CalcGospel 國際數學圖譜
返回

IAL 2020 Oct Q8

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 8

(a) Show that the transformation x=eux = e^u transforms the differential equation

x2d2ydx2+3xdydx−8y=4ln⁡xx>0(I)\begin{align*} x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\frac{\mathrm{d}y}{\mathrm{d}x} - 8y = 4\ln x \qquad x > 0 \qquad \text{(I)} \end{align*}

into the differential equation

d2ydu2+2dydu−8y=4u(II)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} + 2\frac{\mathrm{d}y}{\mathrm{d}u} - 8y = 4u \qquad \text{(II)} \end{align*}
(6)

(b) Determine the general solution of differential equation (II), expressing yy as a function of uu.

(7)

(c) Hence obtain the general solution of differential equation (I).

(1)
解答

(a)

Since x=eux=e^u,

dxdu= eududx= e−u\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}u} = &\,e^u\\[4mm] \frac{\mathrm{d}u}{\mathrm{d}x} = &\,e^{-u} \end{align*}

Hence

dydx= dydududx= e−udydud2ydx2= ddx(dydududx)= ddx(e−udydu)= ddu(e−udydu)dudx= e−uddu(e−udydu)= e−u(−e−udydu+e−ud2ydu2)= e−2u(d2ydu2−dydu)\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{\mathrm{d}y}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}\\[4mm] = &\,e^{-u}\frac{\mathrm{d}y}{\mathrm{d}u}\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}\right)\\[4mm] = &\,\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{-u}\frac{\mathrm{d}y}{\mathrm{d}u}\right)\\[4mm] = &\,\frac{\mathrm{d}}{\mathrm{d}u}\left(e^{-u}\frac{\mathrm{d}y}{\mathrm{d}u}\right)\frac{\mathrm{d}u}{\mathrm{d}x}\\[4mm] = &\,e^{-u}\frac{\mathrm{d}}{\mathrm{d}u}\left(e^{-u}\frac{\mathrm{d}y}{\mathrm{d}u}\right)\\[4mm] = &\,e^{-u}\left(-e^{-u}\frac{\mathrm{d}y}{\mathrm{d}u}+e^{-u}\frac{\mathrm{d}^2y}{\mathrm{d}u^2}\right)\\[4mm] = &\,e^{-2u}\left(\frac{\mathrm{d}^2y}{\mathrm{d}u^2}-\frac{\mathrm{d}y}{\mathrm{d}u}\right) \end{align*}

So

 x2d2ydx2+3xdydx−8y= e2u⋅e−2u(d2ydu2−dydu)+3eu⋅e−udydu−8y= d2ydu2+2dydu−8y\begin{align*} &\,x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\frac{\mathrm{d}y}{\mathrm{d}x} - 8y \\[4mm] = &\,e^{2u}\cdot e^{-2u}\left(\frac{\mathrm{d}^2y}{\mathrm{d}u^2}-\frac{\mathrm{d}y}{\mathrm{d}u}\right) + 3e^u\cdot e^{-u}\frac{\mathrm{d}y}{\mathrm{d}u} - 8y\\[4mm] = &\,\frac{\mathrm{d}^2y}{\mathrm{d}u^2}+2\frac{\mathrm{d}y}{\mathrm{d}u}-8y \end{align*}

Now 4ln⁡x=4u4\ln x = 4u, so differential equation (II) is

d2ydu2+2dydu−8y=4u.\frac{\mathrm{d}^2y}{\mathrm{d}u^2}+2\frac{\mathrm{d}y}{\mathrm{d}u}-8y=4u.
Alternative methods for (a)

A shorter accepted route is to work backwards from (II). Then

dydx= 1xdydud2ydx2= 1xddu(1xdydu)= 1x2(d2ydu2−dydu)\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}u}\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = &\,\frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}u}\right)\\[4mm] = &\,\frac{1}{x^2}\left(\frac{\mathrm{d}^2y}{\mathrm{d}u^2}-\frac{\mathrm{d}y}{\mathrm{d}u}\right) \end{align*}

so substituting into (I) gives the required equation (II).

Another accepted style is to mix xx, yy and uu until the final line, provided the chain rule steps are clear.

(b)

The complementary function is

m2+2m−8=0\begin{align*} m^2+2m-8=0 \end{align*}

so m=2,−4m=2,-4, and

yc=Ae2u+Be−4u\begin{align*} y_c = Ae^{2u}+Be^{-4u} \end{align*}

Take a particular solution of the form yp=au+by_p=au+b. Then

0+2a−8(au+b)=4u\begin{align*} 0+2a-8(au+b)=4u \end{align*}

so a=−12a=-\frac12 and b=−18b=-\frac18.

Hence

y=Ae2u+Be−4u−12u−18\begin{align*} y = Ae^{2u}+Be^{-4u}-\frac12u-\frac18 \end{align*}

(c)

Since u=ln⁡xu=\ln x,

y=Ax2+Bx−4−12ln⁡x−18\begin{align*} y = Ax^2+Bx^{-4}-\frac12\ln x-\frac18 \end{align*}