# Question 8:

## Part (a):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $v = y^{-2}$, $\frac{dv}{dy} = -2y^{-3}$ | B1 | Correct derivative |
| $\frac{dy}{dx} = \frac{dy}{dv} \times \frac{dv}{dx} = -\frac{y^3}{2}\frac{dv}{dx}$ | M1A1 | Attempt $\frac{dy}{dx}$ or $\frac{dv}{dx}$ using the chain rule; correct |
| Substitute into equation to obtain equation in $v$ and $x$ only | dM1 | Substitute in equation (I) to obtain an equation in $v$ and $x$ only |
| $\frac{dv}{dx} - 12vx = -6xe^{x^2}$ | A1* | Correct completion with no errors seen |

**(5 marks)**

## Part (b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| IF: $e^{\int -12x\,dx} = e^{-6x^2}$ | M1A1 | IF of form $e^{\int \pm 12x\,dx}$ and attempt the integration; correct IF |
| $ve^{-6x^2} = \int -6xe^{x^2} \times (e^{-6x^2})\,dx = \int -6xe^{-5x^2}\,dx$ | dM1 | Multiply through by their IF and integrate the LHS. Depends on first M mark of (b) |
| $ve^{-6x^2} = \frac{6}{10}e^{-5x^2}\ (+c)$ | A1 | Correct integration of the complete equation with or without constant |
| $v\,(= y^{-2}) = \frac{6}{10}e^{x^2} + ce^{6x^2}$ | ddM1 | Include the constant and multiply through by $e^{6x^2}$. Depends on both previous M marks of (b) |
| $y^2 = \frac{1}{\frac{6}{10}e^{x^2} + ce^{6x^2}}$ or e.g. $y^2 = \frac{10}{6e^{x^2} + ke^{6x^2}}$ | A1 | Any equivalent form (no need to change letter used for constant when rearranging) |

**(6 marks) [11 total]**

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