# Question 9:

## Part (a)(i):

| Working | Mark | Guidance |
|---------|------|----------|
| $x=t^{\frac{1}{2}} \Rightarrow \frac{dx}{dy}=\frac{1}{2}t^{-\frac{1}{2}}\frac{dt}{dy} \Rightarrow \frac{dy}{dx}=...$ or $t=x^2 \Rightarrow \frac{dt}{dx}=2x \Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=...$ | M1 | Applies chain rule and proceeds to expression for $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = 2t^{\frac{1}{2}}\frac{dy}{dt}$ | A1 | Any correct expression for $\frac{dy}{dx}$ in terms of $y$ and $t$ |

## Part (a)(ii):

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx}=2t^{\frac{1}{2}}\frac{dy}{dt} \Rightarrow \frac{d^2y}{dx^2}=\frac{dy}{dt}t^{-\frac{1}{2}}\frac{dt}{dx}+2t^{\frac{1}{2}}\frac{d^2y}{dt^2}\frac{dt}{dx}$ | dM1A1 | Uses product rule to differentiate equation of form $\frac{dy}{dx}=kt^{\frac{1}{2}}\frac{dy}{dt}$. **Or equivalent expressions where ... is non-zero** |
| $\frac{d^2y}{dx^2} = 2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}$ | A1 | Correct expression in terms of $y$ and $t$ |

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $x\frac{d^2y}{dx^2}-(6x^2+1)\frac{dy}{dx}+9x^3y=x^5 \Rightarrow t^{\frac{1}{2}}\left(2\frac{dy}{dt}+4t\frac{d^2y}{dt^2}\right)-(6t+1)2t^{\frac{1}{2}}\frac{dy}{dt}+9t^{\frac{3}{2}}y=t^{\frac{5}{2}}$ | M1 | Substitutes expressions from part (a) and replaces $x$ with $t^{\frac{1}{2}}$ |
| $\Rightarrow 4\frac{d^2y}{dt^2}-12\frac{dy}{dt}+9y=t$ | A1* | Obtains given answer with no errors and sufficient working shown. **Must follow full marks in (a)** |

## Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| $4m^2-12m+9=0 \Rightarrow m=\frac{3}{2}$ | M1 | Attempts to solve $4m^2-12m+9=0$. Apply general guidance for solving 3TQ |
| $(y=)e^{\frac{3}{2}t}(At+B)$ | A1 | Correct CF. No need for "$y=$". Must be in terms of $t$ in GS. Allow equivalents for the $\frac{3}{2}$ |
| $(y=)at+b \Rightarrow \frac{dy}{dt}=a \Rightarrow \frac{d^2y}{dt^2}=0 \Rightarrow -12a+9(at+b)=t$ | M1 | Starts with correct PI form and differentiates to obtain $\frac{dy}{dt}=a$ and $\frac{d^2y}{dt^2}=0$ and substitutes. NB starting with PI of $y=at$ is M0 |
| $9a=1 \Rightarrow a=...$, $9b-12a=0 \Rightarrow b=...$ | dM1 | Complete method to find $a$ and $b$ by comparing coefficients. **Depends on previous method mark** |
| $y=e^{\frac{3}{2}t}(At+B)+\frac{1}{9}t+\frac{4}{27}$ | A1 | Correct GS including "$y=$", must be in terms of $t$. Allow equivalent exact fractions |

## Part (d):

| Working | Mark | Guidance |
|---------|------|----------|
| $y=e^{\frac{3}{2}x^2}(Ax^2+B)+\frac{1}{9}x^2+\frac{4}{27}$ | B1ft | Correct equation including "$y=$" (follow through from (c)). Answer to (c) must be in terms of $t$; answer to (d) should be same as (c) with $t$ replaced by $x^2$ |

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