# Question 8:

## Part (a):

| Working | Mark | Notes |
|---------|------|-------|
| $t = x^2 \Rightarrow dt = 2x\,dx$ or $dx = \frac{1}{2}t^{-\frac{1}{2}}dt$ oe | M1 | May be implied by subsequent work |
| $\int 2x^5 e^{-x^2}dx = \int t^2 e^{-t}\,dt$ | M1 | Integral in terms of $t$ only; must change $dx$ to $dt$ |
| $= -t^2 e^{-t} + 2\int te^{-t}\,dt$ | M1 | Integration by parts; reduces power of $t$; sign errors allowed |
| $= -t^2 e^{-t} - 2te^{-t} + 2\int e^{-t}\,dt$ | dM1 | Integration by parts again in same direction |
| $= -t^2 e^{-t} - 2te^{-t} - 2e^{-t} (+C)$ oe | A1 | Correct integration; constant not needed |
| $= -x^4 e^{-x^2} - 2x^2 e^{-x^2} - 2e^{-x^2}(+C)$ oe | A1 | Reverse substitution; **this mark cannot be recovered in (b)** |

**(6 marks)**

## Part (b):

| Working | Mark | Notes |
|---------|------|-------|
| Integrating Factor $e^{\int\frac{4}{x}dx} = x^4$ | B1 | Use of $x^4$ seen |
| $\frac{d}{dx}(x^4 y) = 2x^5 e^{-x^2}$ or $x^4 y = \int 2x^5 e^{-x^2}\,dx$ | M1 | Multiply through by their IF |
| $x^4 y = -x^4 e^{-x^2} - 2x^2 e^{-x^2} - 2e^{-x^2}(+C)$ | A1ft | Use answer from (a), which must be a function of $x$ |
| $y = -e^{-x^2} - \frac{2e^{-x^2}}{x^2} - \frac{2e^{-x^2}}{x^4} + \frac{C}{x^4}$ | A1 | Complete $y = \ldots$; include constant and deal with correctly; **not follow through** |

**(4 marks)**

## ALT Method (Substitution $t = x^2$):

**Part (b) - ALT:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = x^2 \Rightarrow dt = 2x\,dx$ or $dx = \frac{1}{2}t^{-\frac{1}{2}}dt$ | — | Setup of substitution |
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$, $x\frac{dy}{dx} = 2t\frac{dy}{dt}$; equation becomes $2t\frac{dt}{dx} + 4y = 2te^{-t}$ | — | Change of variable |
| Integrating Factor $e^{\int \frac{2}{t}dx} = t^2$ | B1 | Use of $t^2$ seen |
| $\frac{d}{dt}(t^2 y) = t^2 e^{-t}$ or $t^2 y = \int t^2 e^{-t}\,dt$ | M1 | Multiply through by IF |
| $t^2 y = -t^2 e^{-t} - 2te^{-t} - 2e^{-t} (+C)$ | A1ft | Use their work in (a) to integrate RHS |
| $y = -e^{-x^2} - \dfrac{2e^{-x^2}}{x^2} - \dfrac{2e^{-x^2}}{x^4} + \dfrac{C}{x^4}$ | A1 | Reverse substitution; complete to $y=$; include constant correctly. **Not follow through** |
| | **(4)** | |

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## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = -e^{-1} - 2e^{-1} - 2e^{-1} + C$ | M1 | Attempt to substitute $x=1$, $y=0$ into their $y$, provided it includes a constant |
| $\Rightarrow C = 5e^{-1}$ | A1 | **Not ft** — must have been obtained using a correct expression for $y$ |
| $y = -e^{-x^2} - \dfrac{2e^{-x^2}}{x^2} - \dfrac{2e^{-x^2}}{x^4} + \dfrac{5e^{-1}}{x^4}$ | A1ft | Must start $y=$; follow through their $C$ and expression for $y$ |
| | **(3)** | |
| | **Total 13** | |

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## Common Alternative Forms:

**Part (a):** $-x^4e^{-x^2} - 2x^2e^{-x^2} - 2e^{-x^2}(+C)$; $e^{-x^2}(-x^4-2x^2-2)(+C)$; $-e^{-x^2}(x^4+2x^2+2)(+C)$; $\dfrac{-(x^4+2x^2+2)}{e^{x^2}}(+C)$

**Part (b):** $y = -e^{-x^2} - \dfrac{2e^{-x^2}}{x^2} - \dfrac{2e^{-x^2}}{x^4} + \dfrac{C}{x^4}$; $y = e^{-x^2}\!\left(-1-\dfrac{2}{x^2}-\dfrac{2}{x^4}\right)+\dfrac{C}{x^4}$; $y = -e^{-x^2}\!\left(1+\dfrac{2}{x^2}+\dfrac{2}{x^4}\right)+\dfrac{C}{x^4}$; $y = \dfrac{-(x^4+2x^2+2)}{x^4 e^{x^2}}+\dfrac{C}{x^4}$

**Part (c):** As above with $C = 5e^{-1}$