CalcGospel 國際數學圖譜
返回

FP2 2026 Jan A Q6

A Level / Edexcel / FP2

IAL 2026 Jan A Paper · Question 6

题目

Problem

6.

(a) Find the general solution of the differential equation

d2ydx2−6dydx+8y= 2x2+x\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} -6\frac{\mathrm{d}y}{\mathrm{d}x} +8y =&\, 2x^2+x \end{align*}
(8)

(b) Find the particular solution of this differential equation for which

y=1anddydx=0when x=0\begin{align*} y=1 \quad\text{and}\quad \frac{\mathrm{d}y}{\mathrm{d}x}=0 \quad\text{when }x=0 \end{align*}
(5)
题目中文翻译

(a) 求下列微分方程的通解：

d2ydx2−6dydx+8y= 2x2+x\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} -6\frac{\mathrm{d}y}{\mathrm{d}x} +8y =&\, 2x^2+x \end{align*}
(8)

(b) 求此微分方程满足下列条件的特解：

y=1且dydx=0当 x=0\begin{align*} y=1 \quad\text{且}\quad \frac{\mathrm{d}y}{\mathrm{d}x}=0 \quad\text{当 }x=0 \end{align*}
(5)

解答

(a)

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；右边是二次多项式，所以特解设为 αx2+βx+γ\alpha x^2+\beta x+\gamma。

答题过程

展开

The auxiliary equation is

m2−6m+8=0\begin{align*} m^2-6m+8=0 \end{align*}

so

(m−2)(m−4)=0\begin{align*} (m-2)(m-4)=0 \end{align*}

Thus

m=2,4\begin{align*} m=2,\quad 4 \end{align*}

The complementary function is

yc=Ae2x+Be4x\begin{align*} y_c=A\mathrm{e}^{2x}+B\mathrm{e}^{4x} \end{align*}

For a particular integral, let

yp=αx2+βx+γ\begin{align*} y_p=\alpha x^2+\beta x+\gamma \end{align*}

Then

yp′=2αx+β,yp′′=2α\begin{align*} y_p'=2\alpha x+\beta, \qquad y_p''=2\alpha \end{align*}

Substitute into the differential equation:

2α−6(2αx+β)+8(αx2+βx+γ)= 2x2+x\begin{align*} 2\alpha-6(2\alpha x+\beta) +8(\alpha x^2+\beta x+\gamma) =&\, 2x^2+x \end{align*}

Compare coefficients:

8α= 2−12α+8β= 12α−6β+8γ= 0\begin{align*} 8\alpha=&\,2\\[2mm] -12\alpha+8\beta=&\,1\\[2mm] 2\alpha-6\beta+8\gamma=&\,0 \end{align*}

So

α=14,β=12,γ=516\begin{align*} \alpha=\frac14, \qquad \beta=\frac12, \qquad \gamma=\frac{5}{16} \end{align*}

Therefore the general solution is

y= Ae2x+Be4x+14x2+12x+516\begin{align*} y =&\, A\mathrm{e}^{2x}+B\mathrm{e}^{4x} +\frac14x^2+\frac12x+\frac{5}{16} \end{align*}

(b)

解法一

思路

展开

把 x=0x=0、y=1y=1 代入通解得到第一个方程；再求导并用 y′(0)=0y'(0)=0 得到第二个方程，联立求 A,BA,B。

答题过程

展开

From part (a),

y= Ae2x+Be4x+14x2+12x+516\begin{align*} y =&\, A\mathrm{e}^{2x}+B\mathrm{e}^{4x} +\frac14x^2+\frac12x+\frac{5}{16} \end{align*}

Using y=1y=1 when x=0x=0,

1=A+B+516\begin{align*} 1=A+B+\frac{5}{16} \end{align*}

so

A+B=1116\begin{align*} A+B=\frac{11}{16} \end{align*}

Differentiate:

y′= 2Ae2x+4Be4x+12x+12\begin{align*} y' =&\, 2A\mathrm{e}^{2x}+4B\mathrm{e}^{4x} +\frac12x+\frac12 \end{align*}

Using y′=0y'=0 when x=0x=0,

0= 2A+4B+12\begin{align*} 0 =&\, 2A+4B+\frac12 \end{align*}

so

2A+4B=−12\begin{align*} 2A+4B=-\frac12 \end{align*}

Solving

A+B=1116,2A+4B=−12,\begin{align*} A+B=\frac{11}{16}, \qquad 2A+4B=-\frac12, \end{align*}

Subtracting twice the first equation from the second gives

2B= −12−2216= −158B= −1516\begin{align*} 2B =&\,-\frac12-\frac{22}{16}\\[3mm] =&\,-\frac{15}{8}\\[3mm] B =&\,-\frac{15}{16} \end{align*}

Then A=1116−B=138A=\dfrac{11}{16}-B=\dfrac{13}{8}. Thus

A=138,B=−1516\begin{align*} A=\frac{13}{8}, \qquad B=-\frac{15}{16} \end{align*}

Therefore

y= 138e2x−1516e4x+14x2+12x+516\begin{align*} y =&\, \frac{13}{8}\mathrm{e}^{2x} -\frac{15}{16}\mathrm{e}^{4x} +\frac14x^2+\frac12x+\frac{5}{16} \end{align*}