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IAL 2025 June Q4

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 4

题目

Problem

Given that y=λxe3xy=\lambda x\mathrm{e}^{3x} is a particular integral of the differential equation

4d2ydx2−11dydx−3y=78e3x\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -11\frac{\mathrm{d}y}{\mathrm{d}x} -3y = 78\mathrm{e}^{3x} \end{align*}

(a) determine the value of the constant λ\lambda

(3)

(b) Hence determine the general solution of the differential equation.

(3)

Given also that y=92y=\frac92 and dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0 at x=0x=0

(c) determine the particular solution of the differential equation.

(4)

解答

(a)

解法一

思路

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把 y=λxe3xy=\lambda x\mathrm{e}^{3x} 代入微分方程。这里需要用乘积法则求一阶和二阶导数；代入后 xe3xx\mathrm{e}^{3x} 项会抵消，剩下可求 λ\lambda。

答题过程

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Given

y=λxe3x\begin{align*} y=\lambda x\mathrm{e}^{3x} \end{align*}

Differentiate:

y′=λe3x+3λxe3xy′′=3λe3x+3λe3x+9λxe3x=6λe3x+9λxe3x\begin{align*} y' ={}& \lambda\mathrm{e}^{3x} +3\lambda x\mathrm{e}^{3x}\\[3mm] y'' ={}& 3\lambda\mathrm{e}^{3x} +3\lambda\mathrm{e}^{3x} +9\lambda x\mathrm{e}^{3x}\\[3mm] ={}& 6\lambda\mathrm{e}^{3x} +9\lambda x\mathrm{e}^{3x} \end{align*}

Substitute into the differential equation:

4(6λe3x+9λxe3x)−11(λe3x+3λxe3x)−3(λxe3x)=78e3x\begin{align*} 4(6\lambda\mathrm{e}^{3x}+9\lambda x\mathrm{e}^{3x}) -11(\lambda\mathrm{e}^{3x}+3\lambda x\mathrm{e}^{3x}) -3(\lambda x\mathrm{e}^{3x}) ={}& 78\mathrm{e}^{3x} \end{align*}

Simplify:

(24λ−11λ)e3x+(36λ−33λ−3λ)xe3x=78e3x13λe3x=78e3x\begin{align*} (24\lambda-11\lambda)\mathrm{e}^{3x} +(36\lambda-33\lambda-3\lambda)x\mathrm{e}^{3x} ={}& 78\mathrm{e}^{3x}\\[3mm] 13\lambda\mathrm{e}^{3x} ={}& 78\mathrm{e}^{3x} \end{align*}

Therefore

λ=6\begin{align*} \boxed{\lambda=6} \end{align*}

(b)

解法一

思路

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通解等于 complementary function 加 particular integral。先解辅助方程，再加上 (a) 中得到的 6xe3x6x\mathrm{e}^{3x}。

答题过程

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The auxiliary equation is

4m2−11m−3=0\begin{align*} 4m^2-11m-3=0 \end{align*}

Factorise:

(4m+1)(m−3)=0\begin{align*} (4m+1)(m-3)=0 \end{align*}

so

m=−14orm=3\begin{align*} m=-\frac14 \quad \text{or} \quad m=3 \end{align*}

The complementary function is

yc=Ae−x/4+Be3x\begin{align*} y_{\mathrm{c}} ={}& A\mathrm{e}^{-x/4}+B\mathrm{e}^{3x} \end{align*}

Using part (a), a particular integral is

yp=6xe3x\begin{align*} y_{\mathrm{p}}=6x\mathrm{e}^{3x} \end{align*}

Therefore the general solution is

y=Ae−x/4+Be3x+6xe3x\begin{align*} \boxed{ y=A\mathrm{e}^{-x/4}+B\mathrm{e}^{3x}+6x\mathrm{e}^{3x} } \end{align*}

(c)

解法一

思路

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把 x=0,y=92x=0,y=\frac92 代入通解得到一个关于 A,BA,B 的方程。再对通解求导，用 x=0,y′=0x=0,y'=0 得第二个方程，联立求出 A,BA,B。

答题过程

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From part (b),

y=Ae−x/4+Be3x+6xe3x\begin{align*} y=A\mathrm{e}^{-x/4}+B\mathrm{e}^{3x}+6x\mathrm{e}^{3x} \end{align*}

Use y=92y=\frac92 when x=0x=0:

A+B=92\begin{align*} A+B=\frac92 \end{align*}

Differentiate:

y′=−14Ae−x/4+3Be3x+6e3x+18xe3x\begin{align*} y' ={}& -\frac14A\mathrm{e}^{-x/4} +3B\mathrm{e}^{3x} +6\mathrm{e}^{3x} +18x\mathrm{e}^{3x} \end{align*}

Use y′=0y'=0 when x=0x=0:

−14A+3B+6=0\begin{align*} -\frac14A+3B+6 ={}& 0 \end{align*}

So

A+B=92,−14A+3B=−6\begin{align*} A+B ={}& \frac92,\\ -\frac14A+3B ={}& -6 \end{align*}

From A=92−BA=\frac92-B,

−14(92−B)+3B=−6−98+14B+3B=−6134B=−398B=−32\begin{align*} -\frac14\left(\frac92-B\right)+3B ={}& -6\\[3mm] -\frac98+\frac14B+3B ={}& -6\\[3mm] \frac{13}{4}B ={}& -\frac{39}{8}\\[3mm] B ={}& -\frac32 \end{align*}

Hence

A=92−(−32)=6\begin{align*} A ={}& \frac92-\left(-\frac32\right)\\[2mm] ={}& 6 \end{align*}

Therefore the particular solution is

y=6e−x/4−32e3x+6xe3x\begin{align*} \boxed{ y=6\mathrm{e}^{-x/4}-\frac32\mathrm{e}^{3x} +6x\mathrm{e}^{3x} } \end{align*}