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IAL 2024 June Q7

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 7

Given that y=exsin⁡xy = \mathrm{e}^x \sin x

(a) show that

d6ydx6=kd2ydx2\begin{align*} \frac{\mathrm{d}^6y}{\mathrm{d}x^6} = k\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

where kk is a constant to be determined.

(4)

(b) Hence determine the first 55 non-zero terms in the Maclaurin series expansion for yy , giving each coefficient in simplest form.

(3)
解答

(a)

Given y=exsin⁡xy = \mathrm{e}^x \sin x. Differentiate with respect to xx using the product rule:

dydx= exsin⁡x+excos⁡x= y+excos⁡x\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\mathrm{e}^x \sin x + \mathrm{e}^x \cos x\\[4mm] = &\,y + \mathrm{e}^x \cos x \end{align*}

Differentiate again to find the second derivative:

d2ydx2= dydx+excos⁡x−exsin⁡x= dydx+(dydx−y)−y= 2dydx−2y\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,\frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{e}^x \cos x - \mathrm{e}^x \sin x\\[4mm] = &\,\frac{\mathrm{d}y}{\mathrm{d}x} + \left( \frac{\mathrm{d}y}{\mathrm{d}x} - y \right) - y\\[4mm] = &\,2\frac{\mathrm{d}y}{\mathrm{d}x} - 2y \end{align*}

Differentiate continuously to find higher derivatives:

d3ydx3= 2d2ydx2−2dydx\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = &\,2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*} d4ydx4= 2d3ydx3−2d2ydx2= 2(2d2ydx2−2dydx)−2d2ydx2= 4d2ydx2−4dydx−2d2ydx2= 2d2ydx2−4dydx= 2(2dydx−2y)−4dydx= 4dydx−4y−4dydx= −4y\begin{align*} \frac{\mathrm{d}^4y}{\mathrm{d}x^4} = &\,2\frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] = &\,2\left( 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} \right) - 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] = &\,4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\frac{\mathrm{d}y}{\mathrm{d}x} - 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] = &\,2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 4\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] = &\,2\left( 2\frac{\mathrm{d}y}{\mathrm{d}x} - 2y \right) - 4\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] = &\,4\frac{\mathrm{d}y}{\mathrm{d}x} - 4y - 4\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] = &\,-4y \end{align*}

Since the fourth derivative is directly proportional to yy:

d5ydx5= −4dydxd6ydx6= −4d2ydx2\begin{align*} \frac{\mathrm{d}^5y}{\mathrm{d}x^5} = &\,-4\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] \frac{\mathrm{d}^6y}{\mathrm{d}x^6} = &\,-4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

Thus, d6ydx6=kd2ydx2\frac{\mathrm{d}^6y}{\mathrm{d}x^6} = k\frac{\mathrm{d}^2y}{\mathrm{d}x^2}, where k=−4k = -4.

(b)

To find the Maclaurin series expansion, evaluate the derivatives at x=0x = 0:

y(0)= e0sin⁡0=0y′(0)= y(0)+e0cos⁡0=0+1=1y′′(0)= 2y′(0)−2y(0)=2(1)−2(0)=2y′′′(0)= 2y′′(0)−2y′(0)=2(2)−2(1)=2y(4)(0)= −4y(0)=−4(0)=0y(5)(0)= −4y′(0)=−4(1)=−4y(6)(0)= −4y′′(0)=−4(2)=−8\begin{align*} y(0) = &\,\mathrm{e}^0 \sin 0 = 0\\[4mm] y'(0) = &\,y(0) + \mathrm{e}^0 \cos 0 = 0 + 1 = 1\\[4mm] y''(0) = &\,2y'(0) - 2y(0) = 2(1) - 2(0) = 2\\[4mm] y'''(0) = &\,2y''(0) - 2y'(0) = 2(2) - 2(1) = 2\\[4mm] y^{(4)}(0) = &\,-4y(0) = -4(0) = 0\\[4mm] y^{(5)}(0) = &\,-4y'(0) = -4(1) = -4\\[4mm] y^{(6)}(0) = &\,-4y''(0) = -4(2) = -8 \end{align*}

The Maclaurin series expansion is given by:

y= y(0)+y′(0)x+y′′(0)2!x2+y′′′(0)3!x3+y(4)(0)4!x4+y(5)(0)5!x5+y(6)(0)6!x6+…= 0+(1)x+22x2+26x3+0+−4120x5+−8720x6+…= x+x2+13x3−130x5−190x6\begin{align*} y = &\,y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \frac{y^{(5)}(0)}{5!}x^5 + \frac{y^{(6)}(0)}{6!}x^6 + \dots\\[4mm] = &\,0 + (1)x + \frac{2}{2}x^2 + \frac{2}{6}x^3 + 0 + \frac{-4}{120}x^5 + \frac{-8}{720}x^6 + \dots\\[4mm] = &\,x + x^2 + \frac{1}{3}x^3 - \frac{1}{30}x^5 - \frac{1}{90}x^6 \end{align*}

These are the first 5 non-zero terms in the expansion.