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Differential Equations
Edexcel A Level Further Mathematics – Core Pure Revision Notes
First Order Differential Equations
Separable Equations
If a first-order ODE can be written as dy dx = f (x)g(y), separate the variables and integrate both
sides:
Z 1 g(y) dy =
Z
f (x) dx
Example: Solve
dy dx = xy, given y = 2 when x = 0. Z 1 y dy =
Z
x dx =⇒ ln |y| = x^2 2
Using y = 2 at x = 0: A = 2. So y = 2ex (^2) / 2 .
Integrating Factor Method (Linear First Order)
For equations of the form dy dx
Integrating Factor Method
- Identify P (x) and compute the integrating factor: μ = e
R (^) P (x) dx .
- Multiply both sides by μ: the left side becomes d dx
(μy).
- Integrate both sides: μy =
R
μQ(x) dx.
- Divide by μ to find y.
Example: Solve
dy dx
2 y x = x^2.
P (x) =
x , so μ = e
R (^2) /x dx = e2 ln^ x^ = x^2.
Multiply: d dx (x^2 y) = x^4. Integrate: x^2 y = x^5 5
C
x^2
Second Order Differential Equations
The general form is:
a d^2 y dx^2
The general solution = complementary function (CF) + particular integral (PI).
The Complementary Function
The CF solves the homogeneous equation (where f (x) = 0). Try y = emx, which gives the auxiliary equation: am^2 + bm + c = 0
Three Cases for the CF Case 1 – Two distinct real roots m 1 ̸= m 2 :
yCF = Aem^1 x^ + Bem^2 x
Case 2 – Repeated root m 1 = m 2 = m:
yCF = (A + Bx)emx
Case 3 – Complex conjugate roots m = p ± qi:
yCF = epx(A cos qx + B sin qx)
The Particular Integral
The PI is a specific solution to the full equation with f (x) ̸= 0. You guess the form of the PI based on f (x):
f (x) type Trial PI form Constant k y = λ Polynomial degree n
y = λnxn^ + · · · + λ 0 (full polynomial)
keαx^ y = λeαx k cos ωx or k sin ωx y = λ cos ωx + μ sin ωx (include both!)
When the Trial PI Clashes with the CF If your trial PI has the same form as a term in the CF, it will give 0 = 0 when substituted
- multiply the trial PI by x. If it still clashes, multiply by x^2. Example: If the CF contains e^2 x^ and f (x) = 3e^2 x, your trial PI is λxe^2 x.
Worked Example – Full Second Order
Solve d^2 y dx^2
dy dx
- 6y = 2ex, with y = 1, dy dx = 0 at x = 0.
CF: Auxiliary equation m^2 − 5 m + 6 = 0 ⇒ (m − 2)(m − 3) = 0 ⇒ m = 2, 3.
yCF = Ae^2 x^ + Be^3 x
PI: Try y = λex. Then y′^ = λex, y′′^ = λex.
Substituting: λex^ − 5 λex^ + 6λex^ = 2ex^ ⇒ 2 λex^ = 2ex^ ⇒ λ = 1.
yP I = ex
General solution: y = Ae^2 x^ + Be^3 x^ + ex
