Challenging +1.2 This is a first-order linear ODE requiring integrating factor method with hyperbolic functions. While the hyperbolic identities and integration require care, the structure is standard: find integrating factor e^(∫tanh x dx) = cosh x, multiply through, integrate using sinh³x = sinh x(cosh²x - 1), then apply initial condition. The 7 marks reflect multiple steps but the approach is methodical without requiring novel insight—harder than average due to hyperbolic manipulation but well within expected Further Maths technique.
Solve the differential equation
$$\frac{dy}{dx} + y\tanh x = \sinh^3 x$$
given that \(y = 3\) when \(x = \ln 2\)
Give your answer in an exact form.
[7 marks]
Question 14:
14 | tanh xdx
e
Writes | 3.1a | M1 | ta n h xd x
Integrating factor = e
t a n h x d x = ln ( c o s h x )
Integrating factor = cosh x
ycosh x = s in h 3 x c o s h x d x
1
= s in h 4 x + c
4
3 5
When x = ln2, sinh x = and cosh x =
4 4
5 1 81
3× = × +c
4 4 256
3759
c=
1024
1 3 7 5 9
ycosh x = s in h 4 x +
4 1 0 2 4
Obtains cosh x | 1.1b | A1
Obtains ycosh x | 1.1b | B1
Writes down
s in h 3 x c o s h x d x
(PI) | 1.1a | M1
Obtains
k s in h 4 x
or
k c o s h 4 x + k c o s h 2 x
1 2
(Accept equivalent
exponential form) | 1.1b | A1
Substitutes x = ln2
and y = 3 into
y c o s h x = k s in h 4 x + c
or
ycoshx
=kcosh 4x+k cosh 2x+c
1 2
(Accept equivalent
exponential form)
and uses to evaluate c | 1.1a | M1
Deduces
1 3759
ycoshx= sinh4 x+
4 1024
Or
ycoshx
1 1 3855
= cosh 4x− cosh 2x+
32 8 1024
ACF.
ISW | 2.2a | A1
Question total | 7
Q | Marking instructions | AO | Marks | Typical solution
Solve the differential equation
$$\frac{dy}{dx} + y\tanh x = \sinh^3 x$$
given that $y = 3$ when $x = \ln 2$
Give your answer in an exact form.
[7 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2024 Q14 [7]}}