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IAL 2022 Jan Q2

A Level / Edexcel / FP2

IAL 2022 Jan Paper · Question 2

题目

Problem

Determine the general solution of the differential equation

2d2ydx2−5dydx−3y=2e3x\begin{align*} 2\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 5\frac{\mathrm{d}y}{\mathrm{d}x} - 3y = 2\mathrm{e}^{3x} \end{align*}
(6)

解答

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。因为右边是 e3x\mathrm{e}^{3x}，而 e3x\mathrm{e}^{3x} 已经出现在 complementary function 中，所以 particular integral 要乘一个 xx，设为

axe3x.\begin{align*} ax\mathrm{e}^{3x}. \end{align*}

答题过程

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For the complementary function, solve the auxiliary equation

2m2−5m−3=0.\begin{align*} 2m^2-5m-3=0. \end{align*}

Factorise:

(2m+1)(m−3)=0.\begin{align*} (2m+1)(m-3)=0. \end{align*}

So

m=−12,m=3.\begin{align*} m=-\frac12,\qquad m=3. \end{align*}

Therefore

yc=Ae−12x+Be3x.\begin{align*} y_c=A\mathrm{e}^{-\frac12x}+B\mathrm{e}^{3x}. \end{align*}

Since e3x\mathrm{e}^{3x} is already part of the complementary function, let

yp=axe3x.\begin{align*} y_p=ax\mathrm{e}^{3x}. \end{align*}

Then

yp′= ae3x+3axe3x,\begin{align*} y_p' =&\,a\mathrm{e}^{3x}+3ax\mathrm{e}^{3x}, \end{align*}

and

yp′′= 3ae3x+3ae3x+9axe3x= 6ae3x+9axe3x.\begin{align*} y_p'' =&\,3a\mathrm{e}^{3x} +3a\mathrm{e}^{3x} +9ax\mathrm{e}^{3x}\\[2mm] =&\,6a\mathrm{e}^{3x} +9ax\mathrm{e}^{3x}. \end{align*}

Substitute into

2y′′−5y′−3y=2e3x.\begin{align*} 2y''-5y'-3y=2\mathrm{e}^{3x}. \end{align*}

This gives

 2(6ae3x+9axe3x) −5(ae3x+3axe3x) −3(axe3x)=2e3x.\begin{align*} &\,2(6a\mathrm{e}^{3x}+9ax\mathrm{e}^{3x})\\[2mm] &\,\hspace{2pt}-5(a\mathrm{e}^{3x}+3ax\mathrm{e}^{3x})\\[2mm] &\,\hspace{4pt}-3(ax\mathrm{e}^{3x}) =2\mathrm{e}^{3x}. \end{align*}

Simplify:

12ae3x+18axe3x−5ae3x−15axe3x−3axe3x= 2e3x.\begin{align*} 12a\mathrm{e}^{3x} +18ax\mathrm{e}^{3x} -5a\mathrm{e}^{3x} -15ax\mathrm{e}^{3x} -3ax\mathrm{e}^{3x} =&\,2\mathrm{e}^{3x}. \end{align*}

The xe3xx\mathrm{e}^{3x} terms cancel, leaving

7ae3x=2e3x.\begin{align*} 7a\mathrm{e}^{3x}=2\mathrm{e}^{3x}. \end{align*}

Hence

a=27.\begin{align*} a=\frac27. \end{align*}

Therefore

yp=27xe3x.\begin{align*} y_p=\frac27x\mathrm{e}^{3x}. \end{align*}

The general solution is

y= Ae−12x+Be3x+27xe3x.\begin{align*} y =&\,A\mathrm{e}^{-\frac12x} +B\mathrm{e}^{3x} +\frac27x\mathrm{e}^{3x}. \end{align*}