FP2 June 2015 Q8
8.
| Scheme | Marks |
|---|---|
| \(x = \mathrm{e}^u \quad \dfrac{\mathrm{d}x}{\mathrm{d}u} = \mathrm{e}^u\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^{-u}\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = x\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{x}\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u}\times\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^{-u}\dfrac{\mathrm{d}y}{\mathrm{d}u}\) | M1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\mathrm{e}^{-u}\dfrac{\mathrm{d}u}{\mathrm{d}x}\dfrac{\mathrm{d}y}{\mathrm{d}u} + \mathrm{e}^{-u}\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^{-2u}\left(-\dfrac{\mathrm{d}y}{\mathrm{d}u} + \dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\right)\) | M1A1 |
| \(x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 7x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 16y = 2\ln x\) | |
| \(\mathrm{e}^{2u}\times\mathrm{e}^{-2u}\left(-\dfrac{\mathrm{d}y}{\mathrm{d}u} + \dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\right) - 7\mathrm{e}^u\times\mathrm{e}^{-u}\dfrac{\mathrm{d}y}{\mathrm{d}u} + 16y = 2\ln\left(\mathrm{e}^u\right)\) | dM1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} - 8\dfrac{\mathrm{d}y}{\mathrm{d}u} + 16y = 2u\) * | A1cso |
| (6) |
Notes
B1: for \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = \mathrm{e}^u\) oe as shown seen explicitly or used
M1: obtaining \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) using chain rule here or seen later
M1: obtaining \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) using product rule (penalise lack of chain rule by the A mark)
A1: a correct expression for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) any equivalent form
dM1: substituting in the equation to eliminate \(x\) Only \(u\) and \(y\) now Depends on the 2nd M mark
A1cso: obtaining the given result from completely correct work
ALTERNATIVE 1
| Scheme | Marks |
|---|---|
| \(x = \mathrm{e}^u \quad \dfrac{\mathrm{d}x}{\mathrm{d}u} = \mathrm{e}^u = x\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}u} = \dfrac{\mathrm{d}y}{\mathrm{d}x}\times\dfrac{\mathrm{d}x}{\mathrm{d}u} = x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} = 1\dfrac{\mathrm{d}x}{\mathrm{d}u}\times\dfrac{\mathrm{d}y}{\mathrm{d}x} + x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\times\dfrac{\mathrm{d}x}{\mathrm{d}u} = x\dfrac{\mathrm{d}y}{\mathrm{d}x} + x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) | M1A1 |
| \(x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} - \dfrac{\mathrm{d}y}{\mathrm{d}u}\) | |
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} - \dfrac{\mathrm{d}y}{\mathrm{d}u}\right) - 7x\times\dfrac{1}{x}\dfrac{\mathrm{d}y}{\mathrm{d}u} + 16y = 2\ln\left(\mathrm{e}^u\right)\) | |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} - 8\dfrac{\mathrm{d}y}{\mathrm{d}u} + 16y = 2u\) * | dM1A1cso |
| (6) |
B1 As above
M1 obtaining \(\dfrac{\mathrm{d}y}{\mathrm{d}u}\) using chain rule here or seen later
M1 obtaining \(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\) using product rule (penalise lack of chain rule by the A mark)
A1 Correct expression for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\) any equivalent form
dM1A1cso As main scheme
ALTERNATIVE 2:
| Scheme | Marks |
|---|---|
| \(u = \ln x \quad \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{x}\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}u}\times\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{x}\dfrac{\mathrm{d}y}{\mathrm{d}u}\) | M1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{1}{x^2}\dfrac{\mathrm{d}y}{\mathrm{d}u} + \dfrac{1}{x}\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\times\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\dfrac{1}{x^2}\dfrac{\mathrm{d}y}{\mathrm{d}u} + \dfrac{1}{x^2}\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\) | M1A1 |
| \(x^2\left(-\dfrac{1}{x^2}\dfrac{\mathrm{d}y}{\mathrm{d}u} + \dfrac{1}{x^2}\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\right) - 7x\times\dfrac{1}{x}\dfrac{\mathrm{d}y}{\mathrm{d}u} + 16y = 2u\) | |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} - 8\dfrac{\mathrm{d}y}{\mathrm{d}u} + 16y = 2u\) * | dM1A1cso |
| (6) |
See the notes for the main scheme.
There are also other solutions which will appear, either starting from equation II and obtaining equation I, or mixing letters \(x\), \(y\) and \(u\) until the final stage.
Mark as follows:
B1 as shown in schemes above
M1 obtaining a first derivative with chain rule
M1 obtaining a second derivative with product rule
A1 correct second derivative with 2 or 3 variables present
dM1 Either substitute in equation I or substitute in equation II according to method chosen and obtain an equation with only \(y\) and \(u\) (following sub in eqn I) or with only \(x\) and \(y\) (following sub in eqn II)
A1cso Obtaining the required result from completely correct work
| Scheme | Marks |
|---|---|
| \(m^2 - 8m + 16 = 0\) | |
| \((m - 4)^2 = 0 \quad m = 4\) | M1A1 |
| (CF \(=\)) \((A + Bu)\mathrm{e}^{4u}\) | A1 |
| PI: try \(y = au + b\) ( or \(y = cu^2 + au + b\) different derivatives, \(c = 0\)) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}u} = a \quad \dfrac{\mathrm{d}^2y}{\mathrm{d}u^2} = 0\) | M1 |
| \(0 - 8a + 16(au + b) = 2u\) | |
| \(a = \dfrac{1}{8} \quad b = \dfrac{1}{16}\) oe (decimals must be 0.125 and 0.0625) | dM1A1 |
| \(\therefore\ y = (A + Bu)\mathrm{e}^{4u} + \dfrac{1}{8}u + \dfrac{1}{16}\) | B1ft |
| (7) |
Notes
M1: writing down the correct aux equation and solving to \(m = \ldots\) (usual rules)
A1: the correct solution \((m = 4)\)
A1: the correct CF – can use any (single) variable
M1: using an appropriate PI and finding \(\dfrac{\mathrm{d}y}{\mathrm{d}u}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}\) Use of \(y = \lambda u\) scores M0
dM1: substitute in the equation to obtain values for the unknowns Dependent on the second M1
A1: correct unknowns two or three (\(c = 0\))
B1ft: a complete solution, follow through their CF and PI. Must have \(y =\) a function of \(u\)
Allow recovery of incorrect variables.
| Scheme | Marks |
|---|---|
| \(y = (A + B\ln x)x^4 + \dfrac{1}{8}\ln x + \dfrac{1}{16}\) | B1 |
| (1) | |
| (14 marks) |
Notes
B1: reverse the substitution to obtain a correct expression for \(y\) in terms of \(x\) No ft here
\(x^4\) or \(\mathrm{e}^{4\ln x}\) allowed. Must start \(y = \ldots\)