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 \(t\).
