(a)     Solve the differential equation \(\frac{{{{\cos }^2}x}}{{{{\text{e}}^y}}} - {{\text{e}}^{{{\text{e}}^y}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) , given that \(y = 0\) when \(x = \pi\).

(b)     Find the value of y when \(x = \frac{\pi }{2}\).

(a)     rearrange \(\frac{{{{\cos }^2}x}}{{{{\text{e}}^y}}} - {{\text{e}}^{{{\text{e}}^y}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) to obtain \({\cos ^2}x{\text{d}}x = {{\text{e}}^y}{{\text{e}}^{{{\text{e}}^y}}}{\text{d}}y\)     (M1)

as \(\int {{{\cos }^2}x{\text{d}}x}  = \int {\frac{{1 + \cos \left( {2x} \right)}}{2}{\text{d}}x = \frac{1}{2}x + \frac{1}{4}\sin } \left( {2x} \right) + {C_1}\)     M1A1

and \(\int {{{\text{e}}^y}{{\text{e}}^{{{\text{e}}^y}}}{\text{d}}y}  = {{\text{e}}^{{{\text{e}}^y}}} + {C_2}\)     A1

Note: The above two integrations are independent and should not be penalized for missing.

a general solution of \(\frac{{{{\cos }^2}x}}{{{{\text{e}}^y}}} - {{\text{e}}^{{{\text{e}}^y}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) is \(\frac{1}{2}x + \frac{1}{4}\sin \left( {2x} \right) - {{\text{e}}^{{{\text{e}}^y}}} = C\)     A1

given that \(y = 0\) when \(x = \pi\), \(C = \frac{\pi }{2} + \frac{1}{4}\sin \left( {2\pi } \right) - {{\text{e}}^{{{\text{e}}^0}}} = \frac{\pi }{2} - {\text{e}}\), (or \( - 1.15\))     (M1)

so, the required solution is defined by the equation

\(\frac{1}{2}x + \frac{1}{4}\sin \left( {2x} \right) - {{\text{e}}^{{{\text{e}}^y}}} = \frac{\pi }{2} - {\text{e}}\) or \(y = \ln \left( {\ln \left( {\frac{1}{2}x + \frac{1}{4}\sin \left( {2x} \right){\text{ + }}{{\text{e}}^{{{\text{e}}^y}}} - \frac{\pi }{2}} \right)} \right)\)     A1     N0

(or equivalent)

(b)     for \(x = \frac{\pi }{2}\), \(y = \ln \left( {\ln \left( {{\text{e}} - \frac{\pi }{4}} \right)} \right)\) (or \( - 0.417\) )     A1

[8 marks]

This was a more difficult question and it was apparent that students did find it so. For those that managed to rearrange the equation to separate the variables, few could manage to successfully integrate both sides. The unfamiliarity of \({{\text{e}}^{{{\text{e}}^y}}}\) seemed to disturb some students.