IAL 2024 June Q8 | 國際數學站

(a) Given that $t = \ln x$ , where $x > 0$ , show that

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \mathrm{e}^{-2t}\left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t}\right) \end{align*}

(3)

(b) Hence show that the transformation $t = \ln x$ , where $x > 0$ , transforms the differential equation

\begin{align*} x^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2y = 1 + 4\ln x - 2(\ln x)^2 \qquad \text{(I)} \end{align*}

into the differential equation

\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} - \frac{\mathrm{d}y}{\mathrm{d}t} - 2y = 1 + 4t - 2t^2 \qquad \text{(II)} \end{align*}

(1)

(5)

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

(1)

解答

(a)

Given $t = \ln x$ , we have $x = \mathrm{e}^t$ and $\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{x} = \mathrm{e}^{-t}$ . Using the chain rule to find $\frac{\mathrm{d}y}{\mathrm{d}x}$ :

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

Now, differentiate again with respect to $x$ using the product rule and chain rule:

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

(b)

Substitute $x = \mathrm{e}^t$ and the result from part (a) into differential equation (I):

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

This matches differential equation (II).

(c)

To solve the differential equation (II), find the complementary function (CF) first. The auxiliary equation is:

\begin{align*} m^2 - m - 2 = &\,0\\[4mm] (m - 2)(m + 1) = &\,0\\[4mm] m = &\,2, -1 \end{align*}

The CF is $y_c = A\mathrm{e}^{2t} + B\mathrm{e}^{-t}$ .

For the particular integral (PI), let $y_p = at^2 + bt + c$ . Differentiating this:

\begin{align*} y_p' = &\,2at + b\\[4mm] y_p'' = &\,2a \end{align*}

Substitute into (II):

\begin{align*} 2a - (2at + b) - 2(at^2 + bt + c) = &\,1 + 4t - 2t^2\\[4mm] -2at^2 + (-2a - 2b)t + (2a - b - 2c) = &\,-2t^2 + 4t + 1 \end{align*}

Compare the coefficients: For $t^2$ : $-2a = -2 \Rightarrow a = 1$ .

For $t$ : $-2a - 2b = 4 \Rightarrow -2(1) - 2b = 4 \Rightarrow -2b = 6 \Rightarrow b = -3$ .

For the constant term: $2a - b - 2c = 1 \Rightarrow 2(1) - (-3) - 2c = 1 \Rightarrow 5 - 2c = 1 \Rightarrow 2c = 4 \Rightarrow c = 2$ .

The particular integral is $y_p = t^2 - 3t + 2$ . Thus, the general solution for $y$ in terms of $t$ is:

\begin{align*} y = &\,A\mathrm{e}^{2t} + B\mathrm{e}^{-t} + t^2 - 3t + 2 \end{align*}

(d)

To find the general solution of (I), replace $t$ with $\ln x$ and $\mathrm{e}^t$ with $x$ :

\begin{align*} y = &\,A(\mathrm{e}^t)^2 + B(\mathrm{e}^t)^{-1} + (\ln x)^2 - 3\ln x + 2\\[4mm] y = &\,Ax^2 + Bx^{-1} + (\ln x)^2 - 3\ln x + 2 \end{align*}