CalcGospel 國際數學圖譜
返回

FP2 2023 Jan Q9

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 9

题目

Problem

9.

(a) Given that x=t12x = t^{\frac{1}{2}} determine, in terms of yy and tt,

(i) dydx\frac{\mathrm{d}y}{\mathrm{d}x}

(ii) d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}

(5)

(b) Hence show that the transformation x=t12x = t^{\frac{1}{2}}, where t>0t > 0, transforms the differential equation

xd2ydx2−(6x2+1)dydx+9x3y= x5(I)\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -(6x^2+1)\frac{\mathrm{d}y}{\mathrm{d}x} +9x^3y =&\, x^5 \qquad \text{(I)} \end{align*}

into the differential equation

4d2ydt2−12dydt+9y= t(II)\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -12\frac{\mathrm{d}y}{\mathrm{d}t} +9y =&\, t \qquad \text{(II)} \end{align*}
(2)

(c) Solve differential equation (II) to determine a general solution for yy in terms of tt.

(5)

(d) Hence determine the general solution of differential equation (I).

(1)
题目中文翻译

(a) 已知 x=t1/2x=t^{1/2}，用 yy、tt 表示

(i) dydx\frac{\mathrm{d}y}{\mathrm{d}x}；

(ii) d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}。

(5)

(b) 由此证明代换 x=t1/2x=t^{1/2}（其中 t>0t>0）将微分方程

xd2ydx2−(6x2+1)dydx+9x3y= x5(I) \begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -(6x^2+1)\frac{\mathrm{d}y}{\mathrm{d}x} +9x^3y=&\,x^5\qquad\text{(I)} \end{align*}

变为微分方程

4d2ydt2−12dydt+9y= t(II). \begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -12\frac{\mathrm{d}y}{\mathrm{d}t}+9y =&\,t\qquad\text{(II)}. \end{align*}
(2)

(c) 解微分方程 (II)，求 yy 关于 tt 的通解。

(5)

(d) 由此求微分方程 (I) 的通解。

(1)

解答

(a)(i)

解法一

思路

展开

由 x=t1/2x=t^{1/2} 可得 dtdx=2t1/2\frac{\mathrm{d}t}{\mathrm{d}x}=2t^{1/2}，再用链式法则把对 xx 的导数写成对 tt 的导数。

答题过程

展开

Since

x=t12,\begin{align*} x=t^{\frac{1}{2}}, \end{align*}

we have

dxdt= 12t−12,\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} =&\,\frac{1}{2}t^{-\frac{1}{2}}, \end{align*}

so

dtdx=2t12.\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}x} =2t^{\frac{1}{2}}. \end{align*}

Therefore

dydx= dydtdtdx= 2t12dydt.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}y}{\mathrm{d}t} \frac{\mathrm{d}t}{\mathrm{d}x}\\[2mm] =&\,2t^{\frac{1}{2}}\frac{\mathrm{d}y}{\mathrm{d}t}. \end{align*}

Thus

dydx=2t12dydt.\begin{align*} \boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =2t^{\frac{1}{2}}\frac{\mathrm{d}y}{\mathrm{d}t}}. \end{align*}

(a)(ii)

解法一

思路

展开

对 (i) 的结果继续关于 xx 求导。外面用链式法则换成对 tt 求导，里面的 2t1/2dydt2t^{1/2}\frac{\mathrm{d}y}{\mathrm{d}t} 用乘积法则。

答题过程

展开

Using dydx=2t1/2dydt\frac{\mathrm{d}y}{\mathrm{d}x}=2t^{1/2}\frac{\mathrm{d}y}{\mathrm{d}t} and dtdx=2t1/2\frac{\mathrm{d}t}{\mathrm{d}x}=2t^{1/2},

d2ydx2= ddx(2t12dydt)= dtdxddt(2t12dydt)= 2t12(t−12dydt+2t12d2ydt2)= 2dydt+4td2ydt2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac{\mathrm{d}}{\mathrm{d}x} \bigg(2t^{\frac{1}{2}}\frac{\mathrm{d}y}{\mathrm{d}t}\bigg)\\[2mm] =&\,\frac{\mathrm{d}t}{\mathrm{d}x} \frac{\mathrm{d}}{\mathrm{d}t} \bigg(2t^{\frac{1}{2}}\frac{\mathrm{d}y}{\mathrm{d}t}\bigg)\\[2mm] =&\,2t^{\frac{1}{2}} \bigg(t^{-\frac{1}{2}}\frac{\mathrm{d}y}{\mathrm{d}t} +2t^{\frac{1}{2}}\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\bigg)\\[2mm] =&\,2\frac{\mathrm{d}y}{\mathrm{d}t} +4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2}. \end{align*}

