CalcGospel 國際數學圖譜
返回

IAL 2020 Oct Q8

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 8

题目

Problem

(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)

解法一

思路

展开

从 x=eux=e^u 出发，先求 dudx\dfrac{\mathrm{d}u}{\mathrm{d}x}，再把对 xx 的一阶、二阶导数都改写成对 uu 的导数。最后代入原方程，并用 ln⁡x=u\ln x=u。

#### 答题过程
展开

Since x=eux=e^u,

dxdu= eu=xdudx= 1x=e−u\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}u} =&\,e^u=x\\[4mm] \frac{\mathrm{d}u}{\mathrm{d}x} =&\,\frac1x=e^{-u} \end{align*}

Hence

dydx= dydududx= e−udydu\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} \end{align*}

For the second derivative,

d2ydx2= ddx(e−udydu)= ddu(e−udydu)dudx= (−e−udydu+e−ud2ydu2)e−u= e−2u(d2ydu2−dydu)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\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] =&\,\left( -e^{-u}\frac{\mathrm{d}y}{\mathrm{d}u} +e^{-u}\frac{\mathrm{d}^2y}{\mathrm{d}u^2} \right)e^{-u}\\[4mm] =&\,e^{-2u} \left( \frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{\mathrm{d}y}{\mathrm{d}u} \right) \end{align*}

Substitute into (I):

x2d2ydx2+3xdydx−8y= e2ue−2u(d2ydu2−dydu) +3eue−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 =&\,e^{2u}e^{-2u} \left( \frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{\mathrm{d}y}{\mathrm{d}u} \right)\\[2mm] &\,\hspace{2pt}+3e^u 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*}

Also,

4ln⁡x=4ln⁡(eu)=4u\begin{align*} 4\ln x=4\ln(e^u)=4u \end{align*}

Therefore

d2ydu2+2dydu−8y=4u\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} +2\frac{\mathrm{d}y}{\mathrm{d}u} -8y=4u \end{align*}

解法二

思路

展开

也可以先建立 dydu\dfrac{\mathrm{d}y}{\mathrm{d}u} 和 dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} 的关系，再反推出 x2d2ydx2x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}。这种写法少用指数，代入时比较直接。

答题过程

展开

Since x=eux=e^u,

dxdu=x\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}u}=x \end{align*}

By the chain rule,

dydu= dydxdxdu= xdydx\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}u} =&\,\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}u}\\[4mm] =&\,x\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

Differentiate again with respect to uu:

d2ydu2= ddu(xdydx)= dxdudydx+xddu(dydx)= xdydx+x(d2ydx2dxdu)= xdydx+x2d2ydx2\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} =&\,\frac{\mathrm{d}}{\mathrm{d}u} \left(x\frac{\mathrm{d}y}{\mathrm{d}x}\right)\\[4mm] =&\,\frac{\mathrm{d}x}{\mathrm{d}u}\frac{\mathrm{d}y}{\mathrm{d}x} +x\frac{\mathrm{d}}{\mathrm{d}u} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\\[4mm] =&\,x\frac{\mathrm{d}y}{\mathrm{d}x} +x\left( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \frac{\mathrm{d}x}{\mathrm{d}u} \right)\\[4mm] =&\,x\frac{\mathrm{d}y}{\mathrm{d}x} +x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

So

x2d2ydx2= d2ydu2−dydu\begin{align*} x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

and

3xdydx= 3dydu\begin{align*} 3x\frac{\mathrm{d}y}{\mathrm{d}x} =&\,3\frac{\mathrm{d}y}{\mathrm{d}u} \end{align*}

Substituting into (I),

(d2ydu2−dydu)+3dydu−8y= 4ln⁡(eu)d2ydu2+2dydu−8y= 4u\begin{align*} \left( \frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{\mathrm{d}y}{\mathrm{d}u} \right) +3\frac{\mathrm{d}y}{\mathrm{d}u} -8y =&\,4\ln(e^u)\\[4mm] \frac{\mathrm{d}^2y}{\mathrm{d}u^2} +2\frac{\mathrm{d}y}{\mathrm{d}u} -8y =&\,4u \end{align*}

(b)

解法一

思路

展开

(II) 是常系数二阶非齐次微分方程。右边是一次多项式 4u4u，所以特解尝试 au+bau+b。

答题过程

展开

For the complementary function,

m2+2m−8= 0(m+4)(m−2)= 0m= −4, 2\begin{align*} m^2+2m-8 =&\,0\\[4mm] (m+4)(m-2) =&\,0\\[4mm] m =&\,-4,\ 2 \end{align*}

Thus

yc=Ae−4u+Be2u\begin{align*} y_{\mathrm{c}}=Ae^{-4u}+Be^{2u} \end{align*}

Try

yp=au+b\begin{align*} y_{\mathrm{p}}=au+b \end{align*}

Then

yp′=a,yp′′=0\begin{align*} y_{\mathrm{p}}'=a,\qquad y_{\mathrm{p}}''=0 \end{align*}

Substitute into (II):

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

Compare coefficients:

−8a= 4,2a−8b= 0\begin{align*} -8a=&\,4,\\[2mm] 2a-8b=&\,0 \end{align*}

So

a=−12,b=−18\begin{align*} a=-\frac12,\qquad b=-\frac18 \end{align*}

Therefore

y=Ae−4u+Be2u−12u−18\begin{align*} y=Ae^{-4u}+Be^{2u}-\frac12u-\frac18 \end{align*}

(c)

解法一

思路

展开

把 u=ln⁡xu=\ln x 代回 (b)。同时 e2u=x2e^{2u}=x^2，e−4u=x−4e^{-4u}=x^{-4}。

答题过程

展开

Since u=ln⁡xu=\ln x,

e−4u= x−4,e2u= x2\begin{align*} e^{-4u}=&\,x^{-4},\\[2mm] e^{2u}=&\,x^2 \end{align*}

Therefore

y=Ax−4+Bx2−12ln⁡x−18\begin{align*} y=Ax^{-4}+Bx^2-\frac12\ln x-\frac18 \end{align*}