Challenging +1.2 This is a standard second-order linear ODE with constant coefficients requiring auxiliary equation solution (roots -9 and 5), particular integral by inspection for three terms (exponential, linear, quadratic), and applying initial conditions. While methodical with multiple steps for 10 marks, it follows a completely routine algorithm taught in Further Maths with no novel problem-solving required—just careful algebraic execution of standard techniques.
Question 19:
19 | Uses a three-
term auxiliary
equation to
obtain the
Complementary
Function. | 3.1a | M1 | Complementary Function
λ2 + 4λ – 45 = 0
λ = 5 or λ = –9
y = Ae–9x + Be5x
Particular Integral
y = Cxe5x + D + Ex + Fx2
PI
y′ = Ce5x + 5Cxe5x + E + 2Fx
PI
y′′ = 10Ce5x + 25Cxe5x + 2F
PI
10Ce5x + 25Cxe5x + 2F + 4(Ce5x + 5Cxe5x + E + 2Fx)
–45(Cxe5x + D + Ex + Fx2) = 21e5x – 0.3x + 27x2
3 –8 –1 –3
⇒C = , D= , E = , F =
2 225 10 5
General Solution
3 8 1 3
y = Ae–9x +Be5x + xe5x – – x– x2
GS
2 225 10 5
3 15 1 6
y′ = –9Ae–9x +5Be5x + e5x + xe5x – – x
GS
2 2 10 5
x=0
37 8 1
= A+B− ⇒ A+B=
225 225 5
3 1 7
0= –9A+5B+ – ⇒9A–5B=
2 10 5
6 1
A= ,B =
35 35
6 1 3 8 1 3
y = e–9x + e5x + xe5x – – x– x2
35 35 2 225 10 5
Obtains the
correct
Complementary
Function. | 1.1b | A1
Deduces correct
exponential
form of
Particular
Integral, Cxe5x | 2.2a | B1
Uses correct
polynomial form
of Particular
Integral, D + Ex
+ Fx2 | 3.1a | B1
Substitutes their
y′ y′′
and ,
PI PI
with their y ,
PI
into the
differential
equation. | 1.1a | M1
Compares
coefficients to
obtain at least
three of their C,
D, E and F (with
at least two
correct). | 1.1a | M1
Obtains correct
values of C, D,
E and F | 1.1b | A1
Forms their
General
Solution, with
exactly two
arbitrary
constants. | 3.1a | M1
Uses their
37
y = and
GS
225
y′ =0 to form
GS
two equations in
their two
arbitrary
constants. | 1.1a | M1
Obtains the
correct final
result. | 3.2a | A1
Question total | 10
Q | Marking
instructions | AO | Marks | Typical solution