Loav · VCE Specialist Mathematics · Units 3 & 4
S34-E01 · Recommended sequence: Unit 4
1. A differential equation specifies a changing relationship
A differential equation links an unknown function with one or more of its derivatives. Instead of being given a formula for the function, you are told how it changes. The equation dy/dx = 3x² describes a rate depending only on x. The equation dy/dx = 2y describes a rate depending on the current value of y. The equation dy/dx = xy depends on both variables.
A solution is a differentiable function that satisfies the equation throughout a stated interval. A general solution describes a family, normally containing an arbitrary constant for a first-order equation. An initial condition such as y(0) = 4 selects a particular member, when the conditions support one. A solution must satisfy both the differential equation and the initial condition; neither check implies the other.
The word “initial” often refers to time zero, but the independent variable need not be time and the supplied point need not have input zero. Write y(x₀) = y₀ accurately. A constant described as arbitrary in a general solution becomes a determined number after applying enough valid conditions. The resulting formula still needs an interval on which it solves the original equation.
2. Verify before solving
To verify a proposed solution, differentiate it, substitute it into both sides of the equation and simplify until the two sides agree. Then check each condition separately. For y = Ce2x, the derivative is 2Ce2x = 2y, so every real C gives a solution of y′ = 2y. Only C = 4 gives y(0) = 4.
Numerical agreement at several points is useful for detecting mistakes, but does not establish an identity throughout an interval. An algebraic derivative check is stronger when one is available. A CAS can help simplify the difference y′ − F(x, y); a result of zero must still be read alongside domain restrictions and any assumptions used by the system.
Units also test a contextual equation. If a quantity Q is measured in litres and time in minutes, dQ/dt must have units litres per minute. In dQ/dt = kQ, the parameter k has units per minute. A sum of a volume and a rate without a conversion factor is not dimensionally meaningful, even if the algebra can be manipulated.
3. Direct integration is the simplest case
When y′ = f(x), antidifferentiate with respect to x: y = ∫f(x) dx. Retain the arbitrary constant and apply the supplied condition. For y′ = 6x − 4 and y(1) = 5, integration gives y = 3x² − 4x + C, then C = 6. The function is therefore 3x² − 4x + 6.
Use the same integration techniques as in the integration guides, including substitution or recognition when appropriate. The presence of derivative notation does not change antiderivative rules. Conversely, do not differentiate the supplied rate again when the task asks you to recover the original quantity. A rate-to-quantity problem requires integration and a starting value.
A definite-integral form can preserve the initial condition neatly: y(x) = y₀ + ∫ from x₀ to x of f(u) du. The letter u is a dummy integration variable, distinct from the upper endpoint x. This form makes the accumulated change interpretation explicit and can remain useful when the integral is evaluated numerically.
4. Separate variables without losing solutions
For y′ = f(x)g(y), first check for constant equilibrium solutions. Any allowed number a with g(a) = 0 gives y = a. These solutions would disappear if you divided by g(y), so record them before proceeding. On a region where g(y) ≠ 0, rearrange to dy/g(y) = f(x) dx and integrate both sides.
The separation notation is a convenient expression of the chain rule. If G′(y) = 1/g(y), then d[G(y(x))]/dx = y′/g(y) = f(x). Integrating therefore yields G(y) = F(x) + C. It is not an unrestricted permission to treat every derivative symbol as an ordinary fraction in unrelated settings.
Keep logarithm absolute values when integrating 1/y: ln|y| is valid on an interval where y does not cross zero. Exponentiation produces |y| = AeF(x) with A > 0. A signed constant can absorb the positive and negative branches, giving y = CeF(x); the previously checked equilibrium y = 0 may then be included by C = 0. Explain when that recovery occurs rather than assuming every division is harmless.
An implicit solution may be a perfectly good final form if it describes the required branch and satisfies the question. If you solve explicitly, square roots and inverse circular functions may introduce branch choices. Use the initial condition and continuity to select the correct branch on a connected interval. Do not switch signs arbitrarily as x varies.
Worked example 1: A separated equation with an initial point away from zero
Solve dy/dx = 2xy with y(1) = 3. First y = 0 is an equilibrium, although it cannot meet the given condition. For y ≠ 0, separate to dy/y = 2x dx. Integration gives ln|y| = x² + C, so y = Aex² for a nonzero signed constant A.
The condition gives 3 = Ae, hence A = 3/e. Thus y = 3ex²−1 for all real x. Its derivative is 6xex²−1, which equals 2xy, and its value at x = 1 is 3.
The solution stays positive because its multiplier and exponential are positive. It decreases for x < 0 and increases for x > 0, even though the function value itself is never negative. The sign of the rate depends on x as well as y; “exponential-looking” does not automatically mean increasing everywhere.
Worked example 2: An equilibrium and a finite interval of validity
Solve y′ = (y − 2)² with y(0) = 3. The equilibrium is y = 2. For other solutions, integrate (y − 2)⁻² dy = dx to obtain −1/(y − 2) = x + C. At the initial point, −1 = C, so y = 2 + 1/(1 − x).