Thus

d2ydx2=2dydt+4td2ydt2.\begin{align*} \boxed{ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =2\frac{\mathrm{d}y}{\mathrm{d}t} +4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2} }. \end{align*}

(b)

解法一

思路

展开

把 x=t1/2x=t^{1/2}、x2=tx^2=t、x3=t3/2x^3=t^{3/2}、x5=t5/2x^5=t^{5/2} 以及 (a) 的导数结果全部代入。每一项都会含有 t1/2t^{1/2}，最后约掉即可。

答题过程

展开

Substitute the results from part (a) into equation (I):

xd2ydx2−(6x2+1)dydx+9x3y=x5.\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &-(6x^2+1)\frac{\mathrm{d}y}{\mathrm{d}x} +9x^3y=x^5. \end{align*}

Since x=t12x=t^{\frac{1}{2}},

t12(2dydt+4td2ydt2) −(6t+1)(2t12dydt)+9t32y=t52.\begin{align*} t^{\frac{1}{2}} \bigg(2\frac{\mathrm{d}y}{\mathrm{d}t} +4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\bigg)\\[2mm] &\,\hspace{2pt}-(6t+1)\bigg(2t^{\frac{1}{2}} \frac{\mathrm{d}y}{\mathrm{d}t}\bigg) +9t^{\frac{3}{2}}y=t^{\frac{5}{2}}. \end{align*}

Divide by t12t^{\frac{1}{2}}:

2dydt+4td2ydt2−2(6t+1)dydt+9ty=t24td2ydt2−12tdydt+9ty= t2.\begin{align*} 2\frac{\mathrm{d}y}{\mathrm{d}t} +4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2} &-2(6t+1)\frac{\mathrm{d}y}{\mathrm{d}t} +9ty=t^2\\[2mm] 4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -12t\frac{\mathrm{d}y}{\mathrm{d}t} +9ty=&\,t^2. \end{align*}

Since t>0t>0, divide by tt:

4d2ydt2−12dydt+9y=t.\begin{align*} \boxed{ 4\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -12\frac{\mathrm{d}y}{\mathrm{d}t} +9y=t }. \end{align*}

This is equation (II).

(c)

解法一

思路

展开

这是常系数二阶非齐次微分方程。先解齐次方程，辅助方程有重根，所以 complementary function 是

(At+B)e3t2.\begin{align*} (At+B)\mathrm{e}^{\frac{3t}{2}}. \end{align*}

右边是一次式 tt，所以 particular integral 试 at+bat+b。

答题过程

展开

Equation (II) is

4d2ydt2−12dydt+9y=t.\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -12\frac{\mathrm{d}y}{\mathrm{d}t} +9y=t. \end{align*}

The auxiliary equation is

4m2−12m+9= 0(2m−3)2= 0.\begin{align*} 4m^2-12m+9=&\,0\\[2mm] (2m-3)^2=&\,0. \end{align*}

So

m=32\begin{align*} m=\frac{3}{2} \end{align*}

is a repeated root. Therefore

yc=(At+B)e3t2.\begin{align*} y_c=(At+B)\mathrm{e}^{\frac{3t}{2}}. \end{align*}

For a particular integral, try

yp=at+b.\begin{align*} y_p=at+b. \end{align*}

Then

yp′=a,yp′′=0.\begin{align*} y_p'=a,\qquad y_p''=0. \end{align*}

Substitute into equation (II):

4(0)−12a+9(at+b)= t9at+(9b−12a)= t.\begin{align*} 4(0)-12a+9(at+b)=&\,t\\[2mm] 9at+(9b-12a)=&\,t. \end{align*}

Compare coefficients:

9a=1⟹a=19,9b−12a=0⟹9b=43.\begin{align*} 9a=1 \quad&\Longrightarrow\quad a=\frac{1}{9},\\[2mm] 9b-12a=0 \quad&\Longrightarrow\quad 9b=\frac{4}{3}. \end{align*}

Thus

b=427.\begin{align*} b=\frac{4}{27}. \end{align*}

So the general solution is

y=(At+B)e3t2+t9+427.\begin{align*} \boxed{ y=(At+B)\mathrm{e}^{\frac{3t}{2}} +\frac{t}{9}+\frac{4}{27} }. \end{align*}

(d)

解法一

思路

展开

由 x=t1/2x=t^{1/2} 得 t=x2t=x^2。把 (c) 中的 tt 换成 x2x^2。

答题过程

展开

Since

t=x2,\begin{align*} t=x^2, \end{align*}

the general solution of equation (I) is

y=(Ax2+B)e3x22+x29+427.\begin{align*} \boxed{ y=(Ax^2+B)\mathrm{e}^{\frac{3x^2}{2}} +\frac{x^2}{9}+\frac{4}{27} }. \end{align*}