Date | May 2018 | Marks available | 8 | Reference code | 18M.3ca.hl.TZ0.5 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Solve | Question number | 5 | Adapted from | N/A |
Question
Consider the differential equation \(x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y = {x^p} + 1\) where \(x \in \mathbb{R},\,x \ne 0\) and \(p\) is a positive integer, \(p > 1\).
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.
Markscheme
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]