IAL 2023 Jan Q9 | 國際數學站

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

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

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

(5)

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

\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)}\\[2mm] \end{align*}

into the differential equation

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

(2)

(5)

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

(1)

解答

(a)(i)

Given $x = t^{\frac{1}{2}}$ , we can differentiate with respect to $t$ :

\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} = &\,\frac{1}{2}t^{-\frac{1}{2}}\\[4mm] \implies \frac{\mathrm{d}t}{\mathrm{d}x} = &\,2t^{\frac{1}{2}} \end{align*}

By the chain rule:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{\mathrm{d}y}{\mathrm{d}t} \frac{\mathrm{d}t}{\mathrm{d}x}\\[4mm] = &\,2t^{\frac{1}{2}}\frac{\mathrm{d}y}{\mathrm{d}t} \end{align*}

(a)(ii)

To find the second derivative, we differentiate $\frac{\mathrm{d}y}{\mathrm{d}x}$ with respect to $x$ :

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

(b)

Substitute $x$ , $\frac{\mathrm{d}y}{\mathrm{d}x}$ , and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in terms of $t$ into differential equation (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\\[4mm] t^{\frac{1}{2}} \left( 2\frac{\mathrm{d}y}{\mathrm{d}t} + 4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2} \right) - (6t + 1)\left( 2t^{\frac{1}{2}}\frac{\mathrm{d}y}{\mathrm{d}t} \right) + 9t^{\frac{3}{2}}y = &\,t^{\frac{5}{2}} \end{align*}

Since $t > 0$ , divide the entire equation by $t^{\frac{1}{2}}$ :

\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\\[4mm] 4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 2\frac{\mathrm{d}y}{\mathrm{d}t} - 12t\frac{\mathrm{d}y}{\mathrm{d}t} - 2\frac{\mathrm{d}y}{\mathrm{d}t} + 9ty = &\,t^2\\[4mm] 4t\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - 12t\frac{\mathrm{d}y}{\mathrm{d}t} + 9ty = &\,t^2 \end{align*}

Divide by $t$ :

\begin{align*} 4\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - 12\frac{\mathrm{d}y}{\mathrm{d}t} + 9y = &\,t \end{align*}

This matches differential equation (II).

(c)

To solve (II), first solve the corresponding homogeneous equation to find the complementary function (CF):

\begin{align*} 4m^2 - 12m + 9 = &\,0\\[4mm] (2m - 3)^2 = &\,0\\[4mm] m = &\,\frac{3}{2} \text{ (repeated root)} \end{align*}

So the complementary function is $y_c = (At + B)\mathrm{e}^{\frac{3}{2}t}$ .

For the particular integral (PI), assume $y_p = at + b$ . Differentiating gives $y_p' = a$ and $y_p'' = 0$ . Substitute into (II):

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

Equate the coefficients: For $t$ : $9a = 1 \implies a = \frac{1}{9}$ . For the constant: $9b - 12a = 0 \implies 9b = 12\left(\frac{1}{9}\right) \implies 9b = \frac{4}{3} \implies b = \frac{4}{27}$ .

So the particular integral is $y_p = \frac{1}{9}t + \frac{4}{27}$ . The general solution in terms of $t$ is:

\begin{align*} y = &\,(At + B)\mathrm{e}^{\frac{3}{2}t} + \frac{1}{9}t + \frac{4}{27} \end{align*}

(d)

To find the general solution for differential equation (I), substitute $t = x^2$ back into the solution from part (c):

\begin{align*} y = &\,(Ax^2 + B)\mathrm{e}^{\frac{3}{2}x^2} + \frac{1}{9}x^2 + \frac{4}{27} \end{align*}