A2 June 2024 Paper 2 Q19
19 Solve the differential equation
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4\frac{\mathrm{d}y}{\mathrm{d}x} - 45y = 21\mathrm{e}^{5x} - 0.3x + 27x^2\]given that \(y = \dfrac{37}{225}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(x = 0\) [10 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses a three-term auxiliary equation to obtain the Complementary Function. | M1 | 3.1a |
| Obtains the correct Complementary Function. | A1 | 1.1b |
| Deduces correct exponential form of Particular Integral, \(Cx\mathrm{e}^{5x}\) | B1 | 2.2a |
| Uses correct polynomial form of Particular Integral, \(D + Ex + Fx^2\) | B1 | 3.1a |
| Substitutes their \(y^{\prime}_{PI}\) and \(y^{\prime\prime}_{PI}\), with their \(y_{PI}\), into the differential equation. | M1 | 1.1a |
| Compares coefficients to obtain at least three of their \(C\), \(D\), \(E\) and \(F\) (with at least two correct). | M1 | 1.1a |
| Obtains correct values of \(C\), \(D\), \(E\) and \(F\) | A1 | 1.1b |
| Forms their General Solution, with exactly two arbitrary constants. | M1 | 3.1a |
| Uses their \(y_{GS} = \dfrac{37}{225}\) and \(y^{\prime}_{GS} = 0\) to form two equations in their two arbitrary constants. | M1 | 1.1a |
| Obtains the correct final result. | A1 | 3.2a |
| (10 marks) |
Typical solution
Complementary Function
\[\lambda^2 + 4\lambda - 45 = 0\]\[\lambda = 5 \text{ or } \lambda = -9\]\[y = A\mathrm{e}^{-9x} + B\mathrm{e}^{5x}\]Particular Integral
\[y_{PI} = Cx\mathrm{e}^{5x} + D + Ex + Fx^2\]\[y^{\prime}_{PI} = C\mathrm{e}^{5x} + 5Cx\mathrm{e}^{5x} + E + 2Fx\]\[y^{\prime\prime}_{PI} = 10C\mathrm{e}^{5x} + 25Cx\mathrm{e}^{5x} + 2F\]\[\begin{aligned} &10C\mathrm{e}^{5x} + 25Cx\mathrm{e}^{5x} + 2F + 4(C\mathrm{e}^{5x} + 5Cx\mathrm{e}^{5x} + E + 2Fx) \\ &\quad - 45(Cx\mathrm{e}^{5x} + D + Ex + Fx^2) = 21\mathrm{e}^{5x} - 0.3x + 27x^2 \end{aligned}\]\[\Rightarrow C = \frac{3}{2},\ D = \frac{-8}{225},\ E = \frac{-1}{10},\ F = \frac{-3}{5}\]General Solution
\[y_{GS} = A\mathrm{e}^{-9x} + B\mathrm{e}^{5x} + \frac{3}{2}x\mathrm{e}^{5x} - \frac{8}{225} - \frac{1}{10}x - \frac{3}{5}x^2\]\[y^{\prime}_{GS} = -9A\mathrm{e}^{-9x} + 5B\mathrm{e}^{5x} + \frac{3}{2}\mathrm{e}^{5x} + \frac{15}{2}x\mathrm{e}^{5x} - \frac{1}{10} - \frac{6}{5}x\]\(x = 0\)
\[\frac{37}{225} = A + B - \frac{8}{225} \Rightarrow A + B = \frac{1}{5}\]\[0 = -9A + 5B + \frac{3}{2} - \frac{1}{10} \Rightarrow 9A - 5B = \frac{7}{5}\]\[A = \frac{6}{35},\ B = \frac{1}{35}\]\[y = \frac{6}{35}\mathrm{e}^{-9x} + \frac{1}{35}\mathrm{e}^{5x} + \frac{3}{2}x\mathrm{e}^{5x} - \frac{8}{225} - \frac{1}{10}x - \frac{3}{5}x^2\]