Solve the differential equation

\[(x - 1)\frac{{{\text{d}}y}}{{{\text{d}}x}} + xy = (x - 1){{\text{e}}^{ - x}}\]

given that y = 1 when x = 0. Give your answer in the form \(y = f(x)\).

writing the differential equation in standard form gives

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{x - 1}}y = {{\text{e}}^{ - x}}\)     M1

\(\int {\frac{x}{{x - 1}}{\text{d}}x = \int {\left( {1 + \frac{1}{{x - 1}}} \right){\text{d}}x = x + \ln (x - 1)} } \)     M1A1

hence integrating factor is \({{\text{e}}^{x + \ln (x - 1)}} = (x - 1){{\text{e}}^x}\)     M1A1

hence, \((x - 1){{\text{e}}^x}\frac{{{\text{d}}y}}{{{\text{d}}x}} + x{{\text{e}}^x}y = x - 1\)     (A1)

\( \Rightarrow \frac{{{\text{d}}\left[ {(x - 1){{\text{e}}^x}y} \right]}}{{{\text{d}}x}} = x - 1\)     (A1)

\( \Rightarrow (x - 1){{\text{e}}^x}y = \int {(x - 1){\text{d}}x} \)     A1

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

substituting (0, 1), c = –1     (M1)A1

\( \Rightarrow (x - 1){{\text{e}}^x}y = \frac{{{x^2} - 2x - 2}}{2}\)     (A1)

hence, \(y = \frac{{{x^2} - 2x - 2}}{{2(x - 1){{\text{e}}^x}}}\) (or equivalent)     A1

[13 marks]

Apart from some candidates who thought the differential equation was homogenous, the others were usually able to make a good start, and found it quite straightforward. Some made errors after identifying the correct integrating factor, and so lost accuracy marks.