| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.8 This is a second-order differential equation requiring finding a particular integral with a resonance case (the forcing term matches the complementary function frequency), then applying initial conditions. While the method is standard for Further Maths, the product rule differentiation of kx sin 2x twice and the resonance phenomenon make it moderately challenging, placing it above average difficulty. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
**Question 7(i)**
$y = kx\sin 2x \Rightarrow \dfrac{\text{d}y}{\text{d}x} = 2kx\cos 2x + k\sin 2x$ **M1** attempt using the Product Rule
and $\dfrac{\text{d}^2y}{\text{d}x^2} = kx.-4\sin 2x + 2k\cos 2x + 2k\cos 2x$ **M1** attempt using the Product Rule
$= -4y + 4k\cos 2x$ **M1** for substitution into given d.e. or comparison
$\Rightarrow k = 2$ **A1**
**Question 7(ii)**
Comp. Fn. from $m^2 + 4 = 0$ **M1**
$\Rightarrow y_C = A\cos 2x + B\sin 2x$ **A1** Or $R\cos(2x-\alpha)$ etc.
Gen. Soln. is thus $y = A\cos 2x + (B+2x)\sin 2x$ **B1** FT
Then $\dfrac{\text{d}y}{\text{d}x} = -2A\sin 2x + 2(B+2x)\cos 2x + 2\sin 2x$ **B1**
OR $= 2(B+2x)\cos 2x$ if found after $A$ (correctly) evaluated
Substitution of given initial conditions **M1**
$A = 1$ from $x=0,\ y=1$ **A1** FT from an incorrect $x\sin 2x$ term in $y$
$B = \dfrac{1}{2}$ from $x=0,\ \dfrac{\text{d}y}{\text{d}x}=1$
i.e. solution is $y = \cos 2x + \left(2x+\dfrac{1}{2}\right)\sin 2x$ **A1** FT from an incorrect $x\cos 2x$ term in $y'$. Withhold final A mark if in $e^\wedge$complex form.
**Total: 11 marks**7 (i) Find the value of the constant $k$ for which $y = k x \sin 2 x$ is a particular integral of the differential equation $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 8 \cos 2 x$.\\
(ii) Solve $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 8 \cos 2 x$, given that $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ when $x = 0$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2017 Q7 [11]}}