FP2 June 2014 Q5
5.
(a) Find the general solution of the differential equation \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} + 10y = 27\mathrm{e}^{-x}\] (6)
(b) Find the particular solution that satisfies \(y = 0\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 0\) (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 10y = 27\mathrm{e}^{-x}\) | |
| \(m^2 + 2m + 10\ (= 0) \Rightarrow m = \ldots\) Form and solve the aux equation | M1 |
| \(m = -1 \pm 3\mathrm{i}\) | A1 |
| \((y =)\ \mathrm{e}^{-x}(A\cos 3x + B\sin 3x)\) or \((y =)\ A\mathrm{e}^{(-1+3\mathrm{i})x} + B\mathrm{e}^{(-1-3\mathrm{i})x}\) \(y =\) not needed May be seen with \(\theta\) instead of \(x\) | A1 |
| \(y = k\mathrm{e}^{-x},\ y' = -k\mathrm{e}^{-x},\ y'' = k\mathrm{e}^{-x}\) \(y = k\mathrm{e}^{-x}\) and attempt to differentiate twice | M1 |
| \(\mathrm{e}^{-x}(k - 2k + 10k) = 27\mathrm{e}^{-x} \Rightarrow k = 3\) | A1 |
| \(y = \mathrm{e}^{-x}(A\cos 3x + B\sin 3x + 3)\) or \(y = A\mathrm{e}^{(-1+3\mathrm{i})x} + B\mathrm{e}^{(-1-3\mathrm{i})x} + 3\mathrm{e}^{-x}\) Must be \(x\) and have \(y = \ldots\) Ignore any attempts to change the second form. (But see note at end about marking (b)) ft, so \(y =\) their CF + their PI | B1ft (NB A1 on e-PEN) |
| (6) |
| Scheme | Marks |
|---|---|
| \(x = 0,\ y = 0 \Rightarrow A = (-3)\) Uses \(x = 0,\ y = 0\) in an attempt to find \(A\) | M1 |
| \(y' = -\mathrm{e}^{-x}(A\cos 3x + B\sin 3x + 3) + \mathrm{e}^{-x}(3B\cos 3x - 3A\sin 3x)\) M1: Attempt to differentiate using the product rule, with \(A\) or their value of \(A\) A1: Correct derivative, with \(A\) or their value of \(A\) | M1A1 |
| \(x = 0,\ y' = 0 \Rightarrow B = 0\) M1: Uses \(x = 0,\ y' = 0\) and their value of \(A\) in an attempt to find \(B\) A1: \(B = 0\) | M1A1 |
| \(y = \mathrm{e}^{-x}(3 - 3\cos 3x)\) oe cao and cso | A1 |
| (6) | |
| (12 marks) |
Alternative for (b) using \(y = A\mathrm{e}^{(-1+3\mathrm{i})x} + B\mathrm{e}^{(-1-3\mathrm{i})x} + 3\mathrm{e}^{-x}\)
| Scheme | Marks |
|---|---|
| \(x = 0,\ y = 0\) to get an equation in \(A\) and \(B\) May come from the real part of their derivative instead | M1 |
| \(y' = (-1 + 3\mathrm{i})A\mathrm{e}^{(-1+3\mathrm{i})x} + (-1 - 3\mathrm{i})B\mathrm{e}^{(-1-3\mathrm{i})x} - 3\mathrm{e}^{-x}\) M1: Attempt differentiation using chain rule A1: Correct differentiation | M1A1 |
| \(x = 0,\ y' = 0 \Rightarrow -A - B - 3 = 0\) from real parts and \(3A - 3B = 0\) from imaginary parts So \(A = B = -\tfrac{3}{2}\) M1: Uses \(x = 0,\ y' = 0\) and equates imaginary parts to obtain a second equation for \(A\) and \(B\) and attempts to solve their equations A1: \(A = B = -\tfrac{3}{2}\) | M1A1 |
| \(y = -\dfrac{3}{2}\mathrm{e}^{(-1+3\mathrm{i})x} - \dfrac{3}{2}\mathrm{e}^{(-1-3\mathrm{i})x} + 3\mathrm{e}^{-x}\) A1: Ignore any attempts to change. | A1 |
Some may change the second form in (a) before proceeding to (b). If their changed form is correct, all marks for (b) are available; if their changed form is incorrect only M marks are available.