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FP2 2021 Jan Q6

A Level / Edexcel / FP2

IAL 2021 Jan Paper · Question 6

题目

Problem

6. (a) Determine the general solution of the differential equation

 d2ydx2 +2dydx+5y= 6cos⁡x.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+2\frac{\mathrm{d}y}{\mathrm{d}x}+5y\\[4mm] =&\,6\cos x. \end{align*}
(7)

(b) Find the particular solution for which y=0y=0 and dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0 at x=0x=0

(5)
题目中文翻译

6. (a) 求下列微分方程的通解:

 d2ydx2 +2dydx+5y= 6cos⁡x.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+2\frac{\mathrm{d}y}{\mathrm{d}x}+5y\\[4mm] =&\,6\cos x. \end{align*}
(7)

(b) 求在 x=0x=0 时满足 y=0y=0 且 dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0 的特解。

(5)

解答

(a)

解法一

思路

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先解 complementary function,再设 particular integral 为 acos⁡x+bsin⁡xa\cos x+b\sin x。因为右边是 6cos⁡x6\cos x,这个形式最直接。

答题过程

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The auxiliary equation is

m2+2m+5=0.\begin{align*} m^2+2m+5=0. \end{align*}

So

m= −2±4−202= −1±2i.\begin{align*} m=&\,\frac{-2\pm\sqrt{4-20}}{2}\\[4mm] =&\,-1\pm2\mathrm{i}. \end{align*}

Hence the complementary function is

yc= e−x(Acos⁡2x +Bsin⁡2x).\begin{align*} y_c=&\,\mathrm{e}^{-x}\big(A\cos2x\\[4mm] &\,\hspace{2pt}+B\sin2x\big). \end{align*}

For a particular integral, let

yp=acos⁡x+bsin⁡x.\begin{align*} y_p=a\cos x+b\sin x. \end{align*}

Then

yp′= −asin⁡x+bcos⁡x,yp′′= −acos⁡x−bsin⁡x.\begin{align*} y_p'=&\,-a\sin x+b\cos x,\\[2mm] y_p''=&\,-a\cos x-b\sin x. \end{align*}

Substitute into the differential equation:

 (−acos⁡x−bsin⁡x) +2(−asin⁡x+bcos⁡x) +5(acos⁡x+bsin⁡x)= 6cos⁡x.\begin{align*} &\,(-a\cos x-b\sin x)\\[4mm] &\,\hspace{2pt}+2(-a\sin x+b\cos x)\\[4mm] &\,\hspace{4pt}+5(a\cos x+b\sin x)\\[4mm] =&\,6\cos x. \end{align*}

Comparing coefficients,

4a+2b= 6,−2a+4b= 0.\begin{align*} 4a+2b=&\,6,\\[2mm] -2a+4b=&\,0. \end{align*}

Solving gives

a=65,b=35.\begin{align*} a=\frac65,\qquad b=\frac35. \end{align*}

Therefore the general solution is

y= e−x(Acos⁡2x +Bsin⁡2x) +65cos⁡x+35sin⁡x.\begin{align*} y =&\,\mathrm{e}^{-x}\big(A\cos2x\\[4mm] &\,\hspace{2pt}+B\sin2x\big)\\[4mm] &\,+\frac65\cos x+\frac35\sin x. \end{align*}

(b)

解法一

思路

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把 x=0x=0、y=0y=0 代入先求 AA。再对通解微分,用 x=0x=0、y′=0y'=0 求 BB。

答题过程

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

0=A+65,\begin{align*} 0=A+\frac65, \end{align*}

so

A=−65.\begin{align*} A=-\frac65. \end{align*}

Differentiate the general solution:

dydx= −e−x(Acos⁡2x +Bsin⁡2x) +e−x(−2Asin⁡2x +2Bcos⁡2x) −65sin⁡x+35cos⁡x.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,-\mathrm{e}^{-x}\big(A\cos2x\\[4mm] &\,\hspace{2pt}+B\sin2x\big)\\[4mm] &\,+\mathrm{e}^{-x}\big(-2A\sin2x\\[4mm] &\,\hspace{2pt}+2B\cos2x\big)\\[4mm] &\,-\frac65\sin x+\frac35\cos x. \end{align*}

Using dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0 at x=0x=0,

0= −A+2B+35.\begin{align*} 0=&\,-A+2B+\frac35. \end{align*}

Since A=−65A=-\dfrac65,

0= 65+2B+352B= −95B= −910.\begin{align*} 0=&\,\frac65+2B+\frac35\\[4mm] 2B=&\,-\frac95\\[4mm] B=&\,-\frac9{10}. \end{align*}

