## Question 8(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=e^u$, $\frac{dx}{du}=e^u$ or $\frac{du}{dx}=e^{-u}$ or $\frac{dx}{du}=x$ | B1 | Seen explicitly or used |
| $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=e^{-u}\frac{dy}{du}$ | M1 | Obtaining $\frac{dy}{dx}$ using chain rule |
| $\frac{d^2y}{dx^2}=-e^{-u}\frac{du}{dx}\frac{dy}{du}+e^{-u}\frac{d^2y}{du^2}\frac{du}{dx}=e^{-2u}\left(-\frac{dy}{du}+\frac{d^2y}{du^2}\right)$ | M1 A1 | Obtaining $\frac{d^2y}{dx^2}$ using product rule; correct expression |
| $e^{2u}\times e^{-2u}\left(-\frac{dy}{du}+\frac{d^2y}{du^2}\right)+3e^u\times e^{-u}\frac{dy}{du}-8y=4\ln(e^u)$ | dM1 | Substituting to eliminate $x$ ($u$ and $y$ only); depends on 2nd M mark |
| $\frac{d^2y}{du^2}+2\frac{dy}{du}-8y=4u$ ✱ | A1*cso (6) | Obtaining given result from completely correct work |

**Alternative Method 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=e^u$, $\frac{dx}{du}=e^u=x$ | B1 | |
| $\frac{dy}{du}=\frac{dy}{dx}\times\frac{dx}{du}=x\frac{dy}{dx}$ | M1 | |
| $\frac{d^2y}{du^2}=1\frac{dx}{du}\times\frac{dy}{dx}+x\frac{d^2y}{dx^2}\times\frac{dx}{du}=x\frac{dy}{dx}+x^2\frac{d^2y}{dx^2}$ | M1 A1 | |
| $x^2\frac{d^2y}{dx^2}=\frac{d^2y}{du^2}-\frac{dy}{du}$ | | |
| $\left(\frac{d^2y}{du^2}-\frac{dy}{du}\right)+3x\times\frac{1}{x}\frac{dy}{du}-8y=4\ln(e^u)$ | | |
| $\frac{d^2y}{du^2}+2\frac{dy}{du}-8y=4u$ ✱ | dM1 A1*cso (6) | |

## Question (a): Substitution / Change of Variable

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{du} = e^u$ (oe) | B1 | Seen explicitly or used |
| Obtaining $\frac{dy}{du}$ | M1 | Using chain rule here or seen later |
| Obtaining $\frac{d^2y}{du^2}$ | M1 | Using product rule (penalise lack of chain rule by the A mark) |
| Correct expression for $\frac{d^2y}{du^2}$ | A1 | Any equivalent form |
| Substituting to eliminate $x$ ($u$ and $y$ **only**) and obtaining the given result | dM1 A1*cso | Depends on 2nd M mark. Obtaining given result from completely correct work |

**Alternative 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \ln x$, $\quad \frac{du}{dx} = \frac{1}{x}$ | B1 | |
| $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{x}\frac{dy}{du}$ | M1 | |
| $\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x}\frac{d^2y}{du^2}\times\frac{du}{dx} = -\frac{1}{x^2}\frac{dy}{du} + \frac{1}{x^2}\frac{d^2y}{du^2}$ | M1A1 | |
| $x^2\!\left(-\frac{1}{x^2}\frac{dy}{du}+\frac{1}{x^2}\frac{d^2y}{du^2}\right)+3x\times\frac{1}{x}\frac{dy}{du}-8y=4u$ leading to $\frac{d^2y}{du^2}+2\frac{dy}{du}-8y=4u$ $\ast$ | M1A1*cso | |

---

## Question (b): Solving the ODE

| Answer/Working | Mark | Guidance |
|---|---|---|
| Auxiliary equation: $m^2+2m-8=0$ | M1 | Writing down correct aux. equation and solving to $m=\ldots$ (usual rules) |
| $(m+4)(m-2)=0$, $\quad m=-4,\,2$ | A1 | Correct solution $m=-4,\,2$ |
| $\text{CF} = Ae^{-4u}+Be^{2u}$ | A1 | Correct CF – can use any (single) variable |
| PI: try $y=au+b$; $\frac{dy}{du}=a$, $\frac{d^2y}{du^2}=0$ | M1 | Using appropriate PI and finding $\frac{dy}{du}$ and $\frac{d^2y}{du^2}$. Use of $y=\lambda u$ scores M0 |
| $0+2a-8(au+b)=4u \Rightarrow a=-\frac{1}{2},\quad b=-\frac{1}{8}$ | dM1 A1 | Substitute to obtain values for unknowns. Depends on second M1. Correct unknowns (two or three with $c=0$) |
| $\therefore y = Ae^{-4u}+Be^{2u}-\frac{1}{2}u-\frac{1}{8}$ | B1ft (7) | Complete solution with CF and non-zero PI. Must have $y$ as function of $u$. Allow recovery of incorrect variables |

---

## Question (c): Back-Substitution

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = Ax^{-4}+Bx^2-\frac{1}{2}\ln x - \frac{1}{8}$ | B1 (1) | Reverse substitution to obtain correct expression for $y$ in terms of $x$. No ft here. $x^{-4}$ or $e^{-4\ln x}$ and $x^2$ or $e^{2\ln x}$ allowed. Must start $y=\ldots$ |

**Total: [14]**