Find the general solution of the differential equation \((1 - {x^2})\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 1 + xy\) , for \(\left| x \right| < 1\) .

(i)     Show that the solution \(y = f(x)\) that satisfies the condition \(f(0) = \frac{\pi }{2}\) is \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 - {x^2}} }}\) .

(ii)     Find \(\mathop {\lim }\limits_{x \to  - 1} f(x)\) .

rewrite in linear form     M1

\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \left( {\frac{{ - x}}{{1 - {x^2}}}} \right)y = \frac{1}{{1 - {x^2}}}\)

attempt to find integrating factor     M1

\(I = {e^{\int {\frac{{ - x}}{{1 - {x^2}}}} {\rm{d}}x}} = {e^{\frac{1}{2}\ln (1 - {x^2})}}\)     A1

\( = \sqrt {1 - {x^2}} \)     A1

multiply by \(I\) and attempt to integrate     (M1)

\(y{(1 - {x^2})^{\frac{1}{2}}} = \int {\frac{1}{{\sqrt {1 - {x^2}} }}} {\rm{d}}x\)     (A1)

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

[7 marks]

(i)     attempt to find c     M1

\(\frac{\pi }{2} = 0 + c\)     A1

so \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 - {x^2}} }}\)     AG

(ii)     \(\mathop {\lim }\limits_{x \to  - 1} f(x) = \frac{0}{0}\) , so attempt l’Hôpital’s rule     (M1)

consider \(\mathop {\lim }\limits_{x \to  - 1} \frac{{\frac{1}{{{{(1 - {x^2})}^{\frac{1}{2}}}}}}}{{\frac{{ - x}}{{{{(1 - {x^2})}^{\frac{1}{2}}}}}}}\)     A1A1

\( = 1\)     A1

[6 marks]

A good number of wholly correct answers were seen to this question and for stronger candidates it proved to be a successful question. A small number of candidates failed to recognise part (a) as needing an integrating factor. More commonly, students left out the negative sign in the integrating factor or were unable to simplify the integrating factor.

A good number of wholly correct answers were seen to this question and for stronger candidates it proved to be a successful question. In part (b) many students recognised the use of L’Hopital’s rule, but in a number of cases made errors in the differentiation.