| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring substitution to transform a non-linear second-order ODE into a linear one, then solving with particular integral and applying initial conditions. While the substitution work requires careful differentiation using chain rule and the algebra is involved, the overall approach is methodical once the substitution is given. The back-substitution and initial conditions add complexity but follow standard procedures for Further Maths students. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dv}{dx} = 4y^3\frac{dy}{dx}\) | B1 | \(v = y^4\) |
| \(\frac{d^2v}{dx^2} = 4y^3\frac{d^2y}{dx^2} + 12y^2\left(\frac{dy}{dx}\right)^2\) | B1 | |
| \(\frac{d^2v}{dx^2} + \frac{dv}{dx} + 4v = 4y^3\frac{d^2y}{dx^2} + 12y^2\left(\frac{dy}{dx}\right)^2 + 4y^3\frac{dy}{dx} + 4y^4\) | M1 | Uses substitution to find \(v\)-\(x\) equation, AG |
| \(= 4\left(y^3\frac{d^2y}{dx^2} + 3y^2\left(\frac{dy}{dx}\right)^2 + y^3\frac{dy}{dx} + y^4\right) = 4e^{-2x}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{4y^3}\frac{dv}{dx} = \frac{1}{4v^{\frac{3}{4}}}\frac{dv}{dx}\) | (B1) | \(y = v^{\frac{1}{4}}\) |
| \(\frac{d^2y}{dx^2} = \frac{1}{4v^{\frac{3}{4}}}\frac{d^2v}{dx^2} - \frac{3}{16v^{\frac{7}{4}}}\left(\frac{dv}{dx}\right)^2\) | (B1) | |
| \(v^{\frac{3}{4}}\left(\frac{1}{4v^{\frac{3}{4}}}\frac{d^2v}{dx^2} - \frac{3}{16v^{\frac{7}{4}}}\left(\frac{dv}{dx}\right)^2\right) + 3v^{\frac{1}{2}}\left(\frac{1}{4v^{\frac{3}{4}}}\frac{dv}{dx}\right)^2 + v^{\frac{1}{4}}\left(\frac{1}{4v^{\frac{3}{4}}}\frac{dv}{dx}\right) + v = e^{-2x}\) | (M1) | Uses substitution to find \(v\)-\(x\) equation, AG |
| \(\frac{1}{4}\frac{d^2v}{dx^2} - \frac{3}{16v}\left(\frac{dv}{dx}\right)^2 + \frac{3}{16v}\left(\frac{dv}{dx}\right)^2 + \frac{1}{4}\frac{dv}{dx} + v = e^{-2x}\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m^2 + m + 4 = 0\) | M1 | Auxiliary equation. |
| \([v =\,] e^{-\frac{1}{2}x}\left(A\cos\frac{\sqrt{15}}{2}x + B\sin\frac{\sqrt{15}}{2}x\right)\) | A1 | Complementary function. Allow '\(v =\)' missing. |
| \(v = ke^{-2x} \Rightarrow v' = -2ke^{-2x} \Rightarrow v'' = 4ke^{-2x}\) | B1 | Particular integral and its derivatives. |
| \(4ke^{-2x} - 2ke^{-2x} + 4ke^{-2x} = 4e^{-2x}\) | M1 | Substitutes and equates coefficients. |
| \(k = \frac{2}{3}\) | A1 | |
| \(v = y^4 = e^{-\frac{1}{2}x}\left(A\cos\frac{\sqrt{15}}{2}x + B\sin\frac{\sqrt{15}}{2}x\right) + \frac{2}{3}e^{-2x}\) | A1 | Substitutes for \(y\) and finds \(v\) in terms of \(x\). Must not see i. |
| \(v' = 4y^3\frac{dy}{dx} = e^{-\frac{1}{2}x}\left(-\frac{\sqrt{15}}{2}A\sin\frac{\sqrt{15}}{2}x + \frac{\sqrt{15}}{2}B\cos\frac{\sqrt{15}}{2}x\right) - \frac{1}{2}e^{-\frac{1}{2}x}\left(A\cos\frac{\sqrt{15}}{2}x + B\sin\frac{\sqrt{15}}{2}x\right) - \frac{4}{3}e^{-2x}\) | B1 | |
| \(1 = A + \frac{2}{3}\), \(\quad -\frac{3}{2} = \frac{\sqrt{15}}{2}B - \frac{1}{2}A - \frac{4}{3}\) \(\Rightarrow A = \frac{1}{3},\; B = 0\) | M1 A1 | Substitutes initial conditions and forms simultaneous equations. Must have used the product rule when differentiating *their* \(v\) for M1. |
| \(y = \left(\frac{1}{3}e^{-\frac{1}{2}x}\cos\frac{\sqrt{15}}{2}x + \frac{2}{3}e^{-2x}\right)^{\frac{1}{4}}\) | A1 | Accept \(y = \pm\left(\frac{1}{3}e^{-\frac{1}{2}x}\cos\frac{\sqrt{15}}{2}x + \frac{2}{3}e^{-2x}\right)^{\frac{1}{4}}\) |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dv}{dx} = 4y^3\frac{dy}{dx}$ | B1 | $v = y^4$ |
