematics, the focus is on first-order differential equations, where the highest derivative present is the first derivative, $\frac{dy}{dx}$.
To verify if a given function is a solution to a differential equation, substitute the function and its derivative into the DE to check if the left-hand side (LHS) equals the right-hand side (RHS).
Example:
Verify that $y = Ae^x - x - 1$ is a solution to $\frac{dy}{dx} = x + y$.
1. Find the derivative: $\frac{dy}{dx} = Ae^x - 1$.
2. Substitute $y$ into the RHS: $x + (Ae^x - x - 1) = Ae^x - 1$.
3. Since LHS = RHS, the solution is verified.
EXAM TIP: When verifying solutions in “Show that” questions, clearly label your LHS and RHS substitutions to ensure full marks for mathematical communication.
A first-order differential equation is separable if it can be written in the form:
$$\frac{dy}{dx} = f(x)g(y)$$
To solve these equations, we rearrange the terms so that all $y$ terms are on one side with $dy$, and all $x$ terms are on the other with $dx$:
1. Rewrite the equation: $\frac{1}{g(y)} dy = f(x) dx$
2. Integrate both sides: $\int \frac{1}{g(y)} dy = \int f(x) dx$
3. Solve for $y$ if possible to express the solution explicitly.
| Form | Solution Method |
|---|---|
| $\frac{dy}{dx} = f(x)$ | Direct integration: $y = \int f(x) dx$ |
| $\frac{dy}{dx} = g(y)$ | Invert and integrate: $x = \int \frac{1}{g(y)} dy$ |
| $\frac{dy}{dx} = f(x)g(y)$ | Separate variables: $\int \frac{1}{g(y)} dy = \int f(x) dx$ |
COMMON MISTAKE: Forgetting the constant of integration $+c$ during the integration step. This constant must be added before rearranging for $y$.
A first-order linear differential equation has the standard form:
$$\frac{dy}{dx} + P(x)y = Q(x)$$
To solve equations in this form, we use an integrating factor, $I(x)$:
1. Ensure the DE is in the standard form (the coefficient of $\frac{dy}{dx}$ must be 1).
2. Calculate the integrating factor: $I(x) = e^{\int P(x) dx}$. (Note: we do not need $+c$ here).
3. Multiply the entire DE by $I(x)$: $I(x)\frac{dy}{dx} + I(x)P(x)y = I(x)Q(x)$.
4. Recognize the LHS as the derivative of a product: $\frac{d}{dx}[I(x)y] = I(x)Q(x)$.
5. Integrate both sides with respect to $x$: $I(x)y = \int I(x)Q(x) dx$.
6. Solve for $y$.
KEY TAKEAWAY: The integrating factor method turns the left-hand side of a non-separable linear equation into the result of a Product Rule expansion, allowing for direct integration.
A slope field is a graphical representation of a differential equation $\frac{dy}{dx} = f(x, y)$. It consists of short line segments at various points $(x, y)$ in the plane, where the gradient of each segment is equal to the value of the derivative at that point.
VCAA FOCUS: You may be asked to identify which differential equation matches a given slope field. Look for regions where the gradient depends only on $x$ (vertical columns of identical slopes) or only on $y$ (horizontal rows of identical slopes).
Euler’s Method is a numerical technique used to find approximate coordinates of points on a solution curve when an exact solution is difficult to find.
Given $\frac{dy}{dx} = f(x, y)$ and an initial point $(x_0, y_0)$ with step size $h$:
$$x_{n+1} = x_n + h$$
$$y_{n+1} = y_n + h \cdot f(x_n, y_n)$$
STUDY HINT: Think of Euler’s method as “New $y$ = Old $y$ + (Step size $\times$ Gradient at old point)”. It is essentially using a sequence of small linear approximations.
Differential equations are used to model rates of change in real-world scenarios.
APPLICATION: In mixing problems (e.g., salt in a tank), the concentration of the outflow changes as the amount of salt $Q$ changes. Therefore, $\text{Rate Out} = \frac{Q(t)}{V(t)} \times \text{Flow Out}$, where $V(t)$ is the volume of liquid at time $t$.
