Write down the general term.

Find the interval of convergence.

Solve the differential equation \((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\) , giving your answer in the form \(u = f(v)\) .

the general term is \(\frac{{{{(x + 2)}^n}}}{{{3^n}n}}\)      A1

[1 mark]

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \times \frac{{{3^n}n}}{{{{(x + 2)}^n}}}} \right]\)     M1A1A1

\( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{(x + 2){n^{}}}}{{3(n + 1)}}} \right]\)     A1

\( = \frac{{(x + 2)}}{3}\) since \( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^{}}}}{{n + 1}}} \right] = 1\)     A1R1

the series is convergent if \(\left| {\frac{{(x + 2)}}{3}} \right| < 1\)    R1

then \( - 3 < x + 2 < 3 \Rightarrow  - 5 < x < 1\)     A1

if \(x = - 5\) , series is \(1 - 1 + \frac{1}{2} - \frac{1}{3} +  \ldots  + \frac{{{{( - 1)}^n}}}{n} +  \ldots \) which converges     M1A1

if \(x = 1\) , series is \(1 + 1 + \frac{1}{2} + \frac{1}{3} +  \ldots  + \frac{1}{n} +  \ldots \) which diverges     M1A1

the interval of convergence is \( - 5 \le x < 1\)     A1

[13 marks]

\((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\)

\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} = \frac{{(u + 3{v^3})}}{{2v}} = \frac{u}{{2v}} + \frac{{3{v^2}}}{2}\)     M1A1

\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} - \frac{u}{{2v}} = \frac{{3{v^2}}}{2}\)     A1

IF is \({{\rm{e}}^{\int {\frac{1}{{2v}}} {\rm{d}}v}} = {{\rm{e}}^{\frac{1}{2}\ln v}}\)     M1

\( = {v^{\frac{1}{2}}}\)     A1

\(\frac{u}{{\sqrt v }} = \int {\frac{{3{v^{\frac{3}{2}}}}}{2}} {\rm{d}}v\)     M1

\( = \frac{3}{5}{v^{\frac{5}{2}}} + c\)     A1

\(u = \frac{3}{5}{v^3} + c\sqrt v \)     A1

[8 marks]

In (a) the general term was usually found.

Part (b) was completed mostly except for testing the ends of the interval of convergence.

A surprising number of candidates started off their solution by saying "let \(x = u\) and \(y = v\) " as if the world suddenly changed when \(x\) and \(y\) were not being used in a differential equation. Some also after seeing \(u\) and \(v\) thought they had a homogeneous equation and got lost in a maze of algebra that lead nowhere. Find \(\frac{{{\rm{d}}u}}{{{\rm{d}}v}}\) by inverting the given expression was also something that only the best candidates were able to do.