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IAL 2022 Oct Q8

A Level / Edexcel / FP2

IAL 2022 Oct Paper · Question 8

题目

Problem

(a) Show that the transformation x=eux = \mathrm{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} -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

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由 x=eux=\mathrm{e}^u 可得 u=ln⁡xu=\ln x，所以

dudx=1x.\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{x}. \end{align*}

用链式法则把 yxy_x 和 yxxy_{xx} 换成关于 uu 的导数，再代回原方程。

答题过程

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Since

x=eu,\begin{align*} x=\mathrm{e}^u, \end{align*}

we have

u=ln⁡xanddudx=1x.\begin{align*} u=\ln x \quad\text{and}\quad \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{x}. \end{align*}

By the chain rule,

dydx= dydududx= 1xdydu.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x}\\[2mm] =&\,\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}u}. \end{align*}

Differentiate again with respect to xx:

d2ydx2= ddx(1xdydu)= −1x2dydu+1xddx(dydu)= −1x2dydu+1xd2ydu2dudx= −1x2dydu+1x2d2ydu2= 1x2(d2ydu2−dydu).\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{x}\frac{\mathrm{d}y}{\mathrm{d}u}\right)\\[2mm] =&\,-\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}u} +\frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\mathrm{d}y}{\mathrm{d}u}\right)\\[2mm] =&\,-\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}u} +\frac{1}{x}\frac{\mathrm{d}^2y}{\mathrm{d}u^2} \frac{\mathrm{d}u}{\mathrm{d}x}\\[2mm] =&\,-\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}u} +\frac{1}{x^2}\frac{\mathrm{d}^2y}{\mathrm{d}u^2}\\[2mm] =&\,\frac{1}{x^2} \left( \frac{\mathrm{d}^2y}{\mathrm{d}u^2} -\frac{\mathrm{d}y}{\mathrm{d}u} \right). \end{align*}

Substitute into equation (I):

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

Also,

4ln⁡x=4u.\begin{align*} 4\ln x=4u. \end{align*}

Therefore equation (I) becomes

d2ydu2+2dydu−8y=4u.\begin{align*} \boxed{ \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

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也可以先把 uu 导数写成 xx 导数：

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

再对 uu 求导一次，得到 yuuy_{uu} 和 yxxy_{xx} 的关系。这条路线常常更短。

答题过程

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Since x=eux=\mathrm{e}^u,

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

Therefore

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}\\[2mm] =&\,x\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

Differentiate again with respect to uu:

d2ydu2= dxdudydx+xddu(dydx)= xdydx+xd2ydx2dxdu= xdydx+x2d2ydx2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}u^2} =&\,\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)\\[2mm] =&\,x\frac{\mathrm{d}y}{\mathrm{d}x} +x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \frac{\mathrm{d}x}{\mathrm{d}u}\\[2mm] =&\,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*}

Also,

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

Substituting into equation (I),

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

Since u=ln⁡xu=\ln x, the right hand side of equation (I) is 4u4u. Hence

d2ydu2+2dydu−8y=4u.\begin{align*} \boxed{ \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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这是二阶常系数非齐次微分方程。先解齐次方程得到 complementary function，再因为右边是一次式 4u4u，所以 particular integral 可以试

y=au+b.\begin{align*} y=au+b. \end{align*}

答题过程

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Equation (II) is

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*}

The auxiliary equation is

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

So

m=−4,m=2.\begin{align*} m=-4,\qquad m=2. \end{align*}

The complementary function is

yc=Ae−4u+Be2u.\begin{align*} y_c=A\mathrm{e}^{-4u}+B\mathrm{e}^{2u}. \end{align*}

For a particular integral, try

yp=au+b.\begin{align*} y_p=au+b. \end{align*}

Then

dypdu=a,d2ypdu2=0.\begin{align*} \frac{\mathrm{d}y_p}{\mathrm{d}u}=a, \qquad \frac{\mathrm{d}^2y_p}{\mathrm{d}u^2}=0. \end{align*}

Substitute into equation (II):

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

Compare coefficients:

−8a=4⟹a=−12,2a−8b=0⟹−1−8b=0.\begin{align*} -8a=4 \quad&\Longrightarrow\quad a=-\frac{1}{2},\\[2mm] 2a-8b=0 \quad&\Longrightarrow\quad -1-8b=0. \end{align*}

So

b=−18.\begin{align*} b=-\frac{1}{8}. \end{align*}

Therefore the general solution of equation (II) is

y=Ae−4u+Be2u−12u−18.\begin{align*} \boxed{ y=A\mathrm{e}^{-4u}+B\mathrm{e}^{2u} -\frac{1}{2}u-\frac{1}{8} }. \end{align*}

(c)

解法一

思路

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由 x=eux=\mathrm{e}^u 得 u=ln⁡xu=\ln x。所以

e−4u=x−4,e2u=x2.\begin{align*} \mathrm{e}^{-4u}=x^{-4}, \qquad \mathrm{e}^{2u}=x^2. \end{align*}

把 (b) 的结果换回 xx 即可。

答题过程

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Since

u=ln⁡x,\begin{align*} u=\ln x, \end{align*}

we have

e−4u=x−4,e2u=x2.\begin{align*} \mathrm{e}^{-4u}=x^{-4}, \qquad \mathrm{e}^{2u}=x^2. \end{align*}

Therefore the general solution of equation (I) is

y=Ax−4+Bx2−12ln⁡x−18.\begin{align*} \boxed{ y=Ax^{-4}+Bx^2-\frac{1}{2}\ln x-\frac{1}{8} }. \end{align*}