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FP2 2020 Oct Q8

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 8

题目

Problem

8.

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

 x2d2ydx2+3xdydx −8y=4ln⁡x x>0(I)\begin{align*} &\,x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +3x\frac{\mathrm{d}y}{\mathrm{d}x}\\[2mm] &\,\hspace{2pt}-8y=4\ln x\\[3mm] &\,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}\\[2mm] &\,\hspace{2pt}-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)
题目中文翻译

(a) 证明变换 x=eux=e^u 将微分方程

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

化为微分方程

 d2ydu2+2dydu −8y=4u(II).\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}u^2} +2\frac{\mathrm{d}y}{\mathrm{d}u}\\[2mm] &\,\hspace{2pt}-8y=4u\qquad\text{(II)}. \end{align*}
(6)

(b) 求微分方程 (II) 的通解，将 yy 表示为 uu 的函数。

(7)

(c) 由此求微分方程 (I) 的通解。

(1)

解答

(a)

解法一

思路

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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

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

Since x2e−2u=1x^2e^{-2u}=1 and xe−u=1xe^{-u}=1, the left-hand side of (I) becomes

 x2d2ydx2 +3xdydx−8y= (d2ydu2−dydu) +3dydu−8y= d2ydu2+2dydu−8y.\begin{align*} &\,x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] &\,+3x\frac{\mathrm{d}y}{\mathrm{d}x}-8y\\[4mm] =&\,\big(\frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{\mathrm{d}y}{\mathrm{d}u}\big)\\[2mm] &\,+3\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)=4u4\ln x=4\ln(e^u)=4u. 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*}

Equivalently, starting with u=ln⁡xu=\ln x gives

dudx= 1x,dydx= 1xdydu.\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x}=&\,\frac1x,\\[4mm] \frac{\mathrm{d}y}{\mathrm{d}x}=&\,\frac1x\frac{\mathrm{d}y}{\mathrm{d}u}. \end{align*}

Differentiating this product with respect to xx,

 d2ydx2= −1x2dydu +1xd2ydu2dudx= −1x2dydu +1x2d2ydu2.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] =&\,-\frac1{x^2}\frac{\mathrm{d}y}{\mathrm{d}u}\\[2mm] &\,+\frac1x\frac{\mathrm{d}^2y}{\mathrm{d}u^2} \frac{\mathrm{d}u}{\mathrm{d}x}\\[4mm] =&\,-\frac1{x^2}\frac{\mathrm{d}y}{\mathrm{d}u}\\[2mm] &\,+\frac1{x^2}\frac{\mathrm{d}^2y}{\mathrm{d}u^2}. \end{align*}

These are the same derivative identities, since 1/x=e−u1/x=e^{-u}, and give the same transformed equation.

解法二

思路

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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

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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 +xd2ydx2dxdu= xdydx+x2d2ydx2\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}u^2}\\[4mm] =&\,\frac{\mathrm{d}}{\mathrm{d}u} \big(x\frac{\mathrm{d}y}{\mathrm{d}x}\big)\\[4mm] =&\,\frac{\mathrm{d}x}{\mathrm{d}u}\frac{\mathrm{d}y}{\mathrm{d}x}\\[2mm] &\,+x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \frac{\mathrm{d}x}{\mathrm{d}u}\\[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)= 4u\begin{align*} &\,\big(\frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{\mathrm{d}y}{\mathrm{d}u}\big)\\[2mm] &\,+3\frac{\mathrm{d}y}{\mathrm{d}u}-8y\\[4mm] =&\,4\ln(e^u)\\[2mm] =&\,4u \end{align*}

Hence

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*}

(b)

解法一

思路

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(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

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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}\\[2mm] &\,-\frac12u-\frac18 \end{align*}

(c)

解法一

思路

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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

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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\\[2mm] &\,-\frac12\ln x-\frac18 \end{align*}