IB — mathsHigher1 min read

Differential Equations

Solve first-order differential equations by separation of variables for IB Maths.

Tutor AI Team
Tutor AI Team

Differential equations model real-world change. IB Maths focuses on separable first-order equations.

Separation of Variables

dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y) → 1g(y) dy=f(x) dx\frac{1}{g(y)}\,dy = f(x)\,dx → integrate both sides.

Worked Example: Example 1

Problem

dydx=xy\frac{dy}{dx} = xy. 1y dy=x dx\frac{1}{y}\,dy = x\,dx → ln⁡∣y∣=x22+C\ln|y| = \frac{x^2}{2} + C → y=Aex2/2y = Ae^{x^2/2}.

Solution

Worked Example: With Initial Condition

Problem

dydx=3y\frac{dy}{dx} = 3y, y(0)=2y(0) = 2. → y=2e3xy = 2e^{3x}.

Solution

Common Models

  • Exponential growth/decay: dydt=ky\frac{dy}{dt} = ky → y=y0ekty = y_0 e^{kt}.
  • Logistic: dPdt=kP(L−P)\frac{dP}{dt} = kP(L-P).
  • Newton's cooling: dTdt=−k(T−Ts)\frac{dT}{dt} = -k(T - T_s).

Practice Problems

    1. Solve dydx=xy\frac{dy}{dx} = \frac{x}{y}, y(0)=3y(0) = 3.
    1. A population satisfies dPdt=0.02P\frac{dP}{dt} = 0.02P, P(0)=500P(0) = 500. Find P(10)P(10).

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Key Takeaways

  • ✓

    Separate variables → integrate → apply initial conditions.

  • ✓

    dydt=ky\frac{dy}{dt} = ky → y=Aekty = Ae^{kt}.

  • ✓

    Always include constant of integration until initial conditions are used.

Tags:
#ib#maths#calculus#differential-equations

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