| $\frac{d^2v}{dx^2} = 4y^3\frac{d^2y}{dx^2} + 12y^2\left(\frac{dy}{dx}\right)^2$ | B1 | |
| $\frac{d^2v}{dx^2} + \frac{dv}{dx} + 4v = 4y^3\frac{d^2y}{dx^2} + 12y^2\left(\frac{dy}{dx}\right)^2 + 4y^3\frac{dy}{dx} + 4y^4$ | M1 | Uses substitution to find $v$-$x$ equation, AG |
| $= 4\left(y^3\frac{d^2y}{dx^2} + 3y^2\left(\frac{dy}{dx}\right)^2 + y^3\frac{dy}{dx} + y^4\right) = 4e^{-2x}$ | A1 | AG |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{4y^3}\frac{dv}{dx} = \frac{1}{4v^{\frac{3}{4}}}\frac{dv}{dx}$ | (B1) | $y = v^{\frac{1}{4}}$ |
| $\frac{d^2y}{dx^2} = \frac{1}{4v^{\frac{3}{4}}}\frac{d^2v}{dx^2} - \frac{3}{16v^{\frac{7}{4}}}\left(\frac{dv}{dx}\right)^2$ | (B1) | |
| $v^{\frac{3}{4}}\left(\frac{1}{4v^{\frac{3}{4}}}\frac{d^2v}{dx^2} - \frac{3}{16v^{\frac{7}{4}}}\left(\frac{dv}{dx}\right)^2\right) + 3v^{\frac{1}{2}}\left(\frac{1}{4v^{\frac{3}{4}}}\frac{dv}{dx}\right)^2 + v^{\frac{1}{4}}\left(\frac{1}{4v^{\frac{3}{4}}}\frac{dv}{dx}\right) + v = e^{-2x}$ | (M1) | Uses substitution to find $v$-$x$ equation, AG |
| $\frac{1}{4}\frac{d^2v}{dx^2} - \frac{3}{16v}\left(\frac{dv}{dx}\right)^2 + \frac{3}{16v}\left(\frac{dv}{dx}\right)^2 + \frac{1}{4}\frac{dv}{dx} + v = e^{-2x}$ | (A1) | |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 + m + 4 = 0$ | M1 | Auxiliary equation. |
| $[v =\,] e^{-\frac{1}{2}x}\left(A\cos\frac{\sqrt{15}}{2}x + B\sin\frac{\sqrt{15}}{2}x\right)$ | A1 | Complementary function. Allow '$v =$' missing. |
| $v = ke^{-2x} \Rightarrow v' = -2ke^{-2x} \Rightarrow v'' = 4ke^{-2x}$ | B1 | Particular integral and its derivatives. |
| $4ke^{-2x} - 2ke^{-2x} + 4ke^{-2x} = 4e^{-2x}$ | M1 | Substitutes and equates coefficients. |
| $k = \frac{2}{3}$ | A1 | |
| $v = y^4 = e^{-\frac{1}{2}x}\left(A\cos\frac{\sqrt{15}}{2}x + B\sin\frac{\sqrt{15}}{2}x\right) + \frac{2}{3}e^{-2x}$ | A1 | Substitutes for $y$ and finds $v$ in terms of $x$. Must not see i. |
| $v' = 4y^3\frac{dy}{dx} = e^{-\frac{1}{2}x}\left(-\frac{\sqrt{15}}{2}A\sin\frac{\sqrt{15}}{2}x + \frac{\sqrt{15}}{2}B\cos\frac{\sqrt{15}}{2}x\right) - \frac{1}{2}e^{-\frac{1}{2}x}\left(A\cos\frac{\sqrt{15}}{2}x + B\sin\frac{\sqrt{15}}{2}x\right) - \frac{4}{3}e^{-2x}$ | B1 | |
| $1 = A + \frac{2}{3}$, $\quad -\frac{3}{2} = \frac{\sqrt{15}}{2}B - \frac{1}{2}A - \frac{4}{3}$ $\Rightarrow A = \frac{1}{3},\; B = 0$ | M1 A1 | Substitutes initial conditions and forms simultaneous equations. Must have used the product rule when differentiating *their* $v$ for M1. |
| $y = \left(\frac{1}{3}e^{-\frac{1}{2}x}\cos\frac{\sqrt{15}}{2}x + \frac{2}{3}e^{-2x}\right)^{\frac{1}{4}}$ | A1 | Accept $y = \pm\left(\frac{1}{3}e^{-\frac{1}{2}x}\cos\frac{\sqrt{15}}{2}x + \frac{2}{3}e^{-2x}\right)^{\frac{1}{4}}$ |8 It is given that $\mathbf { v } = y ^ { 4 }$ and
$$y ^ { 3 } \frac { d ^ { 2 } y } { d x ^ { 2 } } + 3 y ^ { 2 } \left( \frac { d y } { d x } \right) ^ { 2 } + y ^ { 3 } \frac { d y } { d x } + y ^ { 4 } = e ^ { - 2 x }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { d ^ { 2 } v } { d x ^ { 2 } } + \frac { d v } { d x } + 4 v = 4 e ^ { - 2 x }$$
\item Find $y$ in terms of $x$, given that, when $x = 0 , y = 1$ and $\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 3 } { 8 }$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q8 [14]}}