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IAL 2021 Oct Q7

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 7

题目

Problem

(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}^2 y}{\mathrm{d}x^2} + 2(1 + 2\sqrt{x}) \frac{\mathrm{d}y}{\mathrm{d}x} - 15y = 15x \qquad \text{(I)} \end{align*}

into the differential equation

d2ydt2+2dydt−15y=15t2(II)\begin{align*} \frac{\mathrm{d}^2 y}{\mathrm{d}t^2} + 2\frac{\mathrm{d}y}{\mathrm{d}t} - 15y = 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)

解答

(a)

解法一

思路

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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

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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} =&\,2\frac{\mathrm{d}y}{\mathrm{d}x} +2t\frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right). \end{align*}

But

ddt(dydx)= d2ydx2dxdt= 2td2ydx2.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right) =&\,\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} =&\,2\frac{\mathrm{d}y}{\mathrm{d}x} +4t^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] =&\,2\frac{\mathrm{d}y}{\mathrm{d}x} +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} =&\,\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -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} +2(1+2\sqrt{x})\frac{\mathrm{d}y}{\mathrm{d}x} -15y =&\,15x. \end{align*}

This gives

(d2ydt2−2dydx)+2(1+2t)dydx−15y= 15t2.\begin{align*} \left( \frac{\mathrm{d}^2y}{\mathrm{d}t^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \right) +2(1+2t)\frac{\mathrm{d}y}{\mathrm{d}x} -15y =&\,15t^2. \end{align*}

Simplify the derivative terms:

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

This is (II).

(b)

解法一

思路

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(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

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The differential equation is

d2ydt2+2dydt−15y= 15t2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} +2\frac{\mathrm{d}y}{\mathrm{d}t} -15y =&\,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)-15(at^2+bt+c) =&\,15t^2. \end{align*}

Expand:

−15at2+(4a−15b)t+(2a+2b−15c)= 15t2.\begin{align*} -15at^2+(4a-15b)t+(2a+2b-15c) =&\,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⟹b=−415.\begin{align*} 4(-1)-15b=0 \quad\Longrightarrow\quad b=-\frac{4}{15}. \end{align*}

Finally,

2(−1)+2(−415)−15c= 0,−2−815−15c= 0,−3815−15c= 0.\begin{align*} 2\left(-1\right) +2\left(-\frac{4}{15}\right) -15c =&\,0,\\[2mm] -2-\frac{8}{15}-15c =&\,0,\\[2mm] -\frac{38}{15}-15c =&\,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} -t^2-\frac{4}{15}t-\frac{38}{225}. \end{align*}

(c)

解法一

思路

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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

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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} -t^2-\frac{4}{15}t-\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}} +B\mathrm{e}^{3\sqrt{x}} -x-\frac{4}{15}\sqrt{x} -\frac{38}{225}. \end{align*}