The formula is undefined at x = 1. The connected solution interval containing the initial input is (−∞, 1). Its derivative is 1/(1 − x)², exactly (y − 2)², and its value at zero is 3. It increases and becomes unbounded as x approaches 1 from below.
The same algebraic expression is defined for x > 1, but that disconnected branch is not a continuation through the singularity of this initial-value solution. Record the equilibrium separately in the general family: division by (y − 2)² excluded it during separation. The differential equation itself is defined at y = 2, so the equilibrium is genuine.
5. Second derivative equations need two conditions
For y″ = f(x), integrate once to obtain y′ = F(x) + C₁, then again to obtain y = ∫F(x) dx + C₁x + C₂. Two constants appear because a second derivative loses both a constant vertical shift and a linear component. Use two independent conditions to determine them, commonly one value and one gradient.
Do not merge C₁x and C₂ into a single constant: the first changes with x. Also do not apply a condition on y to the expression for y′. Label each condition and substitute it into the appropriate equation. This guide concerns the direct form y″ = f(x); it does not require a general theory of second-order differential equations.
Worked example 3: Recover a curve from curvature and initial information
A curve has y″ = 6x − 4, y′(0) = 2 and y(0) = 1. Integrate once: y′ = 3x² − 4x + C₁. The slope condition gives C₁ = 2. Integrate again: y = x³ − 2x² + 2x + C₂. The value condition gives C₂ = 1.
Therefore y = x³ − 2x² + 2x + 1. Differentiating twice restores 6x − 4, while the two substitutions restore the supplied starting gradient and value. The rate y′ is always positive because 3x² − 4x + 2 = 3(x − 2/3)² + 2/3.
Although the curve increases everywhere, its concavity changes at x = 2/3. The equation for its second derivative describes changes of gradient, not the gradient's sign. This distinction links differential equations to the graph analysis in second derivatives and concavity.
6. State the interval and interpret the solution
Domain checking after integration includes denominator zeros, logarithm arguments, square-root signs and inverse-function branches. The interval should contain the initial input and should not cross a singularity of the original equation or the selected solution. A physical application can impose a narrower interval, such as nonnegative time before a container empties.
As a non-routine check, solve y′ = 2x(1 + y²), y(0) = 1. Separation gives arctan y = x² + π/4 locally, hence y = tan(x² + π/4). The connected interval containing zero is −√π/2 < x < √π/2, because the argument first reaches π/2 at those endpoints. A CAS output without this interval hides an important part of the solution.
For numerical checking, choose test inputs inside the claimed interval, compare the original right-hand side with the derivative, and confirm the initial point exactly. If a graph appears to jump between branches, inspect the singularities. The next guides develop contextual differential-equation models and slope fields and Euler approximation, which make the same solution conditions visible in different ways.
Practise, then check
- Solve y′ = 4x + 3 with y(2) = 1.
- Verify y = 5e−2x solves y′ = −2y and state its initial value at zero.
- Find the general solutions of y′ = 3y, including any equilibrium.
- Solve y′ = x/y with y(0) = −2, stating the selected branch.
- Solve y′ = y² with y(0) = 2 and give the connected interval containing zero.
- Find y when y″ = 12x², y′(0) = −1 and y(1) = 4.
- A student divides y′ = y(y − 4) by y(y − 4). Which solutions must be checked separately, and why?
Fully explained answers
- Integration gives y = 2x² + 3x + C. The condition gives 1 = 8 + 6 + C, so C = −13. Thus y = 2x² + 3x − 13 for all real x. Differentiation and substitution verify both requirements.
- The derivative is −10e−2x, while −2y is the same expression. Hence the equation holds for every real x. The value at zero is 5. A different multiplier would still solve the equation but would change that initial value.
- For y ≠ 0, separation gives ln|y| = 3x + C, hence y = Ae3x. The equilibrium y = 0 is also valid and is included if A is allowed to be any real number, including zero. State this recovery explicitly.
- Multiplying by y gives y dy = x dx, so y² = x² + C. The initial condition gives C = 4 and selects y = −√(x² + 4). It never reaches zero, so the original denominator is valid for all real x. The positive root fails the initial condition.
- Separate and integrate to obtain −1/y = x + C. Since y(0) = 2, C = −1/2, giving y = 2/(1 − 2x). Its singularity is x = 1/2, so the connected interval containing zero is (−∞, 1/2). The equilibrium zero does not meet this condition.
- First y′ = 4x³ + C₁; the slope condition makes C₁ = −1. Then y = x⁴ − x + C₂. Substituting x = 1 gives C₂ = 4, so y = x⁴ − x + 4. Two integrations require two separately determined constants.
- The constant functions y = 0 and y = 4 make the original right-hand side zero, matching their derivatives. Division excludes both, so they must be recorded before separation and included in the full solution discussion. Their exclusion is an algebraic side effect, not a reason to discard them.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4, Areas of Study 4 as applicable, with Outcomes 1–3. Scope follows the VCAA Mathematics Study Design 2023–2027 and the current Specialist Mathematics FAQ. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 11A–11C (pp478–492);11F (pp507–510);11H (pp516–517). All teaching prose, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
