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FP2 2021 June Q8

A Level / Edexcel / FP2

IAL 2021 June Paper · Question 8

题目

Problem

8. (a) Show that the substitution v=y−2v = y^{-2} transforms the differential equation

 dydx+6xy= 3xex2y3, x>0.(I)\begin{align*} &\,\frac{\mathrm{d}y}{\mathrm{d}x}+6xy\\[4mm] =&\,3x\mathrm{e}^{x^2}y^3,\\[4mm] &\,x>0.\qquad\text{(I)} \end{align*}

into the differential equation

 dvdx−12vx= −6xex2, x>0.(II)\begin{align*} &\,\frac{\mathrm{d}v}{\mathrm{d}x}-12vx\\[4mm] =&\,-6x\mathrm{e}^{x^2},\\[4mm] &\,x>0.\qquad\text{(II)} \end{align*}
(5)

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

(6)
题目中文翻译

8. (a) 证明代换 v=y−2v=y^{-2} 可将微分方程

 dydx+6xy= 3xex2y3, x>0.(I)\begin{align*} &\,\frac{\mathrm{d}y}{\mathrm{d}x}+6xy\\[4mm] =&\,3x\mathrm{e}^{x^2}y^3,\\[4mm] &\,x>0.\qquad\text{(I)} \end{align*}

化为微分方程

 dvdx−12vx= −6xex2, x>0.(II)\begin{align*} &\,\frac{\mathrm{d}v}{\mathrm{d}x}-12vx\\[4mm] =&\,-6x\mathrm{e}^{x^2},\\[4mm] &\,x>0.\qquad\text{(II)} \end{align*}
(5)

(b) 由此求微分方程 (I) 的通解，将答案写成 y2=f(x)y^2=f(x) 的形式。

(6)

解答

(a)

解法一

思路

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由 v=y−2v=y^{-2} 可得

dvdx=−2y−3dydx.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x} =-2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

所以要把原方程乘以合适的 yy 的幂，让它变成含 vv 和 dvdx\dfrac{\mathrm{d}v}{\mathrm{d}x} 的方程。

答题过程

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Given

v=y−2,\begin{align*} v=y^{-2}, \end{align*}

differentiate with respect to xx:

dvdx= −2y−3dydx.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x} =&\,-2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

So

y−3dydx= −12dvdx.\begin{align*} y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} =&\,-\frac12\frac{\mathrm{d}v}{\mathrm{d}x}. \end{align*}

Now start with differential equation (I):

dydx+6xy= 3xex2y3.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}+6xy =&\,3x\mathrm{e}^{x^2}y^3. \end{align*}

Divide by y3y^3:

y−3dydx+6xy−2= 3xex2.\begin{align*} y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} +6xy^{-2} =&\,3x\mathrm{e}^{x^2}. \end{align*}

Using v=y−2v=y^{-2},

−12dvdx+6xv= 3xex2.\begin{align*} -\frac12\frac{\mathrm{d}v}{\mathrm{d}x} +6xv =&\,3x\mathrm{e}^{x^2}. \end{align*}

Multiply by −2-2:

dvdx−12xv= −6xex2.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}-12xv =&\,-6x\mathrm{e}^{x^2}. \end{align*}

This is differential equation (II).

解法二

思路

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也可以先把 v=y−2v=y^{-2} 写成

y=v−12.\begin{align*} y=v^{-\frac12}. \end{align*}

然后把 yy 和 dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} 都用 vv 表示，再直接代入原方程。这个方法比较直接，关键是链式法则不要漏掉负号。

答题过程

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Since

v=y−2,\begin{align*} v=y^{-2}, \end{align*}

we have

y=v−12.\begin{align*} y=v^{-\frac12}. \end{align*}

Differentiate with respect to xx:

dydx= −12v−32dvdx.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,-\frac12v^{-\frac32} \frac{\mathrm{d}v}{\mathrm{d}x}. \end{align*}

Also,

y3=v−32.\begin{align*} y^3=v^{-\frac32}. \end{align*}

Substitute these into differential equation (I):

 −12v−32dvdx +6xv−12= 3xex2v−32.\begin{align*} &\,-\frac12v^{-\frac32} \frac{\mathrm{d}v}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}+6xv^{-\frac12}\\[4mm] =&\,3x\mathrm{e}^{x^2}v^{-\frac32}. \end{align*}

Multiply by −2v32-2v^{\frac32}:

dvdx−12xv= −6xex2.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x} -12xv =&\,-6x\mathrm{e}^{x^2}. \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：

dvdx−12xv=−6xex2.\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}x}-12xv=-6x\mathrm{e}^{x^2}. \end{align*}

先求 integrating factor，再解出 vv。最后因为 v=y−2v=y^{-2}，所以 y2=1vy^2=\dfrac1v。

答题过程

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The integrating factor is

I(x)= e∫−12x dx= e−6x2.\begin{align*} I(x) =&\,\mathrm{e}^{\int -12x\,\mathrm{d}x}\\[4mm] =&\,\mathrm{e}^{-6x^2}. \end{align*}

Multiply equation (II) by e−6x2\mathrm{e}^{-6x^2}:

 e−6x2dvdx −12xe−6x2v= −6xex2e−6x2.\begin{align*} &\,\mathrm{e}^{-6x^2}\frac{\mathrm{d}v}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}-12x\mathrm{e}^{-6x^2}v\\[4mm] =&\,-6x\mathrm{e}^{x^2}\mathrm{e}^{-6x^2}. \end{align*}

Using the product rule on the left,

 ddx(ve−6x2)= −6xe−5x2.\begin{align*} &\,\frac{\mathrm{d}}{\mathrm{d}x} \big(v\mathrm{e}^{-6x^2}\big)\\[4mm] =&\,-6x\mathrm{e}^{-5x^2}. \end{align*}

To integrate the right-hand side, put t=−5x2t=-5x^2. Then dt=−10x dx\mathrm{d}t=-10x\,\mathrm{d}x, so

 ∫−6xe−5x2 dx= 35∫et dt= 35e−5x2+C.\begin{align*} &\,\int-6x\mathrm{e}^{-5x^2}\,\mathrm{d}x\\[4mm] =&\,\frac35\int\mathrm{e}^{t}\,\mathrm{d}t\\[4mm] =&\,\frac35\mathrm{e}^{-5x^2}+C. \end{align*}

Integrating the complete equation therefore gives

ve−6x2=35e−5x2+C.v\mathrm{e}^{-6x^2}=\frac35\mathrm{e}^{-5x^2}+C.

Therefore,

v= 35ex2+Ce6x2.\begin{align*} v =&\,\frac35\mathrm{e}^{x^2}+C\mathrm{e}^{6x^2}. \end{align*}

Since v=y−2v=y^{-2},

y2= 1v= 135ex2+Ce6x2.\begin{align*} y^2 =&\,\frac1v\\[4mm] =&\,\frac{1} {\frac35\mathrm{e}^{x^2}+C\mathrm{e}^{6x^2}}. \end{align*}

Multiplying numerator and denominator by 1010 gives an equivalent form:

y2= 106ex2+Ke6x2,\begin{align*} y^2 =&\,\frac{10} {6\mathrm{e}^{x^2}+K\mathrm{e}^{6x^2}}, \end{align*}

where K=10CK=10C is an arbitrary constant.