CalcGospel 國際數學圖譜
返回

FP2 2026 Jan Q8

A Level / Edexcel / FP2

IAL 2026 Jan Paper · Question 8

题目

Problem

8.

(a) Given that u=xyu=xy, where uu is a function of xx (x>0)(x>0), show that

d2ydx2=1x(d2udx2−2dydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac1x \left( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \right) \end{align*}
(3)

(b) Hence show that the transformation u=xyu=xy transforms the differential equation

xd2ydx2+2(2x+1)dydx+13xy=17e3x−4y(I)\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2(2x+1)\frac{\mathrm{d}y}{\mathrm{d}x} +13xy = 17\mathrm{e}^{3x}-4y\qquad\text{(I)} \end{align*}

into the differential equation

d2udx2+4dudx+13u=17e3x(II)\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4\frac{\mathrm{d}u}{\mathrm{d}x} +13u = 17\mathrm{e}^{3x}\qquad\text{(II)} \end{align*}
(2)

(c) Solve differential equation (II) to determine a general solution for uu in terms of xx.

(5)

(d) Hence determine the general solution of differential equation (I).

(1)
题目中文翻译

(a) 已知 u=xyu=xy，其中 uu 是 xx 的函数，且 x>0x>0。证明

d2ydx2=1x(d2udx2−2dydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac1x\Big(\frac{\mathrm{d}^2u}{\mathrm{d}x^2}-2\frac{\mathrm{d}y}{\mathrm{d}x}\Big) \end{align*}
(3)

(b) 由此证明代换 u=xyu=xy 可将微分方程

 xd2ydx2+2(2x+1)dydx+13xy =17e3x−4y(I)\begin{align*} &\,x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2(2x+1)\frac{\mathrm{d}y}{\mathrm{d}x}+13xy\\[2mm] &\,\hspace{2pt}=17\mathrm{e}^{3x}-4y\qquad\text{(I)} \end{align*}

化为微分方程

d2udx2+4dudx+13u=17e3x(II)\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2}+4\frac{\mathrm{d}u}{\mathrm{d}x}+13u=17\mathrm{e}^{3x}\qquad\text{(II)} \end{align*}
(2)

(c) 解微分方程 (II)，求出以 xx 表示的 uu 的通解。

(5)

(d) 由此求微分方程 (I) 的通解。

(1)

解答

(a)

解法一

思路

展开

从 u=xyu=xy 求导两次。第一次会得到 u′=xy′+yu'=xy'+y，第二次再整理，就可以把 y′′y'' 表示出来。

答题过程

展开

Since

u=xy,\begin{align*} u=xy, \end{align*}

differentiating gives

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

Differentiate again:

d2udx2= xd2ydx2+dydx+dydx= xd2ydx2+2dydx\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} =&\, x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +\frac{\mathrm{d}y}{\mathrm{d}x} +\frac{\mathrm{d}y}{\mathrm{d}x}\\[3mm] =&\, x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

Therefore

xd2ydx2= d2udx2−2dydx\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

and since x>0x>0,

d2ydx2= 1x(d2udx2−2dydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac1x \Big( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \Big) \end{align*}

解法二

思路

展开

因为 x>0x>0，我们可以将变换式变形为 y=uxy=\dfrac{u}{x}。然后对 xx 连续求导两次，得到 y′′y'' 关于 uu 及其导数的表达式，最后通过将已知关系代回，证明其与目标等式右边（RHS）完全等价。

答题过程

展开

Since x>0x>0, we can express yy in terms of uu and xx as:

y= ux= ux−1\begin{align*} y =&\, \frac{u}{x}\\[3mm] =&\, ux^{-1} \end{align*}

Differentiate with respect to xx using the Product Rule:

dydx= x−1dudx−ux−2= 1xdudx−ux2\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\, x^{-1}\frac{\mathrm{d}u}{\mathrm{d}x} - ux^{-2}\\[4mm] =&\, \frac1x \frac{\mathrm{d}u}{\mathrm{d}x} - \frac{u}{x^2} \end{align*}

Differentiate again with respect to xx:

d2ydx2= ddx(x−1dudx−ux−2)= (x−1d2udx2−x−2dudx) −(x−2dudx−2ux−3)= 1xd2udx2−2x2dudx+2ux3\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{\mathrm{d}}{\mathrm{d}x}\Big( x^{-1}\frac{\mathrm{d}u}{\mathrm{d}x} - ux^{-2} \Big)\\[4mm] =&\, \Big( x^{-1}\frac{\mathrm{d}^2u}{\mathrm{d}x^2} - x^{-2}\frac{\mathrm{d}u}{\mathrm{d}x} \Big)\\[2mm] &\,\hspace{2pt}-\Big( x^{-2}\frac{\mathrm{d}u}{\mathrm{d}x} - 2ux^{-3} \Big)\\[4mm] =&\, \frac1x \frac{\mathrm{d}^2u}{\mathrm{d}x^2} - \frac2{x^2} \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{2u}{x^3} \end{align*}

Now examine the RHS of the target equation:

