FP2 June 2007 Q7
7. For the differential equation \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3\frac{\mathrm{d}y}{\mathrm{d}x} + 2y = 2x(x + 3),\] find the solution for which at \(x = 0\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) and \(y = 1\).
| Scheme | Marks |
|---|---|
| C.F. \(m^2 + 3m + 2 = 0 \qquad m = -1\) and \(m = -2\) | M1 |
| \(y = A\mathrm{e}^{-x} + B\mathrm{e}^{-2x}\) | A1 2 |
| P.I. \(y = cx^2 + dx + e\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2cx + d,\ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2c \qquad 2c + 3(2cx + d) + 2(cx^2 + dx + e) \equiv 2x^2 + 6x\) | M1 |
| \(2c = 2 \qquad c = 1\) (One correct value) | A1 |
| \(6c + 2d = 6 \qquad d = 0\) \(2c + 3d + 2e = 0 \qquad e = -1\) (Other two correct values) | A1 |
| General soln: \(y = A\mathrm{e}^{-x} + B\mathrm{e}^{-2x} + x^2 - 1\) (Their C.F. + their P.I.) | A1ft 5 |
| \(x = 0, y = 1\): \(1 = A + B - 1 \qquad (A + B = 2)\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -A\mathrm{e}^{-x} - 2B\mathrm{e}^{-2x} + 2x,\ x = 0,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 \qquad 1 = -A - 2B\) | M1 |
| Solving simultaneously: \(A = 5\) and \(B = -3\) | M1A1 |
| Solution: \(y = 5\mathrm{e}^{-x} - 3\mathrm{e}^{-2x} + x^2 - 1\) | A1 5 |
| (12 marks) |
Notes
1st M: Attempt to solve auxiliary equation.
2nd M: Substitute their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) into the D.E. to form an identity in \(x\) with unknown constants.
3rd M: Using \(y = 1\) at \(x = 0\) in their general solution to find an equation in \(A\) and \(B\).
4th M: Differentiating their general solution (condone ‘slips’, but the powers of each term must be correct) and using \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) at \(x = 0\) to find an equation in \(A\) and \(B\).
5th M: Solving simultaneous equations to find both a value of \(A\) and a value of \(B\).