Solve the differential equation given that \(y =  - 1\) when \(x = 1\). Give your answer in the form \(y = f\left( x \right)\).

Show that the \(x\)-coordinate(s) of the points on the curve \(y = f\left( x \right)\) where \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) satisfy the equation \({x^{p - 1}} = \frac{1}{p}\).

Deduce the set of values for \(p\) such that there are two points on the curve \(y = f\left( x \right)\) where \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\). Give a reason for your answer.

METHOD 1

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

integrating factor \( = {{\text{e}}^{\int { - \frac{1}{x}{\text{d}}x} }}\)     M1

\({\text{ = }}{{\text{e}}^{ - {\text{ln}}\,x}}\)     (A1)

= \(\frac{1}{x}\)     A1

\(\frac{1}{x}\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{y}{{{x^2}}} = {x^{p - 2}} + \frac{1}{{{x^2}}}\)     (M1)

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

\(\frac{y}{x} = \frac{1}{{p - 1}}{x^{p - 1}} - \frac{1}{x} + C\)    A1

Note: Condone the absence of C.

\(y = \frac{1}{{p - 1}}{x^p} + Cx - 1\)

substituting \(x = 1\), \(y =  - 1 \Rightarrow C =  - \frac{1}{{p - 1}}\)    M1 

Note: Award M1 for attempting to find their value of C.

\(y = \frac{1}{{p - 1}}\left( {{x^p} - x} \right) - 1\)      A1

[8 marks]

METHOD 2

put \(y = vx\) so that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)    M1(A1)

substituting,       M1 

\(x\left( {v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}} \right) - vx = {x^p} + 1\)     (A1)

\(x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p - 1}} + \frac{1}{x}\)      M1

\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = {x^{p - 2}} + \frac{1}{{{x^2}}}\)

\(v = \frac{1}{{p - 1}}{x^{p - 1}} - \frac{1}{x} + C\)     A1

Note: Condone the absence of C.

\(y = \frac{1}{{p - 1}}{x^p} + Cx - 1\)

substituting \(x = 1\), \(y =  - 1 \Rightarrow C =  - \frac{1}{{p - 1}}\)    M1 

Note: Award M1 for attempting to find their value of C.

\(y = \frac{1}{{p - 1}}\left( {{x^p} - x} \right) - 1\)      A1

[8 marks]

METHOD 1

find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) and solve \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) for \(x\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{p - 1}}\left( {p{x^{p - 1}} - 1} \right)\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow p{x^{p - 1}} - 1 = 0\)     A1

\(p{x^{p - 1}} = 1\)

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

\({x^{p - 1}} = \frac{1}{p}\)     AG

METHOD 2

substitute \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) and their \(y\) into the differential equation and solve for \(x\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow  - \left( {\frac{{{x^p} - x}}{{p - 1}}} \right) + 1 = {x^p} + 1\)     M1

\({x^p} - x = {x^p} - p{x^p}\)     A1

\(p{x^{p - 1}} = 1\)

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

\({x^{p - 1}} = \frac{1}{p}\)     AG

[2 marks]

there are two solutions for \(x\) when \(p\) is odd (and \(p > 1\)     A1

if \(p - 1\) is even there are two solutions (to \({x^{p - 1}} = \frac{1}{p}\))

and if \(p - 1\) is odd there is only one solution (to \({x^{p - 1}} = \frac{1}{p}\))   R1

Note: Only award the R1 if both cases are considered.

[4 marks]