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]