RHS= 1x(d2udx2−2dydx)= 1x[d2udx2 −2(1xdudx−ux2)]= 1xd2udx2−2x2dudx+2ux3\begin{align*} \text{RHS} =&\, \frac1x \Big( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} \Big)\\[4mm] =&\, \frac1x \Big[ \frac{\mathrm{d}^2u}{\mathrm{d}x^2}\\[2mm] &\,\hspace{2pt}-2\Big( \frac1x \frac{\mathrm{d}u}{\mathrm{d}x} - \frac{u}{x^2} \Big) \Big]\\[4mm] =&\, \frac1x \frac{\mathrm{d}^2u}{\mathrm{d}x^2} - \frac2{x^2} \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{2u}{x^3} \end{align*}

Since LHS=d2ydx2=RHS\text{LHS} = \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \text{RHS}, the identity is proven:

d2ydx2= 1x(d2udx2−2dydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac1x \Big( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \Big) \end{align*}

(b)

解法一

思路

展开

把 (a) 的 y′′y'' 代入原方程，同时由 u′=xy′+yu'=xy'+y 得到 xy′=u′−yxy'=u'-y。这样所有含 yy 的项会抵消，留下关于 uu 的二阶方程。

答题过程

展开

From part (a),

xd2ydx2= d2udx2−2dydx\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

Also,

dudx= xdydx+y⟹xdydx= dudx−y\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} =&\, x\frac{\mathrm{d}y}{\mathrm{d}x}+y \quad\Longrightarrow\quad x\frac{\mathrm{d}y}{\mathrm{d}x} =&\, \frac{\mathrm{d}u}{\mathrm{d}x}-y \end{align*}

Substitute these into (I):

(d2udx2−2dydx)+2(2x+1)dydx+13u =17e3x−4y\begin{align*} &\Big( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \Big) +2(2x+1)\frac{\mathrm{d}y}{\mathrm{d}x} +13u\\[3mm] &\,\hspace{2pt}=17\mathrm{e}^{3x}-4y \end{align*}

So

d2udx2+4xdydx+13u= 17e3x−4yd2udx2+4(dudx−y)+13u= 17e3x−4y\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4x\frac{\mathrm{d}y}{\mathrm{d}x} +13u =&\, 17\mathrm{e}^{3x}-4y\\[3mm] \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4\Big( \frac{\mathrm{d}u}{\mathrm{d}x}-y \Big) +13u =&\, 17\mathrm{e}^{3x}-4y \end{align*}

The −4y-4y terms cancel, giving

d2udx2+4dudx+13u= 17e3x\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4\frac{\mathrm{d}u}{\mathrm{d}x} +13u =&\, 17\mathrm{e}^{3x} \end{align*}

(c)

解法一

思路

展开

这是常系数非齐次二阶微分方程。先解辅助方程得到互补函数，再设特解为 λe3x\lambda\mathrm{e}^{3x}。

答题过程

展开

Solve the homogeneous equation using

m2+4m+13=0\begin{align*} m^2+4m+13=0 \end{align*}

Then

m= −4±16−522= −2±3i\begin{align*} m =&\, \frac{-4\pm\sqrt{16-52}}{2}\\[3mm] =&\, -2\pm3\mathrm{i} \end{align*}

So the complementary function is

uc= e−2x(Acos⁡3x+Bsin⁡3x)\begin{align*} u_c =&\, \mathrm{e}^{-2x} (A\cos3x+B\sin3x) \end{align*}

For a particular integral, try

up=λe3x\begin{align*} u_p=\lambda\mathrm{e}^{3x} \end{align*}

Then

dupdx= 3λe3x,d2updx2= 9λe3x\begin{align*} \frac{\mathrm{d}u_p}{\mathrm{d}x} =&\, 3\lambda\mathrm{e}^{3x}, \qquad \frac{\mathrm{d}^2u_p}{\mathrm{d}x^2} =&\, 9\lambda\mathrm{e}^{3x} \end{align*}

Substitute into the differential equation:

9λe3x+4(3λe3x)+13λe3x= 17e3x34λ= 17λ= 12\begin{align*} 9\lambda\mathrm{e}^{3x} +4(3\lambda\mathrm{e}^{3x}) +13\lambda\mathrm{e}^{3x} =&\, 17\mathrm{e}^{3x}\\[3mm] 34\lambda =&\, 17\\[3mm] \lambda =&\, \frac12 \end{align*}

Therefore the general solution for uu is

u= e−2x(Acos⁡3x+Bsin⁡3x)+12e3x\begin{align*} u =&\, \mathrm{e}^{-2x}(A\cos3x+B\sin3x) +\frac12\mathrm{e}^{3x} \end{align*}

(d)

解法一

思路

展开

最后只需要用 u=xyu=xy，所以 y=uxy=\dfrac{u}{x}。

答题过程

展开

Since

u=xy,\begin{align*} u=xy, \end{align*}

we have

y=ux\begin{align*} y=\frac{u}{x} \end{align*}

Therefore

y= 1x[e−2x(Acos⁡3x+Bsin⁡3x)+12e3x]\begin{align*} y =&\, \frac1x \Big[ \mathrm{e}^{-2x}(A\cos3x+B\sin3x) +\frac12\mathrm{e}^{3x} \Big] \end{align*}