Standard +0.3 This is a standard integrating factor problem with hyperbolic functions. While the hyperbolic context adds slight complexity beyond basic exponential cases, the method is routine: identify the linear form, find integrating factor (cosh x)^(-2), integrate, and apply initial conditions. The hyperbolic identities needed are standard Further Maths content, making this slightly above average difficulty but still a textbook exercise.
or 𝑒∫(cid:2879)(cid:2870)(cid:2930)(cid:2911)(cid:2924)(cid:2918)(cid:3051) (cid:3031)(cid:3051). Integral must come from an attempt to get (cid:2914)(cid:3052)
(cid:2914)(cid:3051)
on its own
or multiplying through by their IF
substituting 𝑥 = 1 and 𝑦 = 1 to lead to a value for 𝑐
oe, e.g. cosh𝑥(sinh𝑥+cosh𝑥) or e(cid:3051)cosh𝑥
Question 11:
11 | d𝑦 sinh𝑥
−2𝑦 = 1
d𝑥 cosh𝑥
(cid:3118)(cid:3177)(cid:3167)(cid:3172)(cid:3166)(cid:3299)
e∫(cid:2879) (cid:2914)(cid:3051)
IF is (cid:3161)(cid:3173)(cid:3177)(cid:3166)(cid:3299)
(cid:2870)
= e(cid:2879)(cid:2870)(cid:2922)(cid:2924)(cid:2913)(cid:2925)(cid:2929)(cid:2918)(cid:3051) = (cid:3435)𝑒(cid:2922)(cid:2924) (cid:2929)(cid:2915)(cid:2913)(cid:2918)(cid:3051)(cid:3439) = sech(cid:2870)𝑥
d
(𝑦sech(cid:2870)𝑥)= sech(cid:2870)𝑥
d𝑥
𝑦sech(cid:2870)𝑥 = ∫sech(cid:2870)𝑥d𝑥+𝑐 = tanh𝑥+𝑐
substituting 𝑥 = 0, 𝑦 = 1 gives 𝑐 = 1
𝑦 = cosh(cid:2870)𝑥(tanh𝑥+1) | B1
M1
A1
M1
A1
M1
A1
[7] | 3.1a
2.1
2.2a
2.1
1.1
1.1
2.2a | (cid:2914)(cid:3052)
−2𝑦tanh𝑥 = 1
(cid:2914)(cid:3051)
or 𝑒∫(cid:2879)(cid:2870)(cid:2930)(cid:2911)(cid:2924)(cid:2918)(cid:3051) (cid:3031)(cid:3051). Integral must come from an attempt to get (cid:2914)(cid:3052)
(cid:2914)(cid:3051)
on its own
or multiplying through by their IF
substituting 𝑥 = 1 and 𝑦 = 1 to lead to a value for 𝑐
oe, e.g. cosh𝑥(sinh𝑥+cosh𝑥) or e(cid:3051)cosh𝑥
11 Solve the differential equation $\cosh x \frac { d y } { d x } - 2 y \sinh x = \cosh x$, given that $y = 1$ when $x = 0$.
\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q11 [7]}}