Let \({x^3}y = a\sin nx\) . Using implicit differentiation, show that

\[{x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + ({n^2}{x^2} + 6)xy = 0\] .

\({x^3}y = a\sin nx\)

attempt to differentiate implicitly     M1

\( \Rightarrow 3{x^2}y + {x^3}\frac{{{\text{d}}y}}{{{\text{d}}x}} = an\cos nx\)     A2 

Note: Award A1 for two out of three correct, A0 otherwise.

\( \Rightarrow 6xy + 3{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 3{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = -a{n^2}\sin nx\)     A2 

Note: Award A1 for three or four out of five correct, A0 otherwise.

\( \Rightarrow 6xy + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = -a{n^2}\sin nx\)

\( \Rightarrow {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 6xy + {n^2}{x^3}y = 0\)     A1

\( \Rightarrow {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + ({n^2}{x^2} + 6)xy = 0\)     AG

[6 marks]

Candidates who are comfortable using implicit differentiation found this to be a fairly straightforward question and were able to answer it in just a few lines. Many candidates, however, were unable to differentiate \({x^3}y\) with respect to x and were therefore unable to proceed. Candidates whose first step was to write \(y = \frac{{a\sin nx}}{{{x^3}}}\) were given no credit since the question required the use of implicit differentiation.