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FP2 2021 June Q6

A Level / Edexcel / FP2

IAL 2021 June 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}\\[4mm] &\,-6\frac{\mathrm{d}y}{\mathrm{d}x}+8y\\[4mm] =&\,2x^2+x. \end{align*}
(8)

(b) Find the particular solution of this differential equation for which y=1y = 1 and dydx=0\frac{\mathrm{d}y}{\mathrm{d}x} = 0 when x=0x = 0

(5)
题目中文翻译

6. (a) 求微分方程

 d2ydx2 −6dydx+8y= 2x2+x.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,-6\frac{\mathrm{d}y}{\mathrm{d}x}+8y\\[4mm] =&\,2x^2+x. \end{align*}

的通解。

(8)

(b) 求此微分方程满足下列条件的特解：当 x=0x=0 时，y=1y=1 且 dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0。

(5)

解答

(a)

解法一

思路

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这是二阶常系数非齐次微分方程。先求 complementary function，再设 particular integral。右边是二次多项式 2x2+x2x^2+x，所以 particular integral 也设成二次多项式。

答题过程

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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*}

giving

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

Hence 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=λx2+μx+ν.\begin{align*} y_p=\lambda x^2+\mu x+\nu. \end{align*}

Then

dypdx= 2λx+μ,d2ypdx2= 2λ.\begin{align*} \frac{\mathrm{d}y_p}{\mathrm{d}x}=&\,2\lambda x+\mu,\\[4mm] \frac{\mathrm{d}^2y_p}{\mathrm{d}x^2}=&\,2\lambda. \end{align*}

Substitute into the differential equation:

 2λ −6(2λx+μ) +8(λx2+μx+ν)= 2x2+x.\begin{align*} &\,2\lambda\\[4mm] &\,\hspace{2pt}-6(2\lambda x+\mu)\\[4mm] &\,\hspace{4pt}+8(\lambda x^2+\mu x+\nu)\\[4mm] =&\,2x^2+x. \end{align*}

Comparing the coefficients of x2x^2, xx and the constant term, respectively,

8λ= 2,−12λ+8μ= 1,2λ−6μ+8ν= 0.\begin{align*} 8\lambda=&\,2,\\[4mm] -12\lambda+8\mu=&\,1,\\[4mm] 2\lambda-6\mu+8\nu=&\,0. \end{align*}

The first equation gives λ=14\lambda=\dfrac14. Substituting into the second,

−3+8μ= 1,μ= 12.\begin{align*} -3+8\mu=&\,1,\\[4mm] \mu=&\,\frac12. \end{align*}

The third equation then gives

12−3+8ν= 0,8ν= 52,ν= 516.\begin{align*} \frac12-3+8\nu=&\,0,\\[4mm] 8\nu=&\,\frac52,\\[4mm] \nu=&\,\frac5{16}. \end{align*}

Therefore,

y= Ae2x+Be4x +14x2+12x+516.\begin{align*} y =&\,A\mathrm{e}^{2x}+B\mathrm{e}^{4x}\\[4mm] &\,\hspace{2pt}+\frac14x^2+\frac12x+\frac5{16}. \end{align*}

(b)

解法一

思路

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把 x=0x=0、y=1y=1 代入通解得到一个关于 A,BA,B 的方程。再对通解求导，用 x=0x=0、y′=0y'=0 得到第二个方程。

答题过程

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Using y=1y=1 at x=0x=0,

1=A+B+516.\begin{align*} 1=A+B+\frac5{16}. \end{align*}

So

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

Differentiate the general solution:

 dydx= 2Ae2x+4Be4x +12x+12.\begin{align*} &\,\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] =&\,2A\mathrm{e}^{2x}+4B\mathrm{e}^{4x}\\[4mm] &\,\hspace{2pt}+\frac12x+\frac12. \end{align*}

Using y′=0y'=0 at 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*}

Solve the simultaneous equations:

A+B= 1116,A+2B= −14.\begin{align*} A+B=&\,\frac{11}{16},\\[2mm] A+2B=&\,-\frac14. \end{align*}

Subtracting gives

B=−1516.\begin{align*} B=-\frac{15}{16}. \end{align*}

Then

A=1116+1516=138.\begin{align*} A=\frac{11}{16}+\frac{15}{16}=\frac{13}{8}. \end{align*}

Therefore the particular solution is

y= 138e2x−1516e4x +14x2+12x+516.\begin{align*} y =&\,\frac{13}{8}\mathrm{e}^{2x} -\frac{15}{16}\mathrm{e}^{4x}\\[4mm] &\,\hspace{2pt}+\frac14x^2+\frac12x+\frac5{16}. \end{align*}