| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.8 This is a standard FP2 second-order differential equation requiring both complementary function (solving auxiliary equation with repeated root) and particular integral (using trial solution for sin 3x), followed by applying initial conditions. While methodical, it involves multiple techniques and careful algebraic manipulation, placing it moderately above average difficulty for A-level but routine for Further Maths students. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m^2-2m=0\Rightarrow m=0,\ 2\) | M1 | Solves auxiliary equation |
| \(A+Be^{2x}\) or \(Ae^0+Be^{2x}\) | A1 | Correct CF (CF or \(y=\) not needed) |
| \(a\cos 3x+b\sin 3x\) | B1 | Correct form for PI |
| \(\frac{dy}{dx}=-3a\sin 3x+3b\cos 3x\), \(\frac{d^2y}{dx^2}=-9a\cos 3x-9b\sin 3x\) | M1A1 | M1: Differentiates twice; change of trig functions needed, \(\pm1\) or \(\pm3\) for coefficients for first derivative, \(\pm1,\pm3\) or \(\pm9\) for second derivative; A1: Correct derivatives |
| \(-9a\cos 3x-9b\sin 3x+6a\sin 3x-6b\cos 3x=26\sin 3x\) | — | Substitution |
| \(-9a-6b=0,\ -9b+6a=26\Rightarrow a=...,\ b=...\) | dM1 | Substitutes and forms simultaneous equations (equating coefficients) and attempts to solve for \(a\) and \(b\); depends on second M mark |
| \(a=\frac{4}{3},\ b=-2\) | A1 | Correct \(a\) and \(b\) |
| \(y=A+Be^{2x}+\frac{4}{3}\cos 3x-2\sin 3x\) | A1ft (8) | Forms GS (ft their CF and PI); must start \(y=...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0=A+B+\frac{4}{3}\) | M1 | Substitutes \(x=0\) and \(y=0\) into GS |
| \(\left(\frac{dy}{dx}\right)=2Be^{2x}-4\sin 3x-6\cos 3x\Rightarrow 0=2B-6\) | M1 | Differentiates and substitutes \(x=0\) and \(y'=0\); change of trig functions needed, \(\pm1\) or \(\pm3\) for coefficients |
| \(0=A+B+\frac{4}{3},\ 0=2B-6\Rightarrow A=...,\ B=...\) | dM1 | Solves simultaneously; depends on second M mark |
| \(A=\frac{-13}{3},\ B=3\) | A1 | Correct values |
| \(y=3e^{2x}-\frac{13}{3}+\frac{4}{3}\cos 3x-2\sin 3x\) | A1ft (5) | Follow through their GS and \(A\), \(B\); must start \(y=...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx}-2y=-\frac{26}{3}\cos 3x+c\) | M1A1 | M1: Integrates both sides wrt \(x\); A1: Correct expression |
| \(I=e^{\int-2\,dx}=e^{-2x}\) | B1 | Correct integrating factor |
| \(ye^{-2x}=\int e^{-2x}\left(-\frac{26}{3}\cos 3x+c\right)dx\) | M1A1 | M1: Uses \(yI=\int I(\text{RHS})\,dx\); A1: Correct expression |
| \(=\frac{4}{3}e^{-2x}\cos 3x-2e^{-2x}\sin 3x-\frac{1}{2}ce^{-2x}+B\) | M1A1 | M1: Integration by parts twice; A1: Correct expression |
| \(y=-\frac{1}{2}c+Be^{2x}+\frac{4}{3}\cos 3x-2\sin 3x\) | — | Must start \(y=...\) |
# Question 5:
$$\frac{d^2y}{dx^2}-2\frac{dy}{dx}=26\sin 3x$$
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2-2m=0\Rightarrow m=0,\ 2$ | M1 | Solves auxiliary equation |
| $A+Be^{2x}$ or $Ae^0+Be^{2x}$ | A1 | Correct CF (CF or $y=$ not needed) |
| $a\cos 3x+b\sin 3x$ | B1 | Correct form for PI |
| $\frac{dy}{dx}=-3a\sin 3x+3b\cos 3x$, $\frac{d^2y}{dx^2}=-9a\cos 3x-9b\sin 3x$ | M1A1 | M1: Differentiates twice; change of trig functions needed, $\pm1$ or $\pm3$ for coefficients for first derivative, $\pm1,\pm3$ or $\pm9$ for second derivative; A1: Correct derivatives |
| $-9a\cos 3x-9b\sin 3x+6a\sin 3x-6b\cos 3x=26\sin 3x$ | — | Substitution |
| $-9a-6b=0,\ -9b+6a=26\Rightarrow a=...,\ b=...$ | dM1 | Substitutes and forms simultaneous equations (equating coefficients) and attempts to solve for $a$ and $b$; depends on second M mark |
| $a=\frac{4}{3},\ b=-2$ | A1 | Correct $a$ and $b$ |
| $y=A+Be^{2x}+\frac{4}{3}\cos 3x-2\sin 3x$ | A1ft **(8)** | Forms GS (ft their CF and PI); must start $y=...$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0=A+B+\frac{4}{3}$ | M1 | Substitutes $x=0$ and $y=0$ into GS |
| $\left(\frac{dy}{dx}\right)=2Be^{2x}-4\sin 3x-6\cos 3x\Rightarrow 0=2B-6$ | M1 | Differentiates and substitutes $x=0$ and $y'=0$; change of trig functions needed, $\pm1$ or $\pm3$ for coefficients |
| $0=A+B+\frac{4}{3},\ 0=2B-6\Rightarrow A=...,\ B=...$ | dM1 | Solves simultaneously; depends on second M mark |
| $A=\frac{-13}{3},\ B=3$ | A1 | Correct values |
| $y=3e^{2x}-\frac{13}{3}+\frac{4}{3}\cos 3x-2\sin 3x$ | A1ft **(5)** | Follow through their GS and $A$, $B$; must start $y=...$ |
**Alternative for (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}-2y=-\frac{26}{3}\cos 3x+c$ | M1A1 | M1: Integrates both sides wrt $x$; A1: Correct expression |
| $I=e^{\int-2\,dx}=e^{-2x}$ | B1 | Correct integrating factor |
| $ye^{-2x}=\int e^{-2x}\left(-\frac{26}{3}\cos 3x+c\right)dx$ | M1A1 | M1: Uses $yI=\int I(\text{RHS})\,dx$; A1: Correct expression |
| $=\frac{4}{3}e^{-2x}\cos 3x-2e^{-2x}\sin 3x-\frac{1}{2}ce^{-2x}+B$ | M1A1 | M1: Integration by parts twice; A1: Correct expression |
| $y=-\frac{1}{2}c+Be^{2x}+\frac{4}{3}\cos 3x-2\sin 3x$ | — | Must start $y=...$ |
---5. (a) Find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 26 \sin 3 x$$
(b) Find the particular solution of this differential equation for which $y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$\\
when $x = 0$ when $x = 0$\\
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\hfill \mbox{\textit{Edexcel FP2 2017 Q5 [13]}}