FP2 June 2013 (R) Q4
4. Given that \[y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 5y = 0\]
Given that \(y = 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\) at \(x = 0\)
| Scheme | Marks |
|---|---|
| \(y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 5y = 0\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + y\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 5\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1M1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{-5\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}}{y}\) | A2,1,0 |
| (4) |
Notes
Q4a – Alternative 1
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{-5y - \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}{y} = -5 - \dfrac{1}{y}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\) | |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{1}{y^2}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^3 - \dfrac{2}{y}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)\) | M1M1 A2,1,0 |
Accept the dash notation in this question
M1 for using the product rule to differentiate \(y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\).
M1 for differentiating \(5y\) and using the product rule or chain rule to differentiate \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\)
A2,1,0 for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{-5\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}}{y}\) Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error. If there are two sign errors and no other error then give A1A0.
Do NOT deduct if the two \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) terms are shown separately.
Alternative to Q4a
Can be re-arranged first and then differentiated.
M1M1 for differentiating, product and chain rule both needed (or quotient rule as an alternative to product rule)
A2,1,0 for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{1}{y^2}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^3 - \dfrac{2}{y}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)\) Give A1A1 if fully correct, A1A0 if one error and A0A0 if more than one error
| Scheme | Marks |
|---|---|
| When \(x = 0\quad \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\) and \(y = 2\) | |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{1}{2}(-10 - 4) = -7\) | M1A1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{-10 - 3 \times 2 \times -7}{2} = 16\) | A1 |
| \(y = 2 + 2x - \dfrac{7}{2(!)}x^2 + \dfrac{16}{3!}x^3 + \ldots\) | M1 |
| \(y = 2 + 2x - \dfrac{7}{2}x^2 + \dfrac{8}{3}x^3\) | A1 |
| (5) | |
| (9 marks) |
Notes
Alternative
| Scheme | Marks |
|---|---|
| \(y = 2 + 2x + ax^2 + bx^3\) | M1 |
| \(\left(2 + 2x + ax^2 + bx^3\right)(2a + 6bx) + \left(2 + 2ax + 3bx^2\ldots\right)^2\) \(+ 5\left(2 + 2x + ax^2 + bx^3\right) = 0\) | M1 |
| Coeffs \(x^0\): \(4a + 4 + 10 = 0\quad a = -\dfrac{7}{2}\) | A1 |
| Coeffs \(x\): \(4a + 12b + 8a + 10 = 0 \Rightarrow b = \dfrac{8}{3}\) | A1 |
| \(y = 2 + 2x - \dfrac{7}{2}x^2 + \dfrac{8}{3}x^3\) | A1 |
M1 for substituting \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\) and \(y = 2\) in the equation to obtain a numerical value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)
A1 for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -7\)
A1 for obtaining the correct value, 16, for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\)
M1 for using the series \(y = \mathrm{f}(x) = \mathrm{f}(0) + x\mathrm{f}^{\prime}(0) + \dfrac{x^2}{2!}\mathrm{f}^{\prime\prime}(0) + \dfrac{x^3}{3!}\mathrm{f}^{\prime\prime\prime}(0) + \ldots\) (2! or 2, 3! or 6) (The general series may be shown explicitly or implied by their substitution)
A1 for \(y = 2 + 2x - \dfrac{7}{2}x^2 + \dfrac{8}{3}x^3\) oe Must have \(y = \ldots\) and be in ascending powers of
Alternative to Q4b
M1 for setting \(y = 2 + 2x + ax^2 + bx^3\)
M1 for \(\left(2 + 2x + ax^2 + bx^3\right)(2a + 6bx) + \left(2 + 2ax + 3bx^2\ldots\right)^2 + 5\left(2 + 2x + ax^2 + bx^3\right) = 0\)
A1 for equating constant terms to get \(a = -\dfrac{7}{2}\)
A1 for equating coeffs of \(x\) to get \(b = \dfrac{8}{3}\) (corrected from the printed mark scheme, which says “coeffs of \(x^2\)”; the scheme above equates the coefficients of \(x\))
A1 for \(y = 2 + 2x - \dfrac{7}{2}x^2 + \dfrac{8}{3}x^3\)