Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\cos ^4}x\) given that \(y = 1\) when \(x = 0\).

(a)     Solve the differential equation, giving your answer in the form \(y = f(x)\).

(b)     (i)     By differentiating both sides of the differential equation, show that

\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y =  - 10\sin x{\cos ^3}x\]

(ii)     Hence find the first four terms of the Maclaurin series for \(y\).

(a)     integrating factor \( = {e^{\int {\tan x{\text{d}}x} }}\)     M1

\( = {{\text{e}}^{\ln \sec x}}\)     A1

\( = \sec x\)     A1

\(\sec x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x\tan x = 2{\cos ^3}x\)     (M1)

integrating,

\(y\sec x = 2\int {{{\cos }^3}x{\text{d}}x} \)     A1

\( = 2\int {\cos x(1 - {{\sin }^2}x){\text{d}}x} \)     A1

\( = 2\left( {\sin x - \frac{{{{\sin }^3}x}}{3}} \right) + C\)     A1

 

Note: Condone the absence of \(C\).

 

(substituting \(x = 0,{\text{ }}y = 1\))

\(1 = C\)     M1

the solution is

\(y = 2\cos x\left( {\sin x - \frac{{{{\sin }^3}x}}{3}} \right) + \cos x\)     A1

[9 marks]

 

(b)     (i)     differentiating the equation,

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 8{\cos ^3}x\sin x\)     A1A1

 

Note: A1 for each side.

 

substituting for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\),

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\left( {2{{\cos }^4}x - y\tan x} \right) =  - 8{\cos ^3}x\sin x\)     A1

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y({\sec ^2}x - {\tan ^2}x) =  - 8{\cos ^3}x\sin x - 2\tan x{\cos ^4}x\) (or equivalent)     A1

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y =  - 10\sin x{\cos ^3}x\)     AG

(ii)     differentiating again,

\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 10{\cos ^4}x + {\text{term involving }} \sin x\)     A1

it follows that

\(y(0) = 1,{\text{ }}y'(0) = 2\)     A1

\(y''(0) =  - 1,{\text{ }}y'''(0) =  - 12\)     A1

attempting to use \(y = y(0) + xy'(0) + \frac{{{x^2}}}{2}y''(0) + \frac{{{x^3}}}{6}y'''(0) +  \ldots \)     (M1)

\(y = 1 + 2x - \frac{{{x^2}}}{2} - 2{x^3}\)     A1

[9 marks]

[N/A]