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IAL 2021 Jan Q4

A Level / Edexcel / FP2

IAL 2021 Jan Paper · Question 4

题目

Problem

(a) Show that the substitution y2=1zy^2 = \dfrac{1}{z} transforms the differential equation

dydx+2y=3xy3y≠0(I)\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + 2y = 3xy^3 \qquad y \neq 0 \qquad \text{(I)} \end{align*}

into the differential equation

dzdx−4z=−6x(II)\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - 4z = -6x \qquad \text{(II)} \end{align*}
(3)

(b) Obtain the general solution of differential equation (II).

(5)

(c) Hence obtain the general solution of differential equation (I), giving your answer in the form y2=f(x)y^2 = f(x)

(1)

解答

(a)

解法一

思路

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由 y2=1zy^2=\dfrac1z 可得 2ydydx=−1z2dzdx2y\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac1{z^2}\dfrac{\mathrm{d}z}{\mathrm{d}x}。题目要 show that，所以要把原方程完整变成只含 zz 和 xx 的方程。

答题过程

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From

y2=z−1,\begin{align*} y^2=z^{-1}, \end{align*}

differentiate with respect to xx:

2ydydx= −z−2dzdx.\begin{align*} 2y\frac{\mathrm{d}y}{\mathrm{d}x} =&\,-z^{-2}\frac{\mathrm{d}z}{\mathrm{d}x}. \end{align*}

Multiply differential equation (I) by 2y2y:

2ydydx+4y2= 6xy4.\begin{align*} 2y\frac{\mathrm{d}y}{\mathrm{d}x}+4y^2 =&\,6xy^4. \end{align*}

Using y2=1zy^2=\dfrac1z and y4=1z2y^4=\dfrac1{z^2},

−1z2dzdx+4z= 6xz2.\begin{align*} -\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}+\frac4z =&\,\frac{6x}{z^2}. \end{align*}

Multiplying by −z2-z^2 gives

dzdx−4z=−6x.\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-4z=-6x. \end{align*}

This is differential equation (II).

(b)

解法一

思路

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(II) 是一阶线性微分方程：

dzdx−4z=−6x.\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-4z=-6x. \end{align*}

Integrating factor 是 e∫−4 dx=e−4x\mathrm{e}^{\int -4\,\mathrm{d}x}=\mathrm{e}^{-4x}。右边的积分需要用 integration by parts。这个方法是通用的一阶线性方程方法。

答题过程

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The integrating factor is

I(x)= e∫−4 dx= e−4x.\begin{align*} I(x)=&\,\mathrm{e}^{\int -4\,\mathrm{d}x}\\[4mm] =&\,\mathrm{e}^{-4x}. \end{align*}

Multiplying the equation by e−4x\mathrm{e}^{-4x},

e−4xdzdx−4e−4xz= −6xe−4xddx(ze−4x)= −6xe−4x.\begin{align*} \mathrm{e}^{-4x}\frac{\mathrm{d}z}{\mathrm{d}x}-4\mathrm{e}^{-4x}z =&\,-6x\mathrm{e}^{-4x}\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}\left(z\mathrm{e}^{-4x}\right) =&\,-6x\mathrm{e}^{-4x}. \end{align*}

Integrating,

ze−4x= −6∫xe−4x dx= −6(−14xe−4x−116e−4x)+C= 32xe−4x+38e−4x+C.\begin{align*} z\mathrm{e}^{-4x} =&\,-6\int x\mathrm{e}^{-4x}\,\mathrm{d}x\\[4mm] =&\,-6\left(-\frac14x\mathrm{e}^{-4x}-\frac1{16}\mathrm{e}^{-4x}\right)+C\\[4mm] =&\,\frac32x\mathrm{e}^{-4x}+\frac38\mathrm{e}^{-4x}+C. \end{align*}

Therefore,

z=32x+38+Ce4x.\begin{align*} z=\frac32x+\frac38+C\mathrm{e}^{4x}. \end{align*}

解法二

思路

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也可以把

dzdx−4z=−6x\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-4z=-6x \end{align*}

看成常系数一阶非齐次微分方程。先解 homogeneous part 得到 complementary function，再设一次多项式 particular integral，因为右边是一次函数 −6x-6x。

答题过程

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First solve the homogeneous equation

dzdx−4z=0.\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-4z=0. \end{align*}

This gives

zc=Ae4x.\begin{align*} z_c=A\mathrm{e}^{4x}. \end{align*}

Since the right hand side is linear in xx, let

zp=λ+μx.\begin{align*} z_p=\lambda+\mu x. \end{align*}

Then

dzpdx=μ.\begin{align*} \frac{\mathrm{d}z_p}{\mathrm{d}x}=\mu. \end{align*}

Substitute into

dzdx−4z=−6x.\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-4z=-6x. \end{align*}

This gives

μ−4(λ+μx)= −6x−4μx+(μ−4λ)= −6x.\begin{align*} \mu-4(\lambda+\mu x)=&\,-6x\\[4mm] -4\mu x+(\mu-4\lambda)=&\,-6x. \end{align*}

Comparing coefficients of xx and the constant terms,

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

Hence,

μ=32,λ=38.\begin{align*} \mu=\frac32,\qquad \lambda=\frac38. \end{align*}

Therefore,

z=zc+zp=Ae4x+38+32x.\begin{align*} z=z_c+z_p =A\mathrm{e}^{4x}+\frac38+\frac32x. \end{align*}

So the general solution is

z=32x+38+Ae4x.\begin{align*} z=\frac32x+\frac38+A\mathrm{e}^{4x}. \end{align*}

(c)

解法一

思路

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题目要求用 y2=f(x)y^2=f(x) 的形式，所以把 zz 的通解代回 y2=1zy^2=\dfrac1z 即可。

答题过程

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Since

y2=1z\begin{align*} y^2=\frac1z \end{align*}

and

z=32x+38+Ce4x,\begin{align*} z=\frac32x+\frac38+C\mathrm{e}^{4x}, \end{align*}

the general solution of differential equation (I) is

y2= 132x+38+Ce4x= 812x+3+Ae4x,\begin{align*} y^2 =&\,\frac{1}{\frac32x+\frac38+C\mathrm{e}^{4x}}\\[4mm] =&\,\frac{8}{12x+3+A\mathrm{e}^{4x}}, \end{align*}

where AA is an arbitrary constant.