A2 June 2020 Paper 2 Q13
13 Charlotte is trying to solve this mathematical problem:
Find the general solution of the differential equation
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} - 2y = 10\mathrm{e}^{-2x}\]
Charlotte’s solution starts as follows:
Particular integral: \(\quad y = \lambda\mathrm{e}^{-2x}\)
so
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -2\lambda\mathrm{e}^{-2x}\]and
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4\lambda\mathrm{e}^{-2x}\]
(a) Show that Charlotte’s method will fail to find a particular integral for the differential equation. [2 marks]
(b) Explain how Charlotte should have started her solution differently and find the general solution of the differential equation. [8 marks]
| Scheme | Marks | AO |
|---|---|---|
| Evaluates Charlotte’s method by substituting her particular integral and its derivatives into the differential equation. | M1 | 2.3 |
| Explains why Charlotte’s method fails to find a particular integral for the differential equation. | E1 | 2.3 |
Typical solution
Continuing Charlotte’s method gives
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x} - 2y = 4\lambda\mathrm{e}^{-2x} - 2\lambda\mathrm{e}^{-2x} - 2\lambda\mathrm{e}^{-2x} = 0\]This would make \(10\mathrm{e}^{-2x}\) equal to zero, which is impossible, so the method fails.
| Scheme | Marks | AO |
|---|---|---|
| Evaluates Charlotte’s method by explaining that she should first have found the complementary function or the auxiliary equation. | E1 | 2.3 |
| Obtains the auxiliary equation and its solutions \(u = -2, 1\) | B1 | 1.1b |
| Writes down the complementary function. FT their solutions of their auxiliary equation. | B1F | 1.1b |
| Selects a method to solve the differential equation by stating the correct PI \(y = \lambda x\mathrm{e}^{-2x}\) | B1 | 3.1a |
| Differentiates their PI twice (must be different from Charlotte’s PI) | M1 | 1.1a |
| Obtains correct 1st and 2nd derivative of the correct PI | A1 | 1.1b |
| Substitutes their PI and its derivatives into the differential equation. | M1 | 1.1a |
| Correctly reasons that the general solution of the differential equation is \(y = A\mathrm{e}^x + B\mathrm{e}^{-2x} - \frac{10}{3}x\mathrm{e}^{-2x}\) | R1 | 2.1 |
| (10 marks) |
Typical solution
Charlotte needs to find the complementary function first.
\[m^2 + m - 2 = 0\]\[m = 1 \text{ or } m = -2\]CF: \(y = A\mathrm{e}^x + B\mathrm{e}^{-2x}\)
The RHS has a similar form to the CF and so we need to introduce a factor of \(x\) into the particular integral.
PI: \(y = \lambda x\mathrm{e}^{-2x}\)
\[y^{\prime} = \lambda\mathrm{e}^{-2x}(-2x + 1)\]\[y^{\prime\prime} = \lambda\mathrm{e}^{-2x}(4x - 4)\]\[\lambda\mathrm{e}^{-2x}(4x - 4 - 2x + 1 - 2x) = 10\mathrm{e}^{-2x}\]\[\lambda = -\frac{10}{3}\]General solution:
\[y = A\mathrm{e}^x + B\mathrm{e}^{-2x} - \frac{10}{3}x\mathrm{e}^{-2x}\]