IB Maths AI HL
Coupled & Second Order Differential Equations
Paper 1 & 2
~7 min read
Second Order Differential Equations
A second order equation involves d2xdt2. The key move is the substitution y = dxdt, which turns it into a coupled first-order system β so every technique from this unit applies. From there you can run Euler’s method for an approximation, or find the exact solution via eigenvalues.
π What you need to know
- The form: d2xdt2 = f(x, dxdt, t).
- The substitution: let y = dxdt, so dydt = d2xdt2.
- Coupled system: dxdt = y, dydt = f(x, y, t).
- Euler’s method: apply the coupled recursions with the given step size.
- Exact solution (for d2xdt2 + adxdt + bx = 0): x = CeΞ»βt + DeΞ»βt.
- Use y = dxdt to apply the initial value of the derivative.
Reducing to a coupled system
The substitution y = dx/dt
d2xdt2 = f(x, dxdt, t) βΆ dxdt = y, dydt = f(x, y, t)
β the coupled-system tools are all in the booklet
π€ Why does y = dxdt work?
Naming the first derivative as a new variable y immediately gives one equation, dxdt = y. Differentiating y gives the second derivative, which the original equation already expresses in terms of x, dxdt, and t β i.e. in terms of x, y, t. So the single 2nd-order equation splits cleanly into two 1st-order ones.
π§ “Name the speed, get a pair”
Let y be the “speed” dxdt. One equation says position changes at that speed (dxdt = y); the other rearranges the original for dydt.
Worked example β Euler’s method
WE 1Rewrite d2xdt2 + 2dxdt + x = 50cos t as a coupled system.
Let y = dxdt and rearrange for dydt.
dΒ²xdtΒ² = βx β 2dxdt + 50cos t
dxdt = y, dydt = βx β 2y + 50cos t
WE 2Initially x = 2 and dxdt = β1. With h = 0.1, find x and dxdt at t = 0.5.
Apply the coupled Euler recursions over 5 steps (y = dxdt).
xn+1 = xn + 0.1 yn
yn+1 = yn + 0.1(βxn β 2yn + 50cos tn)
| n | 0 | 1 | 2 | 3 | 4 | 5 |
|---|
| tn | 0 | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 |
| xn | 2 | 1.9 | 2.3 | 3.0985 | 4.2043 | 5.5357 |
| yn | β1 | 4 | 7.985 | 11.058 | 13.313 | 14.836 |
Answer: x(0.5) β 5.54 and dxdt β 14.8 (3 s.f.).
Exact solutions
For the homogeneous form d2xdt2 + adxdt + bx = 0, the same substitution gives the matrix , whose eigenvalues solve the characteristic equation λ² + aλ + b = 0.
Exact solution (real, distinct, non-zero eigenvalues)
x = CeΞ»βt + DeΞ»βt, where λ² + aΞ» + b = 0
β follows from the booklet eigen-solution
Applying the conditions: you’ll usually be given x and dxdt at t = 0. Differentiate x = CeΞ»βt + DeΞ»βt to get dxdt, then form two simultaneous equations for C and D.
Worked example β exact solution
Consider d2xdt2 + 3dxdt β 4x = 0, with x = 3 and dxdt = β2 initially.
WE 3Rewrite it as a coupled system.
Let y = dxdt and rearrange.
dΒ²xdtΒ² = 4x β 3dxdt
dxdt = y, dydt = 4x β 3y
WE 4The matrix has eigenvalues 1 and β4. Write the general solution for x.
Each eigenvalue gives one exponential term.
Ξ» = 1, β4 (also the roots of λ² + 3Ξ» β 4 = 0)
x = C et + D eβ4t
WE 5Use the initial conditions to find the exact solution.
Differentiate for dxdt, then apply x(0) = 3 and dxdt(0) = β2.
dxdt = C et β 4D eβ4t
x(0): C + D = 3; xβ²(0): C β 4D = β2
C = 2, D = 1
x = 2et + eβ4t
π‘ Top tips
- Always substitute y = dxdt first β it unlocks every coupled-system method.
- Rearrange for the second derivative before reading off dydt.
- The initial dxdt is y0 for Euler.
- Characteristic equation λ² + aλ + b = 0 gives the exponents for the exact solution.
- Two conditions, two constants β differentiate to use the derivative value.
- GDC recursion handles the Euler arithmetic on Paper 2.
β Common mistakes
- Sign error rearranging for d2xdt2 before substituting.
- Forgetting dxdt = y0 when setting up Euler.
- Only using x(0) and not the derivative condition for the exact solution.
- Wrong characteristic equation β mismatching a and b signs.
- Not differentiating the general solution before applying dxdt(0).
That wraps up Coupled & Second Order Differential Equations. The unit grew from one matrix idea β αΊ = Mx β into a complete picture: exact solutions from eigenvalues and eigenvectors, phase portraits showing every trajectory, equilibrium points classified by eigenvalue signs, and sketched trajectories from a starting condition. Second order equations join the same family through the substitution y = dxdt β whether you solve them exactly or step through them with Euler, they’re coupled systems in disguise.
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