# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = e^t \Rightarrow \frac{dx}{dy} = e^t\frac{dt}{dy} \Rightarrow \frac{dy}{dx} = e^{-t}\frac{dy}{dt}$ | M1A1 | M1: Attempt first derivative using chain rule to obtain $\frac{dx}{dy} = e^t\frac{dt}{dy}$; A1: $\frac{dy}{dx} = e^{-t}\frac{dy}{dt}$ oe |
| $\frac{dy}{dx} = x^{-1}\frac{dy}{dt} \Rightarrow \frac{d^2y}{dx^2} = -x^{-2}\frac{dy}{dt} + x^{-1}\frac{d^2y}{dt^2}\cdot\frac{dt}{dx}$ | dM1A1 | M1: Attempt product rule and chain rule, dependent on first method mark, fully correct method with sign errors only; A1: Correct second derivative oe |
| $x^2\left(\frac{1}{x^2}\frac{d^2y}{dt^2} - \frac{1}{x^2}\frac{dy}{dt}\right) - 3x\left(\frac{1}{x}\frac{dy}{dt}\right) + 3y = (e^t)^2$ | M1 | Substitutes their $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$ in terms of $t$ into the differential equation |
| $\frac{d^2y}{dt^2} - 4\frac{dy}{dt} + 3y = e^{2t}$ | A1 | cso |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = e^t \Rightarrow \frac{dy}{dt} = e^t\frac{dy}{dx} = x\frac{dy}{dx}$ | M1A1 | M1: $\frac{dy}{dt} = \frac{dx}{dt}\times\frac{dy}{dx}$; A1: $\frac{dy}{dt} = x\frac{dy}{dx}$ oe |
| $\frac{d^2y}{dt^2} = \frac{dx}{dt}\frac{dy}{dx} + x\frac{d^2y}{dx^2}\cdot\frac{dx}{dt} = x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2}$ | dM1A1 | M1: Product rule and chain rule, dependent on first mark; A1: Correct second derivative oe |
| $\frac{d^2y}{dt^2} - x\frac{dy}{dx} - 3x\frac{dy}{dx} + 3y = e^{2t}$ $= \frac{d^2y}{dt^2} - \frac{dy}{dt} - 3\frac{dy}{dt} + 3y = e^{2t}$ | M1 | Substitutes their $\frac{d^2y}{dx^2}$ and $x\frac{dy}{dx}$ in terms of $t$ into differential equation |
| $\frac{d^2y}{dt^2} - 4\frac{dy}{dt} + 3y = e^{2t}$ | A1 | Cso |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 - 4m + 3 = 0 \Rightarrow m = 1,\ 3$ | M1 | Solves correct quadratic (so should be $m = \pm1, \pm3$) |
| $(y =)\ Ae^{3t} + Be^t$ | A1 | Correct CF in terms of $t$ not $x$ (may be seen later in GS) |
| $y = ke^{2t},\ y' = 2ke^{2t},\ y'' = 4ke^{2t}$ | M1 | Correct form for PI and differentiates twice to obtain multiples of $e^{2t}$, do not allow if clearly integrating |
| $4ke^{2t} - 8ke^{2t} + 3ke^{2t} = e^{2t} \Rightarrow k = \ldots$ | M1 | Substitutes $y, y', y''$ of form $\alpha e^{2t}$ into DE, sets $= e^{2t}$ and finds $k$ |
| $(y) = -e^{2t}$ | A1 | Correct PI or $k = -1$ |
| $y = Ae^{3t} + Be^t - e^{2t}$ | B1ft | Correct full GS **in terms of $t$** (CF + PI with non-zero PI). **Must be $y = \ldots$** |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y =)\ Ax^3 + Bx - x^2$ | B1 | Allow equivalent expressions in terms of $x$, e.g. $(y =)\ Ae^{3\ln x} + Be^{\ln x} - e^{2\ln x}$. **Note $y = \ldots$ is not needed here.** |

---