CalcGospel 國際數學圖譜
返回

FP2 2021 Oct Q7

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 7

题目

Problem

7. (a) Show that the transformation x=t2x = t^2 transforms the differential equation

 4xd2ydx2 +2(1+2x)dydx −15y= 15x.(I)\begin{align*} &\,4x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+2(1+2\sqrt{x})\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{4pt}-15y\\[4mm] =&\,15x.\qquad\text{(I)} \end{align*}

into the differential equation

 d2ydt2 +2dydt−15y= 15t2.(II)\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] &\,\hspace{2pt}+2\frac{\mathrm{d}y}{\mathrm{d}t}-15y\\[4mm] =&\,15t^2.\qquad\text{(II)} \end{align*}
(5)

(b) Solve differential equation (II) to determine yy in terms of tt.

(5)

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

(1)
题目中文翻译

7. (a) 证明变换 x=t2x=t^2 可将微分方程

 4xd2ydx2 +2(1+2x)dydx −15y= 15x.(I)\begin{align*} &\,4x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+2(1+2\sqrt{x})\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{4pt}-15y\\[4mm] =&\,15x.\qquad\text{(I)} \end{align*}

化为微分方程

 d2ydt2 +2dydt−15y= 15t2.(II)\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] &\,\hspace{2pt}+2\frac{\mathrm{d}y}{\mathrm{d}t}-15y\\[4mm] =&\,15t^2.\qquad\text{(II)} \end{align*}
(5)

(b) 解微分方程 (II)，用 tt 表示 yy。

(5)

(c) 由此求微分方程 (I) 的通解。

(1)

解答

(a)

解法一

思路

展开

这里 yy 原本是 xx 的函数，而 x=t2x=t^2，所以 yy 也可以看成 tt 的函数。关键是用 chain rule 把

dydx,d2ydx2\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}, \qquad \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

和关于 tt 的导数联系起来。

答题过程

展开

Since

x=t2,\begin{align*} x=t^2, \end{align*}

we have

dxdt=2t.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t}=2t. \end{align*}

By the chain rule,

dydt= dydxdxdt= 2tdydx.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} =&\,\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}t}\\[2mm] =&\,2t\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

So

dydx= 12tdydt.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{1}{2t}\frac{\mathrm{d}y}{\mathrm{d}t}. \end{align*}

Differentiate

dydt=2tdydx\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} =2t\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

with respect to tt:

 d2ydt2= 2dydx +2tddt(dydx).\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] =&\,2\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}+2t\frac{\mathrm{d}}{\mathrm{d}t} \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg). \end{align*}

But

ddt(dydx)= d2ydx2dxdt= 2td2ydx2.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg) =&\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \frac{\mathrm{d}x}{\mathrm{d}t}\\[2mm] =&\,2t\frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Therefore

 d2ydt2= 2dydx +4t2d2ydx2= 2dydx +4xd2ydx2.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] =&\,2\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}+4t^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] =&\,2\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}+4x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Hence

 4xd2ydx2= d2ydt2 −2dydx.\begin{align*} &\,4x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] =&\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] &\,\hspace{2pt}-2\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

Also, since x=t2x=t^2,

x=t.\begin{align*} \sqrt{x}=t. \end{align*}

Now substitute into (I):

 4xd2ydx2 +2(1+2x)dydx −15y= 15x.\begin{align*} &\,4x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+2(1+2\sqrt{x})\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{4pt}-15y\\[4mm] =&\,15x. \end{align*}

This gives

 (d2ydt2−2dydx) +2(1+2t)dydx −15y= 15t2.\begin{align*} &\,\bigg(\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -2\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\\[4mm] &\,\hspace{2pt}+2(1+2t)\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{4pt}-15y\\[4mm] =&\,15t^2. \end{align*}

Simplify the derivative terms:

 d2ydt2 −2dydx+2dydx +4tdydx−15y= 15t2.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] &\,\hspace{2pt}-2\frac{\mathrm{d}y}{\mathrm{d}x} +2\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{4pt}+4t\frac{\mathrm{d}y}{\mathrm{d}x}-15y\\[4mm] =&\,15t^2. \end{align*}

Since

dydt=2tdydx,\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} =2t\frac{\mathrm{d}y}{\mathrm{d}x}, \end{align*}

we have

4tdydx=2dydt.\begin{align*} 4t\frac{\mathrm{d}y}{\mathrm{d}x} =2\frac{\mathrm{d}y}{\mathrm{d}t}. \end{align*}

Therefore

 d2ydt2 +2dydt−15y= 15t2.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] &\,\hspace{2pt}+2\frac{\mathrm{d}y}{\mathrm{d}t}-15y\\[4mm] =&\,15t^2. \end{align*}

This is (II).

