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FP2 2021 Oct Q4

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 4

题目

Problem

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

 (x+1)dydx−xy= e3x, x>−1.\begin{align*} &\,(x+1)\frac{\mathrm{d}y}{\mathrm{d}x}-xy\\[4mm] =&\,\mathrm{e}^{3x},\\[4mm] &\,x>-1. \end{align*}

giving your answer in the form y=f(x)y = f(x).

(7)

(b) Determine the particular solution of the differential equation for which y=5y = 5 when x=0x = 0.

(2)
题目中文翻译

4. (a) 求微分方程

 (x+1)dydx−xy= e3x, x>−1.\begin{align*} &\,(x+1)\frac{\mathrm{d}y}{\mathrm{d}x}-xy\\[4mm] =&\,\mathrm{e}^{3x},\\[4mm] &\,x>-1. \end{align*}

的通解，将答案写成 y=f(x)y=f(x) 的形式。

(7)

(b) 求此微分方程在 x=0x=0 时满足 y=5y=5 的特解。

(2)

解答

(a)

解法一

思路

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先把微分方程写成一阶线性标准形式

dydx+P(x)y=Q(x).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x). \end{align*}

这里 x>−1x>-1，所以 ln⁡(x+1)\ln(x+1) 可以直接使用，不需要写成 ln⁡∣x+1∣\ln|x+1|。

答题过程

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The differential equation is

(x+1)dydx−xy=e3x.\begin{align*} (x+1)\frac{\mathrm{d}y}{\mathrm{d}x}-xy=\mathrm{e}^{3x}. \end{align*}

Divide by x+1x+1:

dydx−xx+1y= e3xx+1.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} -\frac{x}{x+1}y =&\,\frac{\mathrm{e}^{3x}}{x+1}. \end{align*}

The integrating factor is

I(x)= e∫−xx+1 dx.\begin{align*} I(x) =&\,\mathrm{e}^{\int -\frac{x}{x+1}\,\mathrm{d}x}. \end{align*}

Rewrite the integrand:

−xx+1= −x+1−1x+1= −1+1x+1.\begin{align*} -\frac{x}{x+1} =&\,-\frac{x+1-1}{x+1}\\[2mm] =&\,-1+\frac{1}{x+1}. \end{align*}

So

 ∫−xx+1 dx= ∫(−1+1x+1) dx= −x+ln⁡(x+1),\begin{align*} &\,\int -\frac{x}{x+1}\,\mathrm{d}x\\[4mm] =&\,\int\bigg(-1+\frac{1}{x+1}\bigg)\,\mathrm{d}x\\[4mm] =&\,-x+\ln(x+1), \end{align*}

since x>−1x>-1. Hence

I(x)= e−x+ln⁡(x+1)= (x+1)e−x.\begin{align*} I(x) =&\,\mathrm{e}^{-x+\ln(x+1)}\\[2mm] =&\,(x+1)\mathrm{e}^{-x}. \end{align*}

Multiply the linear differential equation by (x+1)e−x(x+1)\mathrm{e}^{-x}:

 (x+1)e−xdydx −xe−xy= e2x.\begin{align*} &\,(x+1)\mathrm{e}^{-x}\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}-x\mathrm{e}^{-x}y\\[4mm] =&\,\mathrm{e}^{2x}. \end{align*}

The left hand side is

ddx[y(x+1)e−x].\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \big[y(x+1)\mathrm{e}^{-x}\big]. \end{align*}

Therefore

ddx[y(x+1)e−x]= e2x.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \big[y(x+1)\mathrm{e}^{-x}\big] =&\,\mathrm{e}^{2x}. \end{align*}

Integrate:

y(x+1)e−x= ∫e2x dx= 12e2x+C.\begin{align*} y(x+1)\mathrm{e}^{-x} =&\,\int \mathrm{e}^{2x}\,\mathrm{d}x\\[2mm] =&\,\frac12\mathrm{e}^{2x}+C. \end{align*}

So

y= 12e2x+C(x+1)e−x= e3x2(x+1)+Cexx+1.\begin{align*} y =&\,\frac{\frac12\mathrm{e}^{2x}+C} {(x+1)\mathrm{e}^{-x}}\\[4mm] =&\,\frac{\mathrm{e}^{3x}}{2(x+1)} +\frac{C\mathrm{e}^{x}}{x+1}. \end{align*}

(b)

解法一

思路

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把 x=0x=0、y=5y=5 代入 (a) 的 general solution，就能求出常数 CC。

答题过程

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From part (a),

y= e3x2(x+1)+Cexx+1.\begin{align*} y =&\,\frac{\mathrm{e}^{3x}}{2(x+1)} +\frac{C\mathrm{e}^{x}}{x+1}. \end{align*}

Use x=0x=0 and y=5y=5:

5= e02(1)+Ce01= 12+C.\begin{align*} 5 =&\,\frac{\mathrm{e}^{0}}{2(1)} +\frac{C\mathrm{e}^{0}}{1}\\[2mm] =&\,\frac12+C. \end{align*}

Thus

C=92.\begin{align*} C=\frac92. \end{align*}

Therefore the particular solution is

y= e3x2(x+1)+9ex2(x+1).\begin{align*} y =&\,\frac{\mathrm{e}^{3x}}{2(x+1)} +\frac{9\mathrm{e}^{x}}{2(x+1)}. \end{align*}