Therefore the particular solution is

y= e−x(−65cos⁡2x −910sin⁡2x) +65cos⁡x+35sin⁡x.\begin{align*} y =&\,\mathrm{e}^{-x} \big(-\frac65\cos2x\\[4mm] &\,\hspace{2pt}-\frac9{10}\sin2x\big)\\[4mm] &\,+\frac65\cos x+\frac35\sin x. \end{align*}

解法二

思路

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如果熟悉复指数形式,也可以把 complementary function 写成

 e−x(Pe2ix +Qe−2ix).\begin{align*} &\,\mathrm{e}^{-x}\big(P\mathrm{e}^{2\mathrm{i}x} \\[4mm] &\,\hspace{2pt}+Q\mathrm{e}^{-2\mathrm{i}x}\big). \end{align*}

这和上述实三角形式是同一个解空间。若熟悉 Euler formula,可以直接求 P,QP,Q,最后再化回实数形式。

答题过程

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Write the general solution as

y= e−x(Pe2ix +Qe−2ix) +65cos⁡x+35sin⁡x.\begin{align*} y =&\,\mathrm{e}^{-x} \big(P\mathrm{e}^{2\mathrm{i}x} \\[4mm] &\,\hspace{2pt}+Q\mathrm{e}^{-2\mathrm{i}x}\big)\\[4mm] &\,+\frac65\cos x+\frac35\sin x. \end{align*}

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

0=P+Q+65.\begin{align*} 0=P+Q+\frac65. \end{align*}

So

P+Q=−65.\begin{align*} P+Q=-\frac65. \end{align*}

Differentiate:

dydx= −e−x(Pe2ix +Qe−2ix) +e−x(2iPe2ix −2iQe−2ix) −65sin⁡x+35cos⁡x.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,-\mathrm{e}^{-x} \big(P\mathrm{e}^{2\mathrm{i}x} \\[4mm] &\,\hspace{2pt}+Q\mathrm{e}^{-2\mathrm{i}x}\big)\\[4mm] &\,+\mathrm{e}^{-x} \big(2\mathrm{i}P\mathrm{e}^{2\mathrm{i}x} \\[4mm] &\,\hspace{2pt}-2\mathrm{i}Q\mathrm{e}^{-2\mathrm{i}x}\big)\\[4mm] &\,-\frac65\sin x+\frac35\cos x. \end{align*}

Using dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0 at x=0x=0,

0= −(P+Q) +2i(P−Q)+35.\begin{align*} 0=&\,-(P+Q)\\[4mm] &\,\hspace{2pt}+2\mathrm{i}(P-Q)+\frac35. \end{align*}

Since P+Q=−65P+Q=-\dfrac65,

0= 65+2i(P−Q)+35.\begin{align*} 0 =&\,\frac65+2\mathrm{i}(P-Q)+\frac35. \end{align*}

Therefore,

2i(P−Q)= −95,\begin{align*} 2\mathrm{i}(P-Q)=&\,-\frac95, \end{align*}

so

P−Q=910i.P-Q=\frac9{10}\mathrm{i}.

Solving the two equations,

P= 12(−65+910i),Q= 12(−65−910i).\begin{align*} P=&\,\frac12\big(-\frac65+\frac9{10}\mathrm{i}\big),\\[2mm] Q=&\,\frac12\big(-\frac65-\frac9{10}\mathrm{i}\big). \end{align*}

Hence,

y= 12e−x(−65+910i) ×e2ix +12e−x(−65−910i) ×e−2ix +65cos⁡x+35sin⁡x.\begin{align*} y =&\,\frac12\mathrm{e}^{-x} \big(-\frac65+\frac9{10}\mathrm{i}\big) \\[4mm] &\,\hspace{2pt}\times\mathrm{e}^{2\mathrm{i}x}\\[4mm] &\,+\frac12\mathrm{e}^{-x} \big(-\frac65-\frac9{10}\mathrm{i}\big) \\[4mm] &\,\hspace{2pt}\times\mathrm{e}^{-2\mathrm{i}x}\\[4mm] &\,+\frac65\cos x+\frac35\sin x. \end{align*}

Combining the conjugate exponential terms,

y= e−x(−65cos⁡2x −910sin⁡2x) +65cos⁡x+35sin⁡x.\begin{align*} y =&\,\mathrm{e}^{-x} \big(-\frac65\cos2x\\[4mm] &\,\hspace{2pt}-\frac9{10}\sin2x\big)\\[4mm] &\,+\frac65\cos x+\frac35\sin x. \end{align*}

This gives the required particular solution.