(b)

解法一

思路

展开

(II) 是二阶常系数非齐次微分方程。先求 complementary function，再设 particular integral。右边是 15t215t^2，所以 particular integral 设为二次多项式。

答题过程

展开

The differential equation is

 d2ydt2 +2dydt−15y= 15t2.\begin{align*} &\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\\[4mm] &\,\hspace{2pt}+2\frac{\mathrm{d}y}{\mathrm{d}t}-15y\\[4mm] =&\,15t^2. \end{align*}

For the complementary function, solve

m2+2m−15=0.\begin{align*} m^2+2m-15=0. \end{align*}

Factorise:

(m+5)(m−3)=0.\begin{align*} (m+5)(m-3)=0. \end{align*}

So

m=−5,m=3.\begin{align*} m=-5,\qquad m=3. \end{align*}

Therefore

yc=Ae−5t+Be3t.\begin{align*} y_c=A\mathrm{e}^{-5t}+B\mathrm{e}^{3t}. \end{align*}

For the particular integral, let

yp=at2+bt+c.\begin{align*} y_p=at^2+bt+c. \end{align*}

Then

dypdt= 2at+b,\begin{align*} \frac{\mathrm{d}y_p}{\mathrm{d}t} =&\,2at+b, \end{align*}

and

d2ypdt2= 2a.\begin{align*} \frac{\mathrm{d}^2y_p}{\mathrm{d}t^2} =&\,2a. \end{align*}

Substitute into the left hand side:

 2a+2(2at+b) −15(at2+bt+c)= 15t2.\begin{align*} &\,2a+2(2at+b)\\[4mm] &\,\hspace{2pt}-15(at^2+bt+c)\\[4mm] =&\,15t^2. \end{align*}

Expand:

 −15at2+(4a−15b)t +(2a+2b−15c)= 15t2.\begin{align*} &\,-15at^2+(4a-15b)t\\[4mm] &\,\hspace{2pt}+(2a+2b-15c)\\[4mm] =&\,15t^2. \end{align*}

Compare coefficients:

−15a= 15,4a−15b= 0,2a+2b−15c= 0.\begin{align*} -15a=&\,15,\\ 4a-15b=&\,0,\\ 2a+2b-15c=&\,0. \end{align*}

Hence

a=−1.\begin{align*} a=-1. \end{align*}

Then

4(−1)−15b= 0,−15b= 4,b= −415.\begin{align*} 4(-1)-15b=&\,0,\\[4mm] -15b=&\,4,\\[4mm] b=&\,-\frac4{15}. \end{align*}

Finally,

 2(−1)+2(−415) −15c= 0, −2−815−15c= 0, −3815−15c= 0.\begin{align*} &\,2(-1)+2\bigg(-\frac4{15}\bigg)\\[4mm] &\,\hspace{2pt}-15c\\[4mm] =&\,0,\\[4mm] &\,-2-\frac8{15}-15c\\[4mm] =&\,0,\\[4mm] &\,-\frac{38}{15}-15c\\[4mm] =&\,0. \end{align*}

So

c=−38225.\begin{align*} c=-\frac{38}{225}. \end{align*}

Therefore

yp=−t2−415t−38225.\begin{align*} y_p=-t^2-\frac{4}{15}t-\frac{38}{225}. \end{align*}

The general solution of (II) is

y= Ae−5t+Be3t −t2−415t −38225.\begin{align*} y=&\,A\mathrm{e}^{-5t}+B\mathrm{e}^{3t}\\[4mm] &\,\hspace{2pt}-t^2-\frac4{15}t\\[4mm] &\,\hspace{4pt}-\frac{38}{225}. \end{align*}

(c)

解法一

思路

展开

由 x=t2x=t^2 得 t=xt=\sqrt{x}。把 (b) 中的 tt 换成 x\sqrt{x}，并把 t2t^2 换成 xx。

答题过程

展开

Since

x=t2,\begin{align*} x=t^2, \end{align*}

we use

t=x.\begin{align*} t=\sqrt{x}. \end{align*}

From part (b),

y= Ae−5t+Be3t −t2−415t −38225.\begin{align*} y=&\,A\mathrm{e}^{-5t}+B\mathrm{e}^{3t}\\[4mm] &\,\hspace{2pt}-t^2-\frac4{15}t\\[4mm] &\,\hspace{4pt}-\frac{38}{225}. \end{align*}

Therefore the general solution of (I) is

y= Ae−5x +Be3x −x−415x −38225.\begin{align*} y=&\,A\mathrm{e}^{-5\sqrt{x}}\\[4mm] &\,\hspace{2pt}+B\mathrm{e}^{3\sqrt{x}}\\[4mm] &\,\hspace{4pt}-x-\frac4{15}\sqrt{x}\\[4mm] &\,\hspace{6pt}-\frac{38}{225}. \